How to use regex with pattern matcher against multiple strings? - java

I'm reading in a list of strings from a List<String>. The strings look like this:
blah1
blah2
blah3
blah4
In java, I'd like to build a regex that checks for a pattern like this (myString/|yourString) and concatenate that to each of the strings in the list above while doing a pattern match against the lines of a file.
So I do this (the code below is just snippits):
String pattern = "(myString/|yourString.)"
private String listAsString;
private void createListAsStrings() {
StringBuilder sb = new StringBuilder();
for(String string : stringList) {
sb.append(string + "|"); # using the pipe hoping it will do an OR in the regex
}
listAsString = sb.toString();
}
To build the pattern, I'm trying to do the following:
Pattern p = Pattern.compile(pattern + listAsString);
But when I get to running the matcher it doesn't go through each string in the list of strings from my stringbuilder. And then the last problem is that my last string will contain a |.
Is there a way to match myString/blah1 or yourString.blah1 or myString/blah2 etc.. using a regex against each line in a file?
There is a lot of code, so I just posted what seemed relevant.

The expression that you are looking to build should be as follows:
myString/(?:\Qblah1\E|\Qblah2\E)
You need to wrap the strings blah1, blah2, etc. in \Q - \E in case the strings contain regex metacharacters. To fix the addition of leading | use a boolean variable that indicates if this is the first iteration through the loop or not:
StringBuilder sb = new StringBuilder();
boolean isFirst = true;
for(String word : stringList) {
if (!isFirst) {
sb.append('|');
} else {
isFirst = false;
}
sb.append("\\Q");
sb.append(word);
sb.append("\\E");
}
String regex = "myString/" + "(?:" + sb + ")";

I think the basic problem is that your pattern (ignoring the trailing | problem) is something like
(myString/|yourString.)blah1|blah2|blah3
which will match one of these
myString/blah1
yourString.blah1
blah2
blah3
That's how the operator precedence works in regexes. You need an extra set of parentheses around the lines from the file (plus see the other answers about \Q..\E and avoiding the bar at the end of the string).

Related

How to split emails which are seperated by comma(,) containing special characters in sender name? [duplicate]

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(',');
}

Split a string by commas except when the comma is part of the sentence [duplicate]

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(',');
}

java create variable from regex findings

I'm pretty new to Java, but I am looking to create a String variable from a regex finding. But I am not too sure how.
Basically I need: previous_identifer = (all the text in nextline up to the third comma);
Something maybe like this?
previous_identifier = line.split("^(.+?),(.+?),(.+?),");
Or:
line = reader.readLine();
Pattern courseColumnPattern = Pattern.compile("^(.+?),(.+?),(.+?),");
previous_identifier = (courseColumnPattern.matcher(line).find());
But I know that won't work. What should I do differently?
You can use split to return an array of Strings, then use a StringBuilder to build your return string. An advantage of this approach is being able to easily return the first four strings, two strings, ten strings, etc.
int limit = 3, current = 0;
StringBuilder sb = new StringBuilder();
// Used as an example of input
String str = "test,west,best,zest,jest";
String[] strings = str.split(",");
for(String s : strings) {
if(++current > limit) {
// We've reached the limit; bail
break;
}
if(current > 1) {
// Add a comma if it's not the first element. Alternative is to
// append a comma each time after appending s and remove the last
// character
sb.append(",");
}
sb.append(s);
}
System.out.println(sb.toString()); // Prints "test,west,best"
If you don't need to use the three elements separately (you truly want just the first three elements in a chunk), you can use a Matcher with the following regular expression:
String str = "test, west, best, zest, jest";
// Matches against "non-commas", then a comma, then "non-commas", then
// a comma, then "non-commas". This way, you still don't have a trailing
// comma at the end.
Matcher match = Pattern.compile("^([^,]*,[^,]*,[^,]*)").matcher(str);
if(match.find())
{
// Print out the output!
System.out.println(match.group(1));
}
else
{
// We didn't have a match. Handle it here.
}
Your regex will work, but could be expressed more briefly. This is how you can "extract" it:
String head = str.replaceAll("((.+?,){3}).*", "$1");
This matches the whole string, while capturing the target, with the replacement being the captured input using a back reference to group 1.
Despite the downvote, here's proof the code works!
String str = "foo,bar,baz,other,stuff";
String head = str.replaceAll("((.+?,){3}).*", "$1");
System.out.println(head);
Output:
foo,bar,baz,
try an online regex tester to work out the regex, i think you need less brackets to get the entire text, i'd guess something like:
([^,+?],[^,+?],[^,+?])
Which says, find everything except a comma, then a comma, then everything but a comma, then a comman, then everything else that isn't a comma. I suspect this can be improved dramatically, i am not a regex expert
Then your java just needs to compile it and match against your string:
line = reader.readLine();
Pattern courseColumnPattern = Pattern.compile("([^,+?],[^,+?],[^,+?])");
if (previous_identifier.matches()) {
previous_identifier = (courseColumnPattern.matcher(line);
}

