I'm having problems with Java RegEx. That's my regex statement "\"730\"\s+{([^}]+)}" and it works on an regex checking website, but I have trouble getting it to work in Java. That's my current code.
String patternString = '\"730\"\s+{([^}]+)}';
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(vdfContentsString);
boolean matches = matcher.matches();
Thanks for advice.
It says "Illegal escape character in character literal".
Single quotes (') declare characters, double quotes (") declare strings, that's why you get the syntax error Illegal escape character in character literal. Second, regex itself syntactically uses the backslash, as in \s for whitespace. Maybe confusing might be the fact that Java also uses \ for character escaping. That's why you need two backslashes (\\s in Java will become \s for the resulting regular expression).
Then you need to take care of special characters in regular expressions: { and } are quantifiers ("repeat n times"), if you want them literally, escape them (\\{ and \\})
So if you want to match a string like "730" {whatever}, use this regular expression:
"730"\s+\{([^}]+)\}
or in Java:
String patternString = "\"730\"\\s+\\{([^}]+)\\}";
Example:
String str = "\"730\" { \"installdir\" \"C:\\Program Files (x86)\\Steam\\steamapps\\common\\Counter-Strike Global Offensive\" \"HasAllLocalContent\" \"1\" \"UpToDate\" \"1\" }";
String patternString = "\"730\"\\s+\\{([^}]+)\\}";
System.out.println(str.matches(patternString)); // true
String line = line.trim();) after reading a line to remove line breaks or add (?s).* or [\r\n]* to the regular expression above in order to match line breaks.Exception in thread "main" java.util.regex.PatternSyntaxException: Illegal repetition
Escape { and } as well because in Java Regex Pattern it has special meaning.
String patternString = "\"730\"\\s+\\{([^\\}]+)\\}";
String#matches() method looks for whole string if you are looking for sub-string of a long string then use Matcher#find() method and get the result from the groups that is captured by enclosing the pattern inside parenthisis (...).
sample code:
String patternString = "(\"730\"\\s+\\{([^\\}]+)\\})";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(vdfContentsString);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
String.matches() that matches whole string not just a part of it. have a look at the update{, } are Metacharacters (See HERE for metacharacters) and need to be escaped with \\, hence, \\{ .. \\}.
\ is an escape character, while \s, \w, \d etc (See HERE for a list) are metacharacters, therefore, as mentioned above, these need to be escaped as well, hence, \\s+
instead of [^\\}], i would suggest (.+?)}
This is working:
String patternString = '\\\"730\\\"\\s+\\{(.+?)\\}';
The above is the required Java string which gets parsed into the following regular expression: \"730\"\s+\{(.+?)\}, and then it can be used to match the input string. Tadan!
two levels of parsing!
Java and see how the regex is being linked to the java string and subsequently matched
vdfContentsString?