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Regex to extract part of string before regex match

Say i have the following line:
My '123? Text – 56x73: Hello World blablabla
I want to extract everything before " - 56x73 ..."
I already found a regex to match the part which I don't want to extract:
\s–\s\d{1,2}x\d{1,2}:\s.+
How can I get only the other part using Java and Regex?

use
String str= ...
String regex= your regex
Pattern pattern;
Matcher matcher;
pattern = Pattern.compile(regex);
matcher = pattern.matcher(str);
if (matcher.find())
{
matcher.group(0, 1, ...)
use () in your regex to deliminate groups

You already got way around but this can be helpful,
Assuming 56 and 73 will NOT be constant.
Use Regex: "(.*)(\\s)(.*)(\\s)([\\d]+[x][\\d]+)"
then use "group(int number)" where a number will be 1 in this case.
I used .* between two \s intentionally to get around with "-" thing I didn't anything about that but I found this. Also noticed from one of the comment.
If anybody wants to edit and improve my answer you are more than welcome.

Java // No match with RegExp and square brackets

I have a string like
Berlin -> Munich [label="590"]
and now I'm searching a regular expression in Java that checks if a given line (like above) is valid or not.
Currently, my RegExp looks like \\w\\s*->\\s*\\w\\s*\\[label=\"\\d\"\\]"
However, it doesn't work and I've found out that \\w\\s*->\\s*\\w\\s* still works but when adding \\[ it can't find the occurence (\\w\\s*->\\s*\\w\\s*\\[).
What I also found out is that when '->' is removed it works (\\w\\s*\\s*\\w\\s*\\[)
Is the arrow the problem? Can hardly imagine that.
I really need some help on this.
Thank you in advance

This is the correct regular expression:
"\\w+\\s*->\\s*\\w+\\s*\\[label=\"\\d+\"\\]"
What you report about matches and non-matches of partial regular expressions is very unlikely, not possible with the Berlin/Munich string.
Also, if you are really into German city names, you might have to consider names like Castrop-Rauxel (which some wit has called the Latin name of Wanne-Eickel ;-) )

Try this
String message = "Berlin -> Munich [label=\"590\"]";
Pattern p = Pattern.compile("\\w+\\s*->\\s*\\w+\\s*\\[label=\"\\d+\"\\]");
Matcher matcher = p.matcher(message);
while(matcher.find()) {
System.out.println(matcher.group());
}
You need to much more than one token of characters and numbers.

Java, regular expression catching multiple occurances of pattern

This is my original String:
String response = "attributes[{"id":50,"name":super},{"id":55,"name":hello}]";
I'm trying to parse the String and extract all the id values e.g
50
55
Pattern idPattern = Pattern.compile("{\"id\":(.*),");
Matcher matcher = idPattern.matcher(response);
while(matcher.find()){
System.out.println(matcher.group(1));
}
When i try to print the value i get an exception:
java.util.regex.PatternSyntaxException: Illegal repetition
Not had much experience with regular expressions in the past but cannot find a simple solution to this online.
Appreciate any help!

Pattern.compile("\"id\":(\\d+)");

Don't use a greedy match operator like * with a . which matches any character. unnecessarily.
If you want the digits extracted, you can use \d.
"id":(\d+)
Within a Java String,
Pattern.compile("\"id\":(\\d+)");

{ is a reserved character in regular expressions and should be escaped.
\{\"id\":(.*?),
Edit : If you're going to be working with JSON, you should consider using a dedicated JSON parser. It will make your life much easier. See Parsing JSON Object in Java

Need regex to match the given string

I need a regex to match a particular string, say 1.4.5 in the below string . My string will be like
absdfsdfsdfc1.4.5kdecsdfsdff
I have a regex which is giving [c1.4.5k] as an output. But I want to match only 1.4.5. I have tried this pattern:
[^\\W](\\d\\.\\d\\.\\d)[^\\d]
But no luck. I am using Java.
Please let me know the pattern.

