Question

I have an input string containing multiple lines(demarcated by \n). I need to search for a pattern in the lines and if its found, then replace the complete line with empty string.

My code looks like this,

Pattern p = Pattern.compile("^.*@@.*$");  
String regex = "This is the first line \n" +  
               "And this is second line\n" +  
               "Thus is @@{xyz} should not appear \n" +  
               "This is 3rd line and should come\n" +  
               "This will not appear @@{abc}\n" +  
               "But this will appear\n";  
Matcher m = p.matcher(regex);  
System.out.println("Output: "+m.group());  

I expect the response as :

Output: This is the first line       
        And this is second line  
        This is 3rd line and should come  
        But this will appear.

I am unable to get it, please help, me out.

Thanks,
Amit

Answer 1

Others mention turning on multiline mode but since Java does not default to DOTALL (single line mode) there is an easier way… just leave the ^ and $ off.

String result = regex.replaceAll( ".*@@.*", "" );

Note that the issue with either this or using:

"(?m)^.*@@.*$" 

…is that it will leave the blank lines in. If it is a requirement to not have them then the regex will be different.

Full regex that does not leave blank lines:

String result = regex.replaceAll( ".*@@.*(\r?\n|\r)?", "" );