CODEWITHSUNDEEP
javaregexparameters
Chris Tien
Chris Tien

Reputation: 393

Java regular expression to match parameters within a function

I would like to write a regular expression to extract parameter1 and parameter2 of func1(parameter1, parameter2), the length of parameter1 and parameter2 ranges from 1 to 64.

(func1) (\() (.{1,64}) (,\\s*) (.{1,64}) (\))

My version can not deal with the following case (nested function)

func2(func1(ef5b, 7dbdd))

I always get a "7dbdd)" for parameter2. How could I solve this?

Upvotes: 2

Views: 2246

Answers (5)

muton
muton

Reputation: 382

Use [^)]{1,64} (match all except )) instead of .{1,64} (match any) to stop right before the first )

(func1) (\() (.{1,64}) (,\\s*) (.{1,64}) (\))
                                ^
                                replace . with [^)]

Example:

// remove whitespace and escape backslash!
String regex = "(func1)(\\()(.{1,64})(,\\s*)([^)]{1,64})(\\))";
String input = "func2(func1(ef5b, 7dbdd))";
Pattern p = Pattern.compile(regex); // java.util.regex.Pattern
Matcher m = p.matcher(input); // java.util.regex.Matcher
if(m.find()) { // use while loop for multiple occurrences
    String param1 = m.group(3);
    String param2 = m.group(5);

    // process the result...
}

If you want to ignore whitespace tokens, use this one:

func1\s*\(\s*([^\s]{1,64})\s*,\s*([^\s\)]{1,64})\s*\)"

Example:

// escape backslash!
String regex = "func1\\s*\\(\\s*([^\\s]{1,64})\\s*,\\s*([^\\s\\)]{1,64})\\s*\\)";
String input = "func2(func1 ( ef5b, 7dbdd ))";
Pattern p = Pattern.compile(regex); // java.util.regex.Pattern
Matcher m = p.matcher(input); // java.util.regex.Matcher
if(m.find()) { // use while loop for multiple occurrences
    String param1 = m.group(1);
    String param2 = m.group(2);

    // process the result...
}

Upvotes: 1

Rudziankoŭ
Rudziankoŭ

Reputation: 11251

^.*(func1)(\()(.{1,64})(,\s*)(.{1,64}[A-Za-z\d])(\))+

Working example: here

Upvotes: 0

Chris Tien
Chris Tien

Reputation: 393

(func1) (\() (.{1,64}) (,\\s*) ([^)]{1,64}) (\))

Upvotes: 0

HugaBuga
HugaBuga

Reputation: 21

Hope this helpful

func1[^\(]*\(\s*([^,]{1,64}),\s*([^\)]{1,64})\s*\)

Upvotes: 0

Dmitry Egorov
Dmitry Egorov

Reputation: 9650

Use "anything but closing parenthesis" ([^)]) instead of simply "anything" (.):

(func1) (\() (.{1,64}) (,\s*) ([^)]{1,64}) (\))

Demo: https://regex101.com/r/sP6eS1/1

Upvotes: 3

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