It seems simple, but I can't get it work.
I have a string which look like 'NNDDDDDAAAA', where 'N' is non digit, 'D' is digit, and 'A' is anything. I need to replace each A with a space character. Number of 'N's, 'D's, and 'A's in an input string is always different.
I know how to do it with two expressions. I can split a string in to two, and then replace everything in second group with spaces. Like this
Pattern pattern = Pattern.compile("(\\D+\\d+)(.+)");
Matcher matcher = pattern.matcher(input);
if (matcher.matches()) {
return matcher.group(1) + matcher.group(2).replaceAll(".", " ");
}
But I was wondering if it is possible with a single regex expression.
Given your description, I'm assuming that after the NNDDDDD portion, the first A will actually be a N rather than an A, since otherwise there's no solid boundary between the DDDDD and AAAA portions.
So, your string actually looks like NNDDDDDNAAA, and you want to replace the NAAA portion with spaces. Given this, the regex can be rewritten as such: (\\D+\\d+)(\\D.+)
Positive lookbehind in Java requires a fixed length pattern; You can't use the + or * patterns. You can instead use the curly braces and specify a maximum length. For instance, you can use {1,9} in place of each +, and it will match between 1 and 9 characters: (?<=\\D{1,9}\\d{1,9})(\\D.+)
The only problem here is you're matching the NAAA sequence as a single match, so using "NNNDDDDNAAA".replaceAll("(?<=\\D{1,9}\\d{1,9})(\\D.+)", " ") will result in replacing the entire NAAA sequence with a single space, rather than multiple spaces.
You could take the beginning delimiter of the match, and the string length, and use that to append the correct number of spaces, but I don't see the point. I think you're better off with your original solution; Its simple and easy to follow.
If you're looking for a little extra speed, you could compile your Pattern outside the function, and use StringBuilder or StringBuffer to create your output. If you're building a large String out of all these NNDDDDDAAAAA elements, work entirely in StringBuilder until you're done appending.
class Test {
public static Pattern p = Pattern.compile("(\\D+\\d+)(\\D.+)");
public static StringBuffer replace( String input ) {
StringBuffer output = new StringBuffer();
Matcher m = Test.p.matcher(input);
if( m.matches() )
output.append( m.group(1) ).append( m.group(2).replaceAll("."," ") );
return output;
}
public static void main( String[] args ) {
String input = args[0];
long startTime;
StringBuffer tests = new StringBuffer();
startTime = System.currentTimeMillis();
for( int i = 0; i < 50; i++)
{
tests.append( "Input -> Output: '" );
tests.append( input );
tests.append( "' -> '" );
tests.append( Test.replace( input ) );
tests.append( "'\n" );
}
System.out.println( tests.toString() );
System.out.println( "\n" + (System.currentTimeMillis()-startTime));
}
}
Update:
I wrote a quick iterative solution, and ran some random data through both. The iterative solution is around 4-5x faster.
public static StringBuffer replace( String input )
{
StringBuffer output = new StringBuffer();
boolean second = false, third = false;
for( int i = 0; i < input.length(); i++ )
{
if( !second && Character.isDigit(input.charAt(i)) )
second = true;
if( second && !third && Character.isLetter(input.charAt(i)) )
third = true;
if( second && third )
output.append( ' ' );
else
output.append( input.charAt(i) );
}
return output;
}
what do you mean by nondigit vs anything?
[^a-zA-Z0-9]
matches everything that is not a letter or digit.
you would want to replace anything that gets matched by the above regex with a space.
is this what you were talking about?
You want to use positive look behind to match the N's and D's then use a normal match for the A's.
Not sure of the positive look behind grammar in Java, but some article on Java regex with look behind
I know you asked for a regex, but why do you even need a regex for this? How about:
StringBuilder sb = new StringBuilder(inputString);
for (int i = sb.length() - 1; i >= 0; i--) {
if (Character.isDigit(sb.charAt(i)))
break;
sb.setCharAt(i, ' ');
}
String output = sb.toString();
You might find this post interesting. Of course, the above code assumes there will be at least one digit in the string - all characters following the last digit are converted to spaces. If there are no digits, every character is converted to a space.
