返回字符串中的Java正则表达式特殊字符

Question

I'm mostly coding in Perl and struggle now with a regex in Java. I have this:

// String dbfPath = "/result_feature/Monocytes-CD14+_H3K4me3_ENCODE_UW_bwa_samse"
// rs.getString = "Monocytes-CD14+_H3K4me3_ENCODE_UW_bwa_samse"

if(! dbfPath.matches(".*" + rs.getString("rs_name") + "/*"))

My problem is that the rs_name contains wildcards (+). I tried putting [] around, but then I have an illegal range (-).

How can I avoid the returning string being interpreted?

Thanks!

Answer 1

Use Pattern.quote(rs.getString("rs_name")) to automatically escape control characters.

Beware of null s.

Answer 2

您可以使用java.util.regex.Pattern.qoute() ：

if(! dbfPath.matches(".*" + java.util.regex.Pattern.qoute(rs.getString("rs_name") ) + "/*"))

Answer 3

Wrap your rs.getString() inside a Pattern.quote :

Pattern.quote(rs.getString("rs_name"))

Returns a literal pattern String for the specified String.

Answer 4

You could add '\\Q' & '\\E' in your string direclty.

\Q  Nothing, but quotes all characters until \E
\E  Nothing, but ends quoting started by \Q





if(! dbfPath.matches(".*\\Q" + rs.getString("rs_name") + "\\E/*"))

One thing, you should make sure your string doesn't contain '\\E', if yes, the safe way is using Pattern.quote() methods.