regex to replace multiple characters with one

Question

Suppose I have a string like this:

"abc%\%%%%"

I want to replace multiple %%% with only one %. I tried something like

String st = "abc%\%%%%".replaceAll("(%)\\1+", "$1");

But that would also turn "\%%" into "\%", which is not what I want. In other words, I want to replace multiple % but leave alone those preceded by a backslash.

What regex should I use? Thanks

KV Prajapati · Accepted Answer · 2011-11-25 03:41:04Z

2

Try,

String st = "abc%\\%%%%".replaceAll("([%])+","$1");
       st = "Not%%%%\\%%%%sure but%%%\\%%%%%try out".replaceAll("(\\\\)([%])+","$1$2");

edited Nov 25, 2011 at 3:41

answered Nov 25, 2011 at 3:13

KV Prajapati

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Comments

Pipalayan Nayak · Accepted Answer · 2011-11-25 17:09:50Z

0

Try this

String st = "abc%\\%%%%";
String regex = "([\\\\]%)%+";
String result= st.replaceAll(regex,"\\\\%%");

result is abc%\%%

If you want to check the correctness, you can use this statement

 String result= st.replaceAll(regex,"\\\\%R");

result is abc%\%R (which confirms that % preceded by backlash is left alone.)

answered Nov 25, 2011 at 17:09

Pipalayan Nayak

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