Related
i want to write a program in java using REGEX that gets 2 strings from the input ( the first one is shorter than the second one ) and then if the characters of the first string was inside the second string with the same order but they do not need to be next to each other ( it is not substring ) it outputs "true" and if not it outputs "false" here's an example:
example1:
input:
phantom
pphvnbajknzxcvbnatopopoim
output:
true
in the above example it is obvious we can see the word "phantom" in the second string (the characters are in the same order)
example2:
input:
apple
fgayiypvbnltsrgte
output:
false
as you can see apple dos not exists in the second string with the conditions i have earlier mentioned so it outputs false
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String word1 = input.next();
String word2 = input.next();
String pattern = "";
int n = word1.length();
char[] word1CharArr = word1.toCharArray();
for ( int i = 0 ; i < n ; i++) {
pattern += "[:alnum:]" +word1CharArr[i]+"[:alnum:]";
// pattern += ".*\\b|\\B" +word1CharArr[i]+"\\b|\\B";
}
pattern = "^" + pattern + "$";
// pattern = "(?s)" + pattern + ".*";
// System.out.println(pattern);
System.out.println(word2.matches(pattern));
}
}
here is what i did . i broke my first string to its characters and want to use REGEX before and after each character to determine the pattern. I have searched much about REGEX and how to use it but still i have problem here. the part i have commented comes out from one of my searches but it did not work
I emphasize that i want to solve it with REGEX not any other way.
[:alnum:] isn't a thing. Even if it is, that would match exactly one character, not 'any number, from 0 to infinitely many of them'.
You just want phantom with .* in the middle: ^.*p.*h.*a.*n.*t.*o.*m.*$' is all you need. After all, phantom` 'fits', and so does paahaanaataaoaamaa -
String pattern = word1.chars()
.mapToObj(c -> ".*" + (char) c)
.collect(Collectors.joining()) + ".*";
should get the job done.
I have a file name with this format yy_MM_someRandomString_originalFileName.
example:
02_01_fEa3129E_my Pic.png
I want replace the first 2 underscores with / so that the example becomes:
02/01/fEa3129E_my Pic.png
That can be done with replaceAll, but the problem is that files may contain underscores as well.
#Test
void test() {
final var input = "02_01_fEa3129E_my Pic.png";
final var formatted = replaceNMatches(input, "_", "/", 2);
assertEquals("02/01/fEa3129E_my Pic.png", formatted);
}
private String replaceNMatches(String input, String regex,
String replacement, int numberOfTimes) {
for (int i = 0; i < numberOfTimes; i++) {
input = input.replaceFirst(regex, replacement);
}
return input;
}
I solved this using a loop, but is there a pure regex way to do this?
EDIT: this way should be able to let me change a parameter and increase the amount of underscores from 2 to n.
You could use 2 capturing groups and use those in the replacement where the match of the _ will be replaced by /
^([^_]+)_([^_]+)_
Replace with:
$1/$2/
Regex demo | Java demo
For example:
String regex = "^([^_]+)_([^_]+)_";
String string = "02_01_fEa3129E_my Pic.png";
String subst = "$1/$2/";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
String result = matcher.replaceFirst(subst);
System.out.println(result);
Result
02/01/fEa3129E_my Pic.png
Your current solution has few problems:
It is inefficient - because each replaceFirst need to start from beginning of string so it needs to iterate over same starting characters many times.
It has a bug - because of point 1. while iterating from beginning instead of last modified place, we can replace value which was inserted previously.
For instance if we want to replace single character two times, each with X like abc -> XXc after code like
String input = "abc";
input = input.replaceFirst(".", "X"); // replaces a with X -> Xbc
input = input.replaceFirst(".", "X"); // replaces X with X -> Xbc
we will end up with Xbc instead of XXc because second replaceFirst will replace X with X instead of b with X.
To avoid that kind of problems you can rewrite your code to use Matcher#appendReplacement and Matcher#appendTail methods which ensures that we will iterate over input once and can replace each matched part with value we want
private static String replaceNMatches(String input, String regex,
String replacement, int numberOfTimes) {
Matcher m = Pattern.compile(regex).matcher(input);
StringBuilder sb = new StringBuilder();
int i = 0;
while(i++ < numberOfTimes && m.find() ){
m.appendReplacement(sb, replacement); // replaces currently matched part with replacement,
// and writes replaced version to StringBuilder
// along with text before the match
}
m.appendTail(sb); //lets add to builder text after last match
return sb.toString();
}
Usage example:
System.out.println(replaceNMatches("abcdefgh", "[efgh]", "X", 2)); //abcdXXgh
I have the following problem which states
Replace all characters in a string with + symbol except instances of the given string in the method
so for example if the string given was abc123efg and they want me to replace every character except every instance of 123 then it would become +++123+++.
