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.
i have the following sample cases :
1) "Sample"
2) "[10,25]"
I want to form a(only one) regular expression pattern, to which the above examples are passed returns me "Sample" and "10,25".
Note: Input strings do not include Quotes.
I came up with the following expression (?<=\[)(.*?)(?=\]), this satisfies the second case and retreives me only "10,25" but when the first case is matched it returns me blank. I want "Sample" to be returned? can anyone help me.
C#.
here you go, a small regex using a positive lookbehind, sometime these are very handy
Regex
(?<=^|\[)([\w,]+)
Test string
Sample
[10,25]
Result
MATCH 1
[0-6] Sample
MATCH 2
[8-13] 10,25
try at regex101.com
if " is included in your original string, use this regex, this will look for " mark as well, you may choose to remove ^| from lookup if " mark is always included or you may choose to leave it as it is if your text has combination of with and without " marks
Regex
(?<=^|\[|\")([\w,]+)
try at regex101.com
As far as I can tell, the below regex should help:
Regex regex = new Regex(#"^\w+|[[](\w)+\,(\w)+[]]$");
This will match multiple words, or 2 words (alphanumeric) separated by commas and inside square brackets.
One Java example:
// String input = "Sample";
String input = "[10,25]";
String text = "[^,\\[\\]]+";
Pattern pMod = Pattern.compile("(" + text + ")|(?>\\[(" + text + "," + text + ")\\])");
Matcher mMod = pMod.matcher(input);
while (mMod.find()) {
if(mMod.group(1) != null) {
System.out.println(mMod.group(1));
}
if(mMod.group(2)!=null) {
System.out.println(mMod.group(2));
}
}
if input is "[hello&bye,25|35]", then the output is hello&bye,25|35
I am trying to extract all of the text (shown as xxxx) in the follow pattern:
Session["xxxx"]
using c#
This may be Request.Querystring["xxxx"] so I am trying to build the expression dynamically. When I do so, I get all sorts of problems about unescaped charecters or no matches :(
an example might be:
string patternstart = "Session[";
string patternend = "]";
string regexexpr = #"\\" + patternstart + #"(.*?)\\" + patternend ;
string sText = "Text to be searched containing Session[\"xxxx\"] the result would be xxxx";
MatchCollection matches = Regex.Matches(sText, #regexexpr);
Can anyone help with this as I am stumped (as I always seem to be with RegEx :) )
With some little modifications to your code.
string patternstart = Regex.Escape("Session[");
string patternend = Regex.Escape("]");
string regexexpr = patternstart + #"(.*?)" + patternend;
The pattern you construct in your example looks something like this:
\\Session[(.*?)\\]
There are a couple of problems with this. First it assumes the string starts with a literal backslash, second, it wraps the entire (.*?) in a character class, that means it will match any single open parenthesis, period, asterisk, question mark, close parenthesis or backslash. You'd need to escape the the brackets in your pattern, if you want to match a literal [.
You could use a pattern like this:
Session\[(.*?)]
For example:
string regexexpr = #"Session\[(.*?)]";
string sText = "Text to be searched containing Session[\"xxxx\"] the result would be xxxx";
MatchCollection matches = Regex.Matches(sText, #regexexpr);
Console.WriteLine(matches[0].Groups[1].Value); // "xxxx"
The characters [ and ] have a special meaning with regular expressions - they define a group where one of the contained characters must match. To work around this, simply 'escape' them with a leading \ character:
string patternstart = "Session\[";
string patternend = "\]";
An example "final string" could then be:
Session\["(.*)"\]
However, you could easily write your RegEx to handle Session, Querystring, etc automatically if you require (without also matching every other array you throw at it), and avoid having to build up the string in the first place:
(Querystring|Session|Form)\["(.*)"\]
and then take the second match.
Is there a way to convert string without spaces to a proper sentence??
E.g. "WhoAmI" needs to be converted to "Who Am I"
A regex replacement would do this, if you're just talking about inserting a space before each capital letter:
using System;
using System.Text.RegularExpressions;
class Test
{
static void Main()
{
var input = "WhoAmI";
var output = Regex.Replace(input, #"\p{Lu}", " $0").TrimStart();
Console.WriteLine(output);
}
}
However, I suspect there will be significant corner cases. Note that the above uses \p{Lu} instead of just [A-Z] to cope with non-ASCII capital letters; you may find A-Z simpler if you only need to deal with ASCII. The TrimStart() call is to remove the leading space you'd get otherwise.
If every word in the string is starting with uppercase you may just convert each part that is starting with uppercase to a space separated string.
You can use LINQ
string words = "WhoAmI";
string sentence = String.Concat(words.Select(letter => Char.IsUpper(letter) ? " " + letter
: letter.ToString()))
.TrimStart();
How can I get the string before the character "-" using regular expressions?
For example, I have "text-1" and I want to return "text".
So I see many possibilities to achieve this.
string text = "Foobar-test";
Regex Match everything till the first "-"
Match result = Regex.Match(text, #"^.*?(?=-)");
^ match from the start of the string
.*? match any character (.), zero or more times (*) but as less as possible (?)
(?=-) till the next character is a "-" (this is a positive look ahead)
Regex Match anything that is not a "-" from the start of the string
Match result2 = Regex.Match(text, #"^[^-]*");
[^-]* matches any character that is not a "-" zero or more times
Regex Match anything that is not a "-" from the start of the string till a "-"
Match result21 = Regex.Match(text, #"^([^-]*)-");
Will only match if there is a dash in the string, but the result is then found in capture group 1.
Split on "-"
string[] result3 = text.Split('-');
Result is an Array the part before the first "-" is the first item in the Array
Substring till the first "-"
string result4 = text.Substring(0, text.IndexOf("-"));
Get the substring from text from the start till the first occurrence of "-" (text.IndexOf("-"))
You get then all the results (all the same) with this
Console.WriteLine(result);
Console.WriteLine(result2);
Console.WriteLine(result21.Groups[1]);
Console.WriteLine(result3[0]);
Console.WriteLine(result4);
I would prefer the first method.
You need to think also about the behavior, when there is no dash in the string. The fourth method will throw an exception in that case, because text.IndexOf("-") will be -1. Method 1 and 2.1 will return nothing and method 2 and 3 will return the complete string.
Here is my suggestion - it's quite simple as that:
[^-]*
This is something like the regular expression you need:
([^-]*)-
Quick tests in JavaScript:
/([^-]*)-/.exec('text-1')[1] // 'text'
/([^-]*)-/.exec('foo-bar-1')[1] // 'foo'
/([^-]*)-/.exec('-1')[1] // ''
/([^-]*)-/.exec('quux')[1] // explodes
I dont think you need regex to achieve this. I would look at the SubString method along with the indexOf method. If you need more help, add a comment showing what you have attempted and I will offer more help.
You could just use another non-regex based method. Someone gave the suggestion of using Substring, but you could also use Split:
string testString = "my-string";
string[] splitString = testString.Split("-");
string resultingString = splitString[0]; //my
See http://msdn.microsoft.com/en-US/library/ms228388%28v=VS.80%29.aspx for another good example.
If you want use RegEx in .NET,
Regex rx = new Regex(#"^([\w]+)(\-)*");
var match = rx.Match("thisis-thefirst");
var text = match.Groups[1].Value;
Assert.AreEqual("thisis", text);
Find all word and space characters up to and including a -
^[\w ]+-