I'm new to regular expressions and would appreciate your help. I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.
Example input:
This is a string that "will be" highlighted when your 'regular expression' matches something.
Desired output:
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something.
Note that "will be" and 'regular expression' retain the space between the words.
I don't understand why all the others are proposing such complex regular expressions or such long code. Essentially, you want to grab two kinds of things from your string: sequences of characters that aren't spaces or quotes, and sequences of characters that begin and end with a quote, with no quotes in between, for two kinds of quotes. You can easily match those things with this regular expression:
[^\s"']+|"([^"]*)"|'([^']*)'
I added the capturing groups because you don't want the quotes in the list.
This Java code builds the list, adding the capturing group if it matched to exclude the quotes, and adding the overall regex match if the capturing group didn't match (an unquoted word was matched).
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"([^\"]*)\"|'([^']*)'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
if (regexMatcher.group(1) != null) {
// Add double-quoted string without the quotes
matchList.add(regexMatcher.group(1));
} else if (regexMatcher.group(2) != null) {
// Add single-quoted string without the quotes
matchList.add(regexMatcher.group(2));
} else {
// Add unquoted word
matchList.add(regexMatcher.group());
}
}
If you don't mind having the quotes in the returned list, you can use much simpler code:
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"[^\"]*\"|'[^']*'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}
There are several questions on StackOverflow that cover this same question in various contexts using regular expressions. For instance:
parsings strings: extracting words and phrases
Best way to parse Space Separated Text
UPDATE: Sample regex to handle single and double quoted strings. Ref: How can I split on a string except when inside quotes?
m/('.*?'|".*?"|\S+)/g
Tested this with a quick Perl snippet and the output was as reproduced below. Also works for empty strings or whitespace-only strings if they are between quotes (not sure if that's desired or not).
This
is
a
string
that
"will be"
highlighted
when
your
'regular expression'
matches
something.
Note that this does include the quote characters themselves in the matched values, though you can remove that with a string replace, or modify the regex to not include them. I'll leave that as an exercise for the reader or another poster for now, as 2am is way too late to be messing with regular expressions anymore ;)
If you want to allow escaped quotes inside the string, you can use something like this:
(?:(['"])(.*?)(?<!\\)(?>\\\\)*\1|([^\s]+))
Quoted strings will be group 2, single unquoted words will be group 3.
You can try it on various strings here: http://www.fileformat.info/tool/regex.htm or http://gskinner.com/RegExr/
The regex from Jan Goyvaerts is the best solution I found so far, but creates also empty (null) matches, which he excludes in his program. These empty matches also appear from regex testers (e.g. rubular.com).
If you turn the searches arround (first look for the quoted parts and than the space separed words) then you might do it in once with:
("[^"]*"|'[^']*'|[\S]+)+
(?<!\G".{0,99999})\s|(?<=\G".{0,99999}")\s
This will match the spaces not surrounded by double quotes.
I have to use min,max {0,99999} because Java doesn't support * and + in lookbehind.
It'll probably be easier to search the string, grabbing each part, vs. split it.
Reason being, you can have it split at the spaces before and after "will be". But, I can't think of any way to specify ignoring the space between inside a split.