I figured a regular expression is probably the best for this and I came up with this.
str.replaceAll("[^str]","+")
where str is a variable, but its not letting me use the method without putting it in quotations. If I just want to replace the variable string str how can I do that? I ran it with the string manually typed and it worked on the method, but can I just input a variable?
as of right now I believe its looking for the string "str" and not the variable string.
Here is the output its right for so many cases except for two :(
List of open test cases:
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
plusOut("abXYabcXYZ", "ab") → "ab++ab++++"
plusOut("abXYabcXYZ", "abc") → "++++abc+++"
plusOut("abXYabcXYZ", "XY") → "++XY+++XY+"
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
plusOut("--++ab", "++") → "++++++"
plusOut("aaxxxxbb", "xx") → "++xxxx++"
plusOut("123123", "3") → "++3++3"
Looks like this is the plusOut problem on CodingBat.
I had 3 solutions to this problem, and wrote a new streaming solution just for fun.
Solution 1: Loop and check
Create a StringBuilder out of the input string, and check for the word at every position. Replace the character if doesn't match, and skip the length of the word if found.
public String plusOut(String str, String word) {
StringBuilder out = new StringBuilder(str);
for (int i = 0; i < out.length(); ) {
if (!str.startsWith(word, i))
out.setCharAt(i++, '+');
else
i += word.length();
}
return out.toString();
}
This is probably the expected answer for a beginner programmer, though there is an assumption that the string doesn't contain any astral plane character, which would be represented by 2 char instead of 1.
Solution 2: Replace the word with a marker, replace the rest, then restore the word
public String plusOut(String str, String word) {
return str.replaceAll(java.util.regex.Pattern.quote(word), "#").replaceAll("[^#]", "+").replaceAll("#", word);
}
Not a proper solution since it assumes that a certain character or sequence of character doesn't appear in the string.
Note the use of Pattern.quote to prevent the word being interpreted as regex syntax by replaceAll method.
Solution 3: Regex with \G
public String plusOut(String str, String word) {
word = java.util.regex.Pattern.quote(word);
return str.replaceAll("\\G((?:" + word + ")*+).", "$1+");
}
Construct regex \G((?:word)*+)., which does more or less what solution 1 is doing:
\G makes sure the match starts from where the previous match leaves off
((?:word)*+) picks out 0 or more instance of word - if any, so that we can keep them in the replacement with $1. The key here is the possessive quantifier *+, which forces the regex to keep any instance of the word it finds. Otherwise, the regex will not work correctly when the word appear at the end of the string, as the regex backtracks to match .
. will not be part of any word, since the previous part already picks out all consecutive appearances of word and disallow backtrack. We will replace this with +
Solution 4: Streaming
public String plusOut(String str, String word) {
return String.join(word,
Arrays.stream(str.split(java.util.regex.Pattern.quote(word), -1))
.map((String s) -> s.replaceAll("(?s:.)", "+"))
.collect(Collectors.toList()));
}
The idea is to split the string by word, do the replacement on the rest, and join them back with word using String.join method.
Same as above, we need Pattern.quote to avoid split interpreting the word as regex. Since split by default removes empty string at the end of the array, we need to use -1 in the second parameter to make split leave those empty strings alone.
Then we create a stream out of the array and replace the rest as strings of +. In Java 11, we can use s -> String.repeat(s.length()) instead.
The rest is just converting the Stream to an Iterable (List in this case) and joining them for the result
This is a bit trickier than you might initially think because you don't just need to match characters, but the absence of specific phrase - a negated character set is not enough. If the string is 123, you would need:
(?<=^|123)(?!123).*?(?=123|$)
https://regex101.com/r/EZWMqM/1/
That is - lookbehind for the start of the string or "123", make sure the current position is not followed by 123, then lazy-repeat any character until lookahead matches "123" or the end of the string. This will match all characters which are not in a "123" substring. Then, you need to replace each character with a +, after which you can use appendReplacement and a StringBuffer to create the result string:
String inputPhrase = "123";
String inputStr = "abc123efg123123hij";
StringBuffer resultString = new StringBuffer();
Pattern regex = Pattern.compile("(?<=^|" + inputPhrase + ")(?!" + inputPhrase + ").*?(?=" + inputPhrase + "|$)");
Matcher m = regex.matcher(inputStr);
while (m.find()) {
String replacement = m.group(0).replaceAll(".", "+");
m.appendReplacement(resultString, replacement);
}
m.appendTail(resultString);
System.out.println(resultString.toString());
Output:
+++123+++123123+++
Note that if the inputPhrase can contain character with a special meaning in a regular expression, you'll have to escape them first before concatenating into the pattern.