Java Split not working as expected

I am trying to use a simple split to break up the following string: 00-00000
My expression is: ^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])
And my usage is:
String s = "00-00000";
String pattern = "^([0-9][0-9])(-)([0-9])([0-9])([0-9])([0-9])([0-9])";
String[] parts = s.split(pattern);
If I play around with the Pattern and Matcher classes I can see that my pattern does match and the matcher tells me my groupCount is 7 which is correct. But when I try and split them I have no luck.
String.split does not use capturing groups as its result. It finds whatever matches and uses that as the delimiter. So the resulting String[] are substrings in between what the regex matches. As it is the regex matches the whole string, and with the whole string as a delimiter there is nothing else left so it returns an empty array.
If you want to use regex capturing groups you will have to use Matcher.group(), String.split() will not do.
for your example, you could simply do this:
String s = "00-00000";
String pattern = "-";
String[] parts = s.split(pattern);
I can not be sure, but I think what you are trying to do is to get each matched group into an array.
Matcher matcher = Pattern.compile(pattern).matcher();
if (matcher.matches()) {
String s[] = new String[matcher.groupCount()) {
for (int i=0;i<matches.groupCount();i++) {
s[i] = matcher.group(i);
}
}
}
From the documentation:
String[] split(String regex) -- Returns: the array of strings computed by splitting this string around matches of the given regular expression
Essentially the regular expression is used to define delimiters in the input string. You can use capturing groups and backreferences in your pattern (e.g. for lookarounds), but ultimately what matters is what and where the pattern matches, because that defines what goes into the returned array.
If you want to split your original string into 7 parts using regular expression, then you can do something like this:
String s = "12-3456";
String[] parts = s.split("(?!^)");
System.out.println(parts.length); // prints "7"
for (String part : parts) {
System.out.println("Part [" + part + "]");
} // prints "[1] [2] [-] [3] [4] [5] [6] "
This splits on zero-length matching assertion (?!^), which is anywhere except before the first character in the string. This prevents the empty string to be the first element in the array, and trailing empty string is already discarded because we use the default limit parameter to split.
Using regular expression to get individual character of a string like this is an overkill, though. If you have only a few characters, then the most concise option is to use foreach on the toCharArray():
for (char ch : "12-3456".toCharArray()) {
System.out.print("[" + ch + "] ");
}
This is not the most efficient option if you have a longer string.
Splitting on -
This may also be what you're looking for:
String s = "12-3456";
String[] parts = s.split("-");
System.out.println(parts.length); // prints "2"
for (String part : parts) {
System.out.print("[" + part + "] ");
} // prints "[12] [3456] "

Regarding String manipulation

I have a String str which can have list of values like below. I want the first letter in the string to be uppercase and if underscore appears in the string then i need to remove it and need to make the letter after it as upper case. The rest all letter i want it to be lower case.
""
"abc"
"abc_def"
"Abc_def_Ghi12_abd"
"abc__de"
"_"
Output:
""
"Abc"
"AbcDef"
"AbcDefGhi12Abd"
"AbcDe"
""
Well, without showing us that you put any effort into this problem this is going to be kinda vague.
I see two possibilities here:
Split the string at underscores, apply the answer from this question to each part and re-combine them.
Create a StringBuilder, walk through the string and keep track of whether you are
at the start of the string
after an underscore or
somewhere else
and act appropriately on the current character before appending it to the StringBuilder instance.
replace _ with space (str.replace("_", " "))
use WordUtils.capitalizeFully(str); (from commons-lang)
replace space with nothing (str.replace(" ", ""))
You can use following regexp based code:
public static String camelize(String input) {
char[] c = input.toCharArray();
Pattern pattern = Pattern.compile(".*_([a-z]).*");
Matcher m = pattern.matcher(input);
while ( m.find() ) {
int index = m.start(1);
c[index] = String.valueOf(c[index]).toUpperCase().charAt(0);
}
return String.valueOf(c).replace("_", "");
}
Use Pattern/Matcher in the java.util.regex package:
for each string that is in your array do the following:
StringBuffer output = new StringBuffer();
Matcher match = Pattern.compile("[^|_](\w)").matcher(inStr);
while(match.find()) {
match.appendReplacement(output, matcher.match(0).ToUpper());
}
match.appendTail(output);
// Will have the properly capitalized string.
String capitalized = output.ToString();
The regular expression looks for either the start of the string or an underscore "[^|_]"
Then puts the following character into a group "(\w)"
The code then goes through each of the matches in the input string capitalizing the first satisfying group.

Categories

Resources