(not actual Java)
string = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
regex = "\"(\\\"|(?!\\\").)+\"|[^ ]+"; // search for a quoted or non-spaced group
final = new Array();
while (string.length > 0) {
string = string.trim();
if (Regex(regex).test(string)) {
final.push(Regex(regex).match(string)[0]);
string = string.replace(regex, ""); // progress to next "word"
}
}
Also, capturing single quotes could lead to issues:
"Foo's Bar 'n Grill"
//=>
"Foo"
"s Bar "
"n"
"Grill"
String.split() is not helpful here because there is no way to distinguish between spaces within quotes (don't split) and those outside (split). Matcher.lookingAt() is probably what you need:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
str = str + " "; // add trailing space
int len = str.length();
Matcher m = Pattern.compile("((\"[^\"]+?\")|('[^']+?')|([^\\s]+?))\\s++").matcher(str);
for (int i = 0; i < len; i++)
{
m.region(i, len);
if (m.lookingAt())
{
String s = m.group(1);
if ((s.startsWith("\"") && s.endsWith("\"")) ||
(s.startsWith("'") && s.endsWith("'")))
{
s = s.substring(1, s.length() - 1);
}
System.out.println(i + ": \"" + s + "\"");
i += (m.group(0).length() - 1);
}
}
which produces the following output:
0: "This"
5: "is"
8: "a"
10: "string"
17: "that"
22: "will be"
32: "highlighted"
44: "when"
49: "your"
54: "regular expression"
75: "matches"
83: "something."
I liked Marcus's approach, however, I modified it so that I could allow text near the quotes, and support both " and ' quote characters. For example, I needed a="some value" to not split it into [a=, "some value"].
(?<!\\G\\S{0,99999}[\"'].{0,99999})\\s|(?<=\\G\\S{0,99999}\".{0,99999}\"\\S{0,99999})\\s|(?<=\\G\\S{0,99999}'.{0,99999}'\\S{0,99999})\\s"
Jan's approach is great but here's another one for the record.
If you actually wanted to split as mentioned in the title, keeping the quotes in "will be" and 'regular expression', then you could use this method which is straight out of Match (or replace) a pattern except in situations s1, s2, s3 etc
The regex:
'[^']*'|\"[^\"]*\"|( )
The two left alternations match complete 'quoted strings' and "double-quoted strings". We will ignore these matches. The right side matches and captures spaces to Group 1, and we know they are the right spaces because they were not matched by the expressions on the left. We replace those with SplitHere then split on SplitHere. Again, this is for a true split case where you want "will be", not will be.
Here is a full working implementation (see the results on the online demo).
import java.util.*;
import java.io.*;
import java.util.regex.*;
import java.util.List;
class Program {
public static void main (String[] args) throws java.lang.Exception {
String subject = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
Pattern regex = Pattern.compile("\'[^']*'|\"[^\"]*\"|( )");
Matcher m = regex.matcher(subject);
StringBuffer b= new StringBuffer();
while (m.find()) {
if(m.group(1) != null) m.appendReplacement(b, "SplitHere");
else m.appendReplacement(b, m.group(0));
}
m.appendTail(b);
String replaced = b.toString();
String[] splits = replaced.split("SplitHere");
for (String split : splits) System.out.println(split);
} // end main
} // end Program
If you are using c#, you can use
string input= "This is a string that \"will be\" highlighted when your 'regular expression' matches <something random>";
List<string> list1 =
Regex.Matches(input, #"(?<match>\w+)|\""(?<match>[\w\s]*)""|'(?<match>[\w\s]*)'|<(?<match>[\w\s]*)>").Cast<Match>().Select(m => m.Groups["match"].Value).ToList();
foreach(var v in list1)
Console.WriteLine(v);
I have specifically added "|<(?[\w\s]*)>" to highlight that you can specify any char to group phrases. (In this case I am using < > to group.
Output is :
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something random
1st one-liner using String.split()
String s = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
String[] split = s.split( "(?<!(\"|').{0,255}) | (?!.*\\1.*)" );
[This, is, a, string, that, "will be", highlighted, when, your, 'regular expression', matches, something.]
don't split at the blank, if the blank is surrounded by single or double quotes
split at the blank when the 255 characters to the left and all characters to the right of the blank are neither single nor double quotes
adapted from original post (handles only double quotes)
I'm reasonably certain this is not possible using regular expressions alone. Checking whether something is contained inside some other tag is a parsing operation. This seems like the same problem as trying to parse XML with a regex -- it can't be done correctly. You may be able to get your desired outcome by repeatedly applying a non-greedy, non-global regex that matches the quoted strings, then once you can't find anything else, split it at the spaces... that has a number of problems, including keeping track of the original order of all the substrings. Your best bet is to just write a really simple function that iterates over the string and pulls out the tokens you want.