You can do it in one line:
input = input.replaceAll("((?:" + str + ")+)?(?!" + str + ").((?:" + str + ")+)?", "$1+$2");
This optionally captures "123" either side of each character and puts them back (a blank if there's no "123"):
So instead of coming up with a regular expression that matches the absence of a string. We might as well just match the selected phrase and append + the number of skipped characters.
StringBuilder sb = new StringBuilder();
Matcher m = Pattern.compile(Pattern.quote(str)).matcher(input);
while (m.find()) {
for (int i = 0; i < m.start(); i++) sb.append('+');
sb.append(str);
}
int remaining = input.length() - sb.length();
for (int i = 0; i < remaining; i++) {
sb.append('+');
}
Absolutely just for the fun of it, a solution using CharBuffer (unexpectedly it took a lot more that I initially hoped for):
private static String plusOutCharBuffer(String input, String match) {
int size = match.length();
CharBuffer cb = CharBuffer.wrap(input.toCharArray());
CharBuffer word = CharBuffer.wrap(match);
int x = 0;
for (; cb.remaining() > 0;) {
if (!cb.subSequence(0, size < cb.remaining() ? size : cb.remaining()).equals(word)) {
cb.put(x, '+');
cb.clear().position(++x);
} else {
cb.clear().position(x = x + size);
}
}
return cb.clear().toString();
}
To make this work you need a beast of a pattern. Let's say you you are operating on the following test case as an example:
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
What you need to do is build a series of clauses in your pattern to match a single character at a time:
Any character that is NOT "X", "Y" or "Z" -- [^XYZ]
Any "X" not followed by "YZ" -- X(?!YZ)
Any "Y" not preceded by "X" -- (?<!X)Y
Any "Y" not followed by "Z" -- Y(?!Z)
Any "Z" not preceded by "XY" -- (?<!XY)Z
An example of this replacement can be found here: https://regex101.com/r/jK5wU3/4
Here is an example of how this might work (most certainly not optimized, but it works):
import java.util.regex.Pattern;
public class Test {
public static void plusOut(String text, String exclude) {
StringBuilder pattern = new StringBuilder("");
for (int i=0; i<exclude.length(); i++) {
Character target = exclude.charAt(i);
String prefix = (i > 0) ? exclude.substring(0, i) : "";
String postfix = (i < exclude.length() - 1) ? exclude.substring(i+1) : "";
// add the look-behind (?<!X)Y
if (!prefix.isEmpty()) {
pattern.append("(?<!").append(Pattern.quote(prefix)).append(")")
.append(Pattern.quote(target.toString())).append("|");
}
// add the look-ahead X(?!YZ)
if (!postfix.isEmpty()) {
pattern.append(Pattern.quote(target.toString()))
.append("(?!").append(Pattern.quote(postfix)).append(")|");
}
}
// add in the other character exclusion
pattern.append("[^" + Pattern.quote(exclude) + "]");
System.out.println(text.replaceAll(pattern.toString(), "+"));
}
public static void main(String [] args) {
plusOut("12xy34", "xy");
plusOut("12xy34", "1");
plusOut("12xy34xyabcxy", "xy");
plusOut("abXYabcXYZ", "ab");
plusOut("abXYabcXYZ", "abc");
plusOut("abXYabcXYZ", "XY");
plusOut("abXYxyzXYZ", "XYZ");
plusOut("--++ab", "++");
plusOut("aaxxxxbb", "xx");
plusOut("123123", "3");
}
}
UPDATE: Even this doesn't quite work because it can't deal with exclusions that are just repeated characters, like "xx". Regular expressions are most definitely not the right tool for this, but I thought it might be possible. After poking around, I'm not so sure a pattern even exists that might make this work.
The problem in your solution that you put a set of instance string str.replaceAll("[^str]","+") which it will exclude any character from the variable str and that will not solve your problem
EX: when you try str.replaceAll("[^XYZ]","+") it will exclude any combination of character X , character Y and character Z from your replacing method so you will get "++XY+++XYZ".
