regex exclude a character not work

Question

i want to get class name from java file .for example

class Mango {

now i want to get mango as class name.

this is the regex i used

\s*class\s+(\S+)

it works and i captured the class name .the problem is if clasname has no spaces between classname and closing curly bracket i get name as mango{

like below one.

class Mango{

so i want to exclude { from group.so i modyfy to following

\s*class\s+(\S+|[^{])

but it doesn't work and still capture classname with closing bracket. how can i get only class name .

score 4 · Answer 1 · edited May 23 '17 at 12:15

4

try with this regex:

class\s+([\w$]+)

\w - is a word character (a-zA-Z0-9_)

This regex will capture only allowed characters for class names. However if we assume that coder used valid characters, you can also try:

(?<=class\s)\s*(.+?)(?=\s*\{)

DEMO

which is:

(?<=class\s) - positive lookbehind for word class and whitespace,
\s*(.+?) - zero or more whitespaces, and a one or more character,
(?=\s*\{) - positive lookahead for whitspace(s) and curly bracket {

to directly get class name. This regex allow all characters, however it could be useful if coder would use one of rarely used characters allowed in Java names.

edited May 23 '17 at 12:15

Community

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answered Aug 03 '15 at 12:55

m.cekiera

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Even better would be a regex which matches the Java identifier rules, IIRC: `class\s+([a-zA-Z_]\w*)` (identifiers start with an alphabetic or underscore character and follow with zero or more alphanumerics (or underscore). (May better better using Unicode character classes: I cannot recall if Java identifiers can use characters beyond ASCII). – Richard Aug 03 '15 at 13:00
Problem with `\w` is that it will not match all possible characters which can be used in class name like `$`. – Pshemo Aug 03 '15 at 13:00
@m.cekiera thanks but `(?<=class\s)\s*(.+)(?=\s+\{)` doesn't capture mango from `class mango{` – while true Aug 03 '15 at 13:09
@whiletrue sorry, my obvius mistake, I assumed that there would be always space after name, I updated answer – m.cekiera Aug 03 '15 at 13:12

Pshemo · Answer 2 · 2015-08-03T13:25:21.230

To accept only proper characters which can be used as names of classes or variables we may want to use method Character.isJavaIdentifierPart, which can be referred via \p{javaJavaIdentifierPart} as explained in Pattern class documentation (emphasis mine):

Categories that behave like the java.lang.Character boolean isMethodName methods (except for the deprecated ones) are available through the same \p{property} syntax where the specified property has the name javaMethodName.

Demo:

String text = "class Mango{";

Pattern p = Pattern.compile("\\s*class\\s+(\\p{javaJavaIdentifierPart}+)");
Matcher m = p.matcher(text);

if (m.find()){
    System.out.println(m.group(1));
}else{
    System.out.println("no match found");
}

Output: Mango

If you are not going to use this regex in Java engine, then you can use [^\s{] instead of \S like

\s*class\s+([^\s{]+)

It will accept any character except whitespaces and {.

@whiletrue Updated answer again to include `\p{javaJavaIdentifierPart}` which seems best tool in Java's regex to handle this case. — Pshemo, Aug 03 '15 at 13:31

score 2 · Answer 3 · answered Aug 03 '15 at 13:03

Not sure what the question is: do you want an explanation for (a) why your approach does not work or would you like to know (b) what a correct regex for this problem looks like?

If it's the latter, other answers and comments have provided some correct expressions. If it's the former, then consider what

(\S+|[^{])

actually matches. What this basically says is: match everything that is not a whitespace or is not a left curly bracket. Note the "or" in that sentence.

The reason your expression still matches the { after "Mango" is that it satisfies the first part of the sentence: a { is not a whitespace, so your regex matches. The same argument applies to the space between "Mango" and the {: it satisfies the "not a left curly bracket" part, and thus is matched as well.

What you want is an expression that encodes: match everything that is not a whitespace and is not a left curly bracket. As mentioned above, other answers/comments to this question show examples on how to achieve this.

regex exclude a character not work

3 Answers3