A couple hopefully helpful tweaks on Jan's accepted answer:
(['"])((?:\\\1|.)+?)\1|([^\s"']+)
Allows escaped quotes within quoted strings
Avoids repeating the pattern for the single and double quote; this also simplifies adding more quoting symbols if needed (at the expense of one more capturing group)
You can also try this:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something";
String ss[] = str.split("\"|\'");
for (int i = 0; i < ss.length; i++) {
if ((i % 2) == 0) {//even
String[] part1 = ss[i].split(" ");
for (String pp1 : part1) {
System.out.println("" + pp1);
}
} else {//odd
System.out.println("" + ss[i]);
}
}
The following returns an array of arguments. Arguments are the variable 'command' split on spaces, unless included in single or double quotes. The matches are then modified to remove the single and double quotes.
using System.Text.RegularExpressions;
var args = Regex.Matches(command, "[^\\s\"']+|\"([^\"]*)\"|'([^']*)'").Cast<Match>
().Select(iMatch => iMatch.Value.Replace("\"", "").Replace("'", "")).ToArray();
When you come across this pattern like this :
String str = "2022-11-10 08:35:00,470 RAV=REQ YIP=02.8.5.1 CMID=caonaustr CMN=\"Some Value Pyt Ltd\"";
//this helped
String[] str1= str.split("\\s(?=(([^\"]*\"){2})*[^\"]*$)\\s*");
System.out.println("Value of split string is "+ Arrays.toString(str1));
This results in :[2022-11-10, 08:35:00,470, PLV=REQ, YIP=02.8.5.1, CMID=caonaustr, CMN="Some Value Pyt Ltd"]
This regex matches spaces ONLY if it is followed by even number of double quotes.
Related
Problem description
I am trying to split a into separate strings, with the split() method that the String class provides. The documentation tells me that it will split around matches of the argument, which is a regular expression. The delimiter that I use is a comma, but commas can also be escaped. Escaping character that I use is a forward slash / (just to make things easier by not using a backslash, because that requires additional escaping in string literals in both Java and the regular expressions).
For instance, the input might be this:
a,b/,b//,c///,//,d///,
And the output should be:
a
b,b/
c/,/
d/,
So, the string should be split at each comma, unless that comma is preceded by an odd number of slashes (1, 3, 5, 7, ..., ∞) because that would mean that the comma is escaped.
Possible solutions
My initial guess would be to split it like this:
String[] strings = longString.split("(?<*/),");
but that is not allowed because Java doesn't allow infinite look-behind groups. I could limit the recurrence to, say, 2000 by replacing the * with {0,2000}:
String[] strings = longString.split("(?<{0,2000}/),");
but that still puts constraints on the input. So I decided to take the recurrence out of the look-behind group, and came up with this:
String[] strings = longString.split("(?<!/)(?:(//)*),");
However, its output is the following list of strings:
a
b,b (the final slash is lacking in the output)
c/, (the final slash is lacking in the output)
d/,
Why are those slashes omitted in the 2nd and 3rd string, and how can I solve it (in Java)?
You are pretty close. To overcome lookbehind error you can use this workaround:
String[] strings = longString.split("(?<{0,99}/),")
You can achieve the split using a positive look behind for an even number of slashes preceding the comma:
String[] strings = longString.split("(?<=[^/](//){0,999999999}),");
But to display the output you want, you need a further step of removing the remaining escapes:
String longString = "a,b/,b//,c///,//,d///,";
String[] strings = longString.split("(?<=[^/](//){0,999999999}),");
for (String s : strings)
System.out.println(s.replaceAll("/(.)", "$1"));
Output:
a
b,b/
c/,/
d/,
If you don't mind another method with regex, I suggest using .matcher:
Pattern pattern = Pattern.compile("(?:[^,/]+|/.)+");
String test = "a,b/,b//,c///,//,d///,";
Matcher matcher = pattern.matcher(test);
while (matcher.find()) {
System.out.println(matcher.group().replaceAll("/(.)", "$1"));
}
Output:
a
b,b/
c/,/
d/,
ideone demo
This method will match everything except the delimiting commas (kind of the reverse). The advantage is that it doesn't rely on lookarounds.