Actually you should exclude a sequence of characters instead in str.replaceAll.
You can do it by using capture group of characters like (XYZ) then use a negative lookahead to match a string which does not contain characters sequence : ^((?!XYZ).)*$
Check this solution for more info about this problem but you should know that it may be complicated to find regular expression to do that directly.
I have found two simple solutions for this problem :
Solution 1:
You can implement a method to replace all characters with '+' except the instance of given string:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
for(int i = 0; i < str.length(); i++){
// exclude any instance string of exWord from replacing process in str
if(str.substring(i, str.length()).indexOf(exWord) + i == i){
i = i + exWord.length()-1;
}
else{
str = str.substring(0,i) + "+" + str.substring(i+1);//replace each character with '+' symbol
}
}
Note : str.substring(i, str.length()).indexOf(exWord) + i this if statement will exclude any instance string of exWord from replacing process in str.
Output:
+++++++XYZ
Solution 2:
You can try this Approach using ReplaceAll method and it doesn't need any complex regular expression:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
str = str.replaceAll(exWord,"*"); // replace instance string with * symbol
str = str.replaceAll("[^*]","+"); // replace all characters with + symbol except *
str = str.replaceAll("\\*",exWord); // replace * symbol with instance string
Note : This solution will work only if your input string str doesn't contain any * symbol.
Also you should escape any character with a special meaning in a regular expression in phrase instance string exWord like : exWord = "++".
I have a string vaguely like this:
foo,bar,c;qual="baz,blurb",d;junk="quux,syzygy"
that I want to split by commas -- but I need to ignore commas in quotes. How can I do this? Seems like a regexp approach fails; I suppose I can manually scan and enter a different mode when I see a quote, but it would be nice to use preexisting libraries. (edit: I guess I meant libraries that are already part of the JDK or already part of a commonly-used libraries like Apache Commons.)
the above string should split into:
foo
bar
c;qual="baz,blurb"
d;junk="quux,syzygy"
note: this is NOT a CSV file, it's a single string contained in a file with a larger overall structure
Try:
public class Main {
public static void main(String[] args) {
String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String[] tokens = line.split(",(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)", -1);
for(String t : tokens) {
System.out.println("> "+t);
}
}
}
Output:
> foo
> bar
> c;qual="baz,blurb"
> d;junk="quux,syzygy"
In other words: split on the comma only if that comma has zero, or an even number of quotes ahead of it.
Or, a bit friendlier for the eyes:
public class Main {
public static void main(String[] args) {
String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String otherThanQuote = " [^\"] ";
String quotedString = String.format(" \" %s* \" ", otherThanQuote);
String regex = String.format("(?x) "+ // enable comments, ignore white spaces
", "+ // match a comma
"(?= "+ // start positive look ahead
" (?: "+ // start non-capturing group 1
" %s* "+ // match 'otherThanQuote' zero or more times
" %s "+ // match 'quotedString'
" )* "+ // end group 1 and repeat it zero or more times
" %s* "+ // match 'otherThanQuote'
" $ "+ // match the end of the string
") ", // stop positive look ahead
otherThanQuote, quotedString, otherThanQuote);
String[] tokens = line.split(regex, -1);
for(String t : tokens) {
System.out.println("> "+t);
}
}
}
which produces the same as the first example.
EDIT
As mentioned by #MikeFHay in the comments:
I prefer using Guava's Splitter, as it has saner defaults (see discussion above about empty matches being trimmed by String#split(), so I did:
Splitter.on(Pattern.compile(",(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)"))
While I do like regular expressions in general, for this kind of state-dependent tokenization I believe a simple parser (which in this case is much simpler than that word might make it sound) is probably a cleaner solution, in particular with regards to maintainability, e.g.:
String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
List<String> result = new ArrayList<String>();
int start = 0;
boolean inQuotes = false;
for (int current = 0; current < input.length(); current++) {
if (input.charAt(current) == '\"') inQuotes = !inQuotes; // toggle state
else if (input.charAt(current) == ',' && !inQuotes) {
result.add(input.substring(start, current));
start = current + 1;
}
}
result.add(input.substring(start));
If you don't care about preserving the commas inside the quotes you could simplify this approach (no handling of start index, no last character special case) by replacing your commas in quotes by something else and then split at commas:
String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
StringBuilder builder = new StringBuilder(input);
boolean inQuotes = false;
for (int currentIndex = 0; currentIndex < builder.length(); currentIndex++) {
char currentChar = builder.charAt(currentIndex);
if (currentChar == '\"') inQuotes = !inQuotes; // toggle state
if (currentChar == ',' && inQuotes) {
builder.setCharAt(currentIndex, ';'); // or '♡', and replace later
}
}
List<String> result = Arrays.asList(builder.toString().split(","));
http://sourceforge.net/projects/javacsv/
https://github.com/pupi1985/JavaCSV-Reloaded
(fork of the previous library that will allow the generated output to have Windows line terminators \r\n when not running Windows)
http://opencsv.sourceforge.net/
CSV API for Java
Can you recommend a Java library for reading (and possibly writing) CSV files?
