Develop Reference

Repeating capture group in a regular expression

What would be the best way to parse the following string in Java using a single regex?
String:
someprefix foo=someval baz=anotherval baz=somethingelse
I need to extract someprefix, someval, anotherval and somethingelse. The string always contains a prefix value (someprefix in the example) and can have from 0 to 4 key-value pairs (foo=someval baz=anotherval baz=somethingelse in the example)

You can use this regex for capturing your intended text,
(?<==|^)\w+
Which captures a word that is preceded by either an = character or is at ^ start of string.
Sample java code for same,
Pattern p = Pattern.compile("(?<==|^)\\w+");
String s = "someprefix foo=someval baz=anotherval baz=somethingelse";
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println(m.group());
}
Prints,
someprefix
someval
anotherval
somethingelse
Live Demo

What is wrong in regexp in Java

I want to get the word text2, but it returns null. Could you please correct it ?
String str = "Text SETVAR((&&text1 '&&text2'))";
Pattern patter1 = Pattern.compile("SETVAR\\w+&&(\\w+)'\\)\\)");
Matcher matcher = patter1.matcher(str);
String result = null;
if (matcher.find()) {
result = matcher.group(1);
}
System.out.println(result);

One way to do it is to match all possible pattern in parentheses:
String str = "Text SETVAR((&&text1 '&&text2'))";
Pattern patter1 = Pattern.compile("SETVAR[(]{2}&&\\w+\\s*'&&(\\w+)'[)]{2}");
Matcher matcher = patter1.matcher(str);
String result = "";
if (matcher.find()) {
result = matcher.group(1);
}
System.out.println(result);
See IDEONE demo
You can also use [^()]* inside the parentheses to just get to the value inside single apostrophes:
Pattern patter1 = Pattern.compile("SETVAR[(]{2}[^()]*'&&(\\w+)'[)]{2}");
^^^^^^
See another demo
Let me break down the regex for you:
SETVAR - match SETVAR literally, then...
[(]{2} - match 2 ( literally, then...
[^()]* - match 0 or more characters other than ( or ) up to...
'&& - match a single apostrophe and two & symbols, then...
(\\w+) - match and capture into Group 1 one or more word characters
'[)]{2} - match a single apostrophe and then 2 ) symbols literally.

Your regex doesn't match your string, because you didn't specify the opened parenthesis also \\w+ will match any combinations of word character and it won't match space and &.
Instead you can use a negated character class [^']+ which will match any combinations of characters with length 1 or more except one quotation :
String str = "Text SETVAR((&&text1 '&&text2'))";
"SETVAR\\(\\([^']+'&&(\\w+)'\\)\\)"
Debuggex Demo

multiple regex matches in a string

i have the following text:
bla [string1] bli [string2]
I like to match string1 and string2 with regex in a loop in java.
Howto do ?
my code so far, which only matches the first string1, but not also string 2.
String sRegex="(?<=\\[).*?(?=\\])";
Pattern p = Pattern.compile(sRegex); // create the pattern only once,
Matcher m = p.matcher(sFormula);
if (m.find())
{
String sString1 = m.group(0);
String sString2 = m.group(1); // << no match
}

Your regex is not using any captured groups hence this call with throw exceptions:
m.group(1);
You can use just use:
String sRegex="(?<=\\[)[^]]*(?=\\])";
Pattern p = Pattern.compile(sRegex); // create the pattern only once,
Matcher m = p.matcher(sFormula);
while (m.find()) {
System.out.println( m.group() );
}
Also if should be replaced by while to match multiple times to return all matches.

Your approach is confused. Either write your regex so that it matches two [....] sequences in the one pattern, or call find multiple times. Your current attempt has a regex that "finds" just one [...] sequence.
Try something like this:
Pattern p = Pattern.compile("\\[([^\\]]+)]");
Matcher m = p.matcher(formula);
if (m.find()) {
String string1 = m.group(0);
if (m.find(m.end()) {
String string2 = m.group(0);
}
}
Or generalize using a loop and an array of String for the extracted strings.
(You don't need any fancy look-behind patterns in this case. And ugly "hungarian notation" is frowned in Java, so get out of the habit of using it.)

Why does this pattern matching code not work?

I'm trying to do some pattern matching in Java:
Pattern p = Pattern.compile("(\\d+) (\\.+)");
Matcher m = p.matcher("5 soy milk");
String qty = m.group(1);
String name = m.group(2);
I want to end up with one string that contains "5" and one string that contains "soy milk". However, this pattern matching code gives me an IllegalStateException.

You have to call matches() before you attempt to get the groups.
http://docs.oracle.com/javase/7/docs/api/java/util/regex/Matcher.html#matches()
public boolean matches()
Attempts to match the entire region against the pattern.
If the match succeeds then more information can be obtained via the start, end, and group methods.
Try this:
Pattern p = Pattern.compile("(\\d+) (\\.+)");
Matcher m = p.matcher("5 soy milk");
if (m.matches())
{
String qty = m.group(1);
String name = m.group(2);
}

This is because you don't initiate your Matcher. You should p.matcher(...).matches() (or .find(), or .lookingAt(), depending on the desired behaviour -- real regex matching is done with .find()).
And check the result of .matches() since in your case it returns false: \.+ ("\\.+" in a Java string) will try and match a dot one or more times; you should use .+ (".+" in a Java string) to match "any character, one or more times".

How do I capture values from a regex?

How do extract values from a regex where any placeholder can be rederenced by a $number_of_occurance value?
For example, I have a string final_0.25Seg1-Frag1 and I want to find all matches of this string in a file with 0.25 as a wildcard, which I can do using
Pattern regex = Pattern.compile( "/vod/final_\\d+\\.\\d+Seg1-Frag1" );
Matcher regexMatcher = regex.matcher(data2[0]);
I want to retain the value of the value in \\d+\\.\\d and find which among all the matched lines has the biggest value in this position.

Have you looked at Pattern groups ? You can iterate through these to identify matched subexpressions.
From the linked example. Matcher.group(0) is the complete expression.
CharSequence inputStr = "abbabcd"; // could be a String type
String patternStr = "(a(b*))+(c*)";
// Compile and use regular expression
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(inputStr);
boolean matchFound = matcher.find();
if (matchFound) {
// Get all groups for this match
for (int i=0; i<=matcher.groupCount(); i++) {
String groupStr = matcher.group(i);
}
}

Please give an example. I guess you need to read Mathcer docs http://docs.oracle.com/javase/7/docs/api/index.html?java/util/regex/Matcher.html. You can access capturing groups via group method.