I love regexes, but wouldn't it be easy to write the code manually here, i.e.
boolean escaped = false;
for(int i = 0, len = s.length() ; i < len ; i++){
switch(s.charAt(i)){
case "/": escaped = !escaped; break;
case ",":
if(!escaped){
//found a segment, do something with it
}
//Fallthrough!
default:
escaped = false;
}
}
// handle last segment
I want to do a startStr.replaceAll(searchStr, replaceStr) and I have two requirements.
The searchStr must be a whole word, meaning it must have a space, beginning of string or end of string character around it.
e.g.
startStr = "ON cONfirmation, put ON your hat"
searchStr = "ON"
replaceStr = ""
expected = " cONfirmation, put your hat"
The searchStr may contain a regex pattern
e.g.
startStr = "remove this * thing"
searchStr = "*"
replaceStr = ""
expected = "remove this thing"
For requirement 1, I've found that this works:
startStr.replaceAll("\\b"+searchStr+"\\b",replaceStr)
For requirement 2, I've found that this works:
startStr.replaceAll(Pattern.quote(searchStr), replaceStr)
But I can't get them to work together:
startStr.replaceAll("\\b"+Pattern.quote(searchStr)+"\\b", replaceStr)
Here is the simple test case that's failing
startStr = "remove this * thing but not this*"
searchStr = "*"
replaceStr = ""
expected = "remove this thing but not this*"
actual = "remove this * thing but not this*"
What am I missing?
Thanks in advance
First off, the \b, or word boundary, is not going to work for you with the asterisks. The reason is that \b only detects boundaries of word characters. A regex parser won't acknowledge * as a word character, so a wildcard-endowed word that begins or ends with a regex won't be surrounded by valid word boundaries.
Reference pages:
http://www.regular-expressions.info/wordboundaries.html
http://docs.oracle.com/javase/tutorial/essential/regex/bounds.html
An option you might like is to supply wildcard permutations in a regex:
(?<=\s|^)(ON|\*N|O\*|\*)(?=\s|$)
Here's a Java example:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class RegExTest
{
public static void main(String[] args){
String sourcestring = "ON cONfirmation, put * your hat";
sourcestring = sourcestring.replaceAll("(?<=\\s|^)(ON|\\*N|O\\*|\\*)(?=\\s|$)","").replaceAll(" "," ").trim();
System.out.println("sourcestring=["+sourcestring+"]");
}
}
You can write a little function to generate the wildcard permutations automatically. I admit I cheated a little with the spaces, but I don't think that was a requirement anyway.
Play with it online here: http://ideone.com/7uGfIS
The pattern "\\b" matches a word boundary, with a word character on one side and a non-word character on the other. * is not a word character, so \\b\\*\\b won't work. Look-behind and look-ahead match but do not consume patterns. You can specify that the beginning of the string or whitespace must come before your pattern and that whitespace or the end of the string must follow:
startStr.replaceAll("(?<=^|\\s)"+Pattern.quote(searchStr)+"(?=\\s|$)", replaceStr)
Try this,
For removing "ON"
StringBuilder stringBuilder = new StringBuilder();
String[] splittedValue = startStr.split(" ");
for (String value : splittedValue)
{
if (!value.equalsIgnoreCase("ON"))
{
stringBuilder.append(value);
stringBuilder.append(" ");
}
}
System.out.println(stringBuilder.toString().trim());
For removing "*"
String startStr1 = "remove this * thing";
System.out.println(startStr1.replaceAll("\\*[\\s]", ""));
You can use (^| )\*( |$) instead of using \\b
Try this startStr.replaceAll("(^| )youSearchString( |$)", replaceStr);
I need to split a String based on comma as seperator, but if the part of string is enclosed with " the splitting has to stop for that portion from starting of " to ending of it even it contains commas in between.