Java lib or app to convert CSV to XML file?
I would not advise a regex answer from Bart, I find parsing solution better in this particular case (as Fabian proposed). I've tried regex solution and own parsing implementation I have found that:
Parsing is much faster than splitting with regex with backreferences - ~20 times faster for short strings, ~40 times faster for long strings.
Regex fails to find empty string after last comma. That was not in original question though, it was mine requirement.
My solution and test below.
String tested = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\",";
long start = System.nanoTime();
String[] tokens = tested.split(",(?=([^\"]*\"[^\"]*\")*[^\"]*$)");
long timeWithSplitting = System.nanoTime() - start;
start = System.nanoTime();
List<String> tokensList = new ArrayList<String>();
boolean inQuotes = false;
StringBuilder b = new StringBuilder();
for (char c : tested.toCharArray()) {
switch (c) {
case ',':
if (inQuotes) {
b.append(c);
} else {
tokensList.add(b.toString());
b = new StringBuilder();
}
break;
case '\"':
inQuotes = !inQuotes;
default:
b.append(c);
break;
}
}
tokensList.add(b.toString());
long timeWithParsing = System.nanoTime() - start;
System.out.println(Arrays.toString(tokens));
System.out.println(tokensList.toString());
System.out.printf("Time with splitting:\t%10d\n",timeWithSplitting);
System.out.printf("Time with parsing:\t%10d\n",timeWithParsing);
Of course you are free to change switch to else-ifs in this snippet if you feel uncomfortable with its ugliness. Note then lack of break after switch with separator. StringBuilder was chosen instead to StringBuffer by design to increase speed, where thread safety is irrelevant.
You're in that annoying boundary area where regexps almost won't do (as has been pointed out by Bart, escaping the quotes would make life hard) , and yet a full-blown parser seems like overkill.
If you are likely to need greater complexity any time soon I would go looking for a parser library. For example this one
I was impatient and chose not to wait for answers... for reference it doesn't look that hard to do something like this (which works for my application, I don't need to worry about escaped quotes, as the stuff in quotes is limited to a few constrained forms):
final static private Pattern splitSearchPattern = Pattern.compile("[\",]");
private List<String> splitByCommasNotInQuotes(String s) {
if (s == null)
return Collections.emptyList();
List<String> list = new ArrayList<String>();
Matcher m = splitSearchPattern.matcher(s);
int pos = 0;
boolean quoteMode = false;
while (m.find())
{
String sep = m.group();
if ("\"".equals(sep))
{
quoteMode = !quoteMode;
}
else if (!quoteMode && ",".equals(sep))
{
int toPos = m.start();
list.add(s.substring(pos, toPos));
pos = m.end();
}
}
if (pos < s.length())
list.add(s.substring(pos));
return list;
}
(exercise for the reader: extend to handling escaped quotes by looking for backslashes also.)
Try a lookaround like (?!\"),(?!\"). This should match , that are not surrounded by ".
The simplest approach is not to match delimiters, i.e. commas, with a complex additional logic to match what is actually intended (the data which might be quoted strings), just to exclude false delimiters, but rather match the intended data in the first place.
The pattern consists of two alternatives, a quoted string ("[^"]*" or ".*?") or everything up to the next comma ([^,]+). To support empty cells, we have to allow the unquoted item to be empty and to consume the next comma, if any, and use the \\G anchor:
Pattern p = Pattern.compile("\\G\"(.*?)\",?|([^,]*),?");
The pattern also contains two capturing groups to get either, the quoted string’s content or the plain content.
