Java―escape string to use in RegExp

Question

I am a bit new to Java, so I am just unsure if I am doing things like they should or not. I am developing a Program, that is more for learning Java purpose and which is thought for commandline use. One of the Features should be a kind of search an replace.

Starting the program from the commandline like this:

java -jar pro.jar -s ${ftp://www.stuff.com} -r foo

means: search for the exact String ${ftp://www.stuff.com} and replace it with foo. I want to search with an Regexp, so I have to escape the escape-characters ($,(,{,},\,…) in the search string.

${ftp://www.stuff.com} --> \$\{ftp:\/\/www\.stuff\.com\}

Therefore I wrote that function:

private static Pattern getSearchPattern()
{
    String searchArg = cli.getOptionValue( "s" ).trim();
    StringBuffer escapedSearch = new StringBuffer();
    Pattern metas = Pattern.compile( "([\\.\\?\\*\\{\\}\\[\\]\\(\\)\\^\\$\\+\\|\\\\])" );
    Matcher metaMatcher = metas.matcher( searchArg );

    while( metaMatcher.find() )
    {
        metaMatcher.appendReplacement(escapedSearch, "\\\\\\\\$0" );
    }
    metaMatcher.appendTail( escapedSearch );

    return Pattern.compile( escapedSearch.toString() );
}

all in all does this work, but there are so many backslashes. Does this escape all metachrackters? Is that a »robust« solution?

dont know if its a possibility for you but maybe you could make sure there's just a single argument that accepts patterns and provide it always as the last one. then the remainder ofthe commandline is always your argument and there's less escaping to do — radai
– radai, Commented May 4, 2013 at 9:20

flavian · Accepted Answer · 2013-05-04 09:20:53Z

6

You can do something much better than use them. You can use a PatternQuote.

Pattern metas = Pattern.quote("no escaped regex chars in here");

One other way to get rid of \\ is to use character classes []. So if you put [.] instead of \\. the results will be the same and hopefully you get some readability.

answered May 4, 2013 at 9:20

flavian

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2 Comments

Ian Roberts Over a year ago

+1 for Pattern.quote, which is designed for precisely this task. There's also Matcher.quoteReplacement which you can use on the replacement string if you want it to be treated literally (rather than $1 being treated as a backreference to a capturing group in the pattern).

philipp Over a year ago

Thank you! Works just like expected! Nice!

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