Can anyone please help me to solve this using regex with look around.
Resurrecting this question because it had a simple regex solution that wasn't mentioned. This situation sounds very similar to ["regex-match a pattern unless..."][4]
\"[^\"]*\"|(,)
The left side of the alternation matches complete double-quoted strings. We will ignore these matches. The right side matches and captures commas to Group 1, and we know they are the right ones because they were not matched by the expression on the left.
Here is working code (see online demo):
import java.util.regex.*;
import java.util.List;
class Program {
public static void main (String[] args) {
String subject = "\"Messages,Hello\",World,Hobbies,Java\",Programming\"";
Pattern regex = Pattern.compile("\"[^\"]*\"|(,)");
Matcher m = regex.matcher(subject);
StringBuffer b = new StringBuffer();
while (m.find()) {
if(m.group(1) != null) m.appendReplacement(b, "SplitHere");
else m.appendReplacement(b, m.group(0));
}
m.appendTail(b);
String replaced = b.toString();
String[] splits = replaced.split("SplitHere");
for (String split : splits)
System.out.println(split);
} // end main
} // end Program
Reference
How to match pattern except in situations s1, s2, s3
Please try this:
(?<!\G\s*"[^"]*),
If you put this regex in your program, it should be:
String regex = "(?<!\\G\\s*\"[^\"]*),";
But 2 things are not clear:
Does the " only start near the ,, or it can start in the middle of content, such as AAA, BB"CC,DD" ? The regex above only deal with start neer , .
If the content has " itself, how to escape? use "" or \"? The regex above does not deal any escaped " format.
i have a problem to build following regex:
[1,2,3,4]
i found a work-around, but i think its ugly
String stringIds = "[1,2,3,4]";
stringIds = stringIds.replaceAll("\\[", "");
stringIds = stringIds.replaceAll("\\]", "");
String[] ids = stringIds.split("\\,");
Can someone help me please to build one regex, which i can use in the split function
Thanks for help
edit:
i want to get from this string "[1,2,3,4]" to an array with 4 entries. the entries are the 4 numbers in the string, so i need to eliminate "[","]" and ",". the "," isn't the problem.
the first and last number contains [ or ]. so i needed the fix with replaceAll. But i think if i use in split a regex for ",", i also can pass a regex which eliminates "[" "]" too. But i cant figure out, who this regex should look like.
This is almost what you're looking for:
String q = "[1,2,3,4]";
String[] x = q.split("\\[|\\]|,");
The problem is that it produces an extra element at the beginning of the array due to the leading open bracket. You may not be able to do what you want with a single regex sans shenanigans. If you know the string always begins with an open bracket, you can remove it first.
The regex itself means "(split on) any open bracket, OR any closed bracket, OR any comma."
Punctuation characters frequently have additional meanings in regular expressions. The double leading backslashes... ugh, the first backslash tells the Java String parser that the next backslash is not a special character (example: \n is a newline...) so \\ means "I want an honest to God backslash". The next backslash tells the regexp engine that the next character ([ for example) is not a special regexp character. That makes me lol.