Then, with Java 9, we can get an array as
String[] a = p.matcher(input).results()
.map(m -> m.group(m.start(1)<0? 2: 1))
.toArray(String[]::new);
whereas older Java versions need a loop like
for(Matcher m = p.matcher(input); m.find(); ) {
String token = m.group(m.start(1)<0? 2: 1);
System.out.println("found: "+token);
}
Adding the items to a List or an array is left as an excise to the reader.
For Java 8, you can use the results() implementation of this answer, to do it like the Java 9 solution.
For mixed content with embedded strings, like in the question, you can simply use
Pattern p = Pattern.compile("\\G((\"(.*?)\"|[^,])*),?");
But then, the strings are kept in their quoted form.
what about a one-liner using String.split()?
String s = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String[] split = s.split( "(?<!\".{0,255}[^\"]),|,(?![^\"].*\")" );
A regular expression is not capable of handling escaped characters. For my application, I needed the ability to escape quotes and spaces (my separator is spaces, but the code is the same).
Here is my solution in Kotlin (the language from this particular application), based on the one from Fabian Steeg:
fun parseString(input: String): List<String> {
val result = mutableListOf<String>()
var inQuotes = false
var inEscape = false
val current = StringBuilder()
for (i in input.indices) {
// If this character is escaped, add it without looking
if (inEscape) {
inEscape = false
current.append(input[i])
continue
}
when (val c = input[i]) {
'\\' -> inEscape = true // escape the next character, \ isn't added to result
',' -> if (inQuotes) {
current.append(c)
} else {
result += current.toString()
current.clear()
}
'"' -> inQuotes = !inQuotes
else -> current.append(c)
}
}
if (current.isNotEmpty()) {
result += current.toString()
}
return result
}
I think this is not a place to use regular expressions. Contrary to other opinions, I don't think a parser is overkill. It's about 20 lines and fairly easy to test.
Rather than use lookahead and other crazy regex, just pull out the quotes first. That is, for every quote grouping, replace that grouping with __IDENTIFIER_1 or some other indicator, and map that grouping to a map of string,string.
After you split on comma, replace all mapped identifiers with the original string values.
I would do something like this:
boolean foundQuote = false;
if(charAtIndex(currentStringIndex) == '"')
{
foundQuote = true;
}
if(foundQuote == true)
{
//do nothing
}
else
{
string[] split = currentString.split(',');
}
I have a few strings which are like this:
text (255)
varchar (64)
...
I want to find out the number between ( and ) and store that in a string. That is, obviously, store these lengths in strings.
I have the rest of it figured out except for the regex parsing part.
I'm having trouble figuring out the regex pattern.
How do I do this?
The sample code is going to look like this:
Matcher m = Pattern.compile("<I CANT FIGURE OUT WHAT COMES HERE>").matcher("text (255)");
Also, I'd like to know if there's a cheat sheet for regex parsing, from where one can directly pick up the regex patterns
I would use a plain string match
String s = "text (255)";
int start = s.indexOf('(')+1;
int end = s.indexOf(')', start);
if (end < 0) {
// not found
} else {
int num = Integer.parseInt(s.substring(start, end));
}
You can use regex as sometimes this makes your code simpler, but that doesn't mean you should in all cases. I suspect this is one where a simple string indexOf and substring will not only be faster, and shorter but more importantly, easier to understand.
You can use this pattern to match any text between parentheses:
\(([^)]*)\)
Or this to match just numbers (with possible whitespace padding):
\(\s*(\d+)\s*\)
Of course, to use this in a string literal, you have to escape the \ characters:
Matcher m = Pattern.compile("\\(\\s*(\\d+)\\s*\\)")...
Here is some example code:
import java.util.regex.*;
class Main
{
public static void main(String[] args)
{
String txt="varchar (64)";
String re1=".*?"; // Non-greedy match on filler
String re2="\\((\\d+)\\)"; // Round Braces 1
Pattern p = Pattern.compile(re1+re2,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(txt);
if (m.find())
{
String rbraces1=m.group(1);
System.out.print("("+rbraces1.toString()+")"+"\n");
}
}
}
This will print out any (int) it finds in the input string, txt.
The regex is \((\d+)\) to match any numbers between ()
int index1 = string.indexOf("(")
int index2 = string.indexOf(")")
String intValue = string.substring(index1+1, index2-1);
Matcher m = Pattern.compile("\\((\\d+)\\)").matcher("text (255)");
if (m.find()) {
int len = Integer.parseInt (m.group(1));
System.out.println (len);
}