When I read your expression [^\\W](\\d\\.\\d\\.\\d)[^\\d] correctly, then you want a word character before and not a digit ahead. Is that correct?
For that you can use lookbehind and lookahead assertions. Those assertions do only check their condition, but they do not match, therefore that stuff is not included in the result.
(?<=\\w)(\\d\\.\\d\\.\\d)(?!\\d)
Because of that, you can remove the capturing group. You are also repeating yourself in the pattern, you can simplify that, too:
(?<=\\w)\\d(?:\\.\\d){2}(?!\\d)
Would be my pattern for that. (The ?: is a non capturing group)

Your requirements are vague. Do you need to match a series of exactly 3 numbers with exactly two dots?
[0-9]+\.[0-9]+\.[0-9]+
Which could be written as
([0-9]+\.){2}[0-9]+
Do you need to match x many cases of a number, seperated by x-1 dots in between?
([0-9]+\.)+[0-9]+

Use look ahead and look behind.
(?<=c)[\d\.]+(?=k)
Where c is the character that would be immediately before the 1.4.5 and k is the character immediately after 1.4.5. You can replace c and k with any regular expression that would suit your purposes

I think this one should do it : ([0-9]+\\.?)+

Regular Expression
((?<!\d)\d(?:\.\d(?!\d))+)
As a Java string:
"((?<!\\d)\\d(?:\\.\\d(?!\\d))+)"

String str= "absdfsdfsdfc**1.4.5**kdec456456.567sdfsdff22.33.55ffkidhfuh122.33.44";
String regex ="[0-9]{1}\\.[0-9]{1}\\.[0-9]{1}";
Matcher matcher = Pattern.compile( regex ).matcher( str);
if (matcher.find())
{
String year = matcher.group(0);
System.out.println(year);
}
else
{
System.out.println("no match found");
}

Regular expression not extracting the exact pattern

I am working in Java to read a string of over 100000 characters.
I have a list of keywords, that I search the string for, and if the string is present I call a function which does some internal processing.
The kind of keyword I have is "face", for example - I wish to get all the patterns where I have matches for "faces" not "facebook". I can accept a space character behind the face in the string so if in a string I have a match like " face" or " faces" or "face " or " faces" i can accept that too. However I can not accept "duckface" or "duckface " etc.
I have written the regex
Pattern p = Pattern.compile("\\s+"+keyword+"s\\s+|\\s+");
where keyword is my list of keywords, but I am not getting the desired results. Can you read my description and please suggest what might be issue and how I can fix it?
Also if a pointer to a really good regex for Java page is shared I would appreciate that as well.
Thank you Contributers ..
Edit
The reason I know it is not working is I have used the following code:
Pattern p = Pattern.compile("\\s+"+keyword+"s\\s+|\\s+");
Matcher m = p.matcher(myInputDataSting);
if(m.find())
{
System.out.println("Its a Match: "+m.group());
}
This returns a blank string...

If keyword is "face", then your current regex is
\s+faces\s+|\s+
which matches either one or more whitespace characters, followed by faces, followed by one or more whitespace characters, or one or more whitespace characters. (The pipe | has very low precedence.)
What you really want is
\bfaces?\b
which matches a word boundary, followed by face, optionally followed by s, followed by a word boundary.
So, you can write:
Pattern p = Pattern.compile("\\b"+keyword+"s?\\b");
(though obviously this will only work for words like face that form their plurals by simply adding s).
You can find a comprehensive listing of Java's regular-expression support at http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html, but it's not much of a tutorial. For that, I'd recommend just Googling "regular expression tutorial", and finding one that suits you. (It doesn't have to be Java-specific: most of the tutorials you'll find are for flavors of regular-expression that are very similar to Java's.)

You should use
Pattern p = Pattern.compile("\b"+keyword+"s?\b");
, where keyword is not plural. \\b means that keyword must be as a complete word in searched string. s? means that keyword's value may end with s.
If you are not familar enough with regular expressions I recommend reading http://docs.oracle.com/javase/tutorial/essential/regex/index.html, because there are examples and explanations.