Maybe substring [ and ] from beginning and end, then split the rest by ,
String stringIds = "[1,2,3,4]";
String[] ids = stringIds.substring(1,stringIds.length()-1).split(",");
Looks to me like you're trying to make an array (not sure where you got 'regex' from; that means something different). In this case, you want:
String[] ids = {"1","2","3","4"};
If it's specifically an array of integer numbers you want, then instead use:
int[] ids = {1,2,3,4};
Your problem is not amenable to splitting by delimiter. It is much safer and more general to split by matching the integers themselves:
static String[] nums(String in) {
final Matcher m = Pattern.compile("\\d+").matcher(in);
final List<String> l = new ArrayList<String>();
while (m.find()) l.add(m.group());
return l.toArray(new String[l.size()]);
}
public static void main(String args[]) {
System.out.println(Arrays.toString(nums("[1, 2, 3, 4]")));
}
If the first line your code is following:
String stringIds = "[1,2,3,4]";
and you're trying to iterate over all number items, then the follwing code-frag only could work:
try {
Pattern regex = Pattern.compile("\\b(\\d+)\\b", Pattern.MULTILINE);
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
for (int i = 1; i <= regexMatcher.groupCount(); i++) {
// matched text: regexMatcher.group(i)
// match start: regexMatcher.start(i)
// match end: regexMatcher.end(i)
}
}
} catch (PatternSyntaxException ex) {
// Syntax error in the regular expression
}
I'm having a little trouble constructing the regular expression using java.
The constraint is, I need to split a string seperated by !. The two strings will be enclosed in double quotes.
For example:
"value"!"value"
If I performed a java split() on the string above, I want to get:
value
value
However the catch is value can be any characters/punctuations/numerical character/spaces/etc..
So here's a more concrete example. Input:
""he! "l0"!"wor!"d1"
Java's split() should return:
"he! "l0
wor!"d1
Any help is much appreciated. Thanks!
Try this expression: (".*")\s*!\s*(".*")
Although it would not work with split, it should work with Pattern and Matcher and return the 2 strings as groups.
String input = "\" \"he\"\"\"\"! \"l0\" ! \"wor!\"d1\"";
Pattern p = Pattern.compile("(\".*\")\\s*!\\s*(\".*\")");
Matcher m = p.matcher(input);
if(m.matches())
{
String s1 = m.group(1); //" "he""""! "l0"
String s2 = m.group(2); //"wor!"d1"
}
Edit:
This would not work for all cases, e.g. "he"!"llo" ! "w" ! "orld" would get the wrong groups. In that case it would be really hard to determine which ! should be the separator. That's why often rarely used characters are used to separate parts of a string, like # in email addresses :)
have the value split on "!" instead of !
String REGEX = "\"!\"";
String INPUT = "\"\"he! \"l0\"!\"wor!\"d1\"";
String[] items = p.split(INPUT);
It feels like you need to parse on:
DOUBLEQUOTE = "
OTHER = anything that isn't a double quote
EXCLAMATION = !
ITEM = (DOUBLEQUOTE (OTHER | (DOUBLEQUOTE OTHER DOUBLEQUOTE))* DOUBLEQUOTE
LINE = ITEM (EXCLAMATION ITEM)*
It feels like it's possible to create a regular expression for the above (assuming the double quotes in an ITEM can't be nested even further) BUT it might be better served by a very simple grammer.
This might work... excusing missing escapes and the like
^"([^"]*|"[^"]*")*"(!"([^"]*|"[^"]*")*")*$
Another option would be to match against the first part, then, if there's a !and more, prune off the ! and keep matching (excuse the no-particular-language, I'm just trying to illustrate the idea):
resultList = []
while(string matches \^"([^"]*|"[^"]*")*(.*)$" => match(1)) {
resultList += match
string = match(2)
if(string.beginsWith("!")) {
string = string[1:end]
} elseif(string.length > 0) {
// throw an error, since there was no exclamation and the string isn't done
}
}
if(string.length > 0) {
// throw an exception since the string isn't done
}
resultsList == the list of items in the string
EDIT: I realized that my answer doesn't really work. You can have a single doublequote inside the strings, as well as exclamation marks. As such, you really CAN'T have "!" inside one of the strings. As such, the idea of 1) pull quotes off the ends, 2) split on '"!"' is really the right way to go.