使用正则表达式替换字符

Question

String tempprop="(kfsdk)#";
tempprop = tempprop.replaceAll("[^\\s]\\)\\#", "\"?if_exists}");
System.out.println("1"+tempprop+"2");

I want the output as

1(kfsdk"?if_exists}2

but the output of this regular expression is

1(kfsd"?if_exists}2

Last k is getting trimmed, and I don't know why.

if tempprop is ( )#, then output should be 1( )#2 only without "?if_exists This regular expression adds "?if_exists if no space is present else it returns the string as it is

Answer 1

You could use a negative lookbehind instead of [^\\\\s] because it causes some effect in the final output. That is, lookarounds does a zero width match.

String tempprop="(kfsdk)#";
tempprop = tempprop.replaceAll("(?<!\\s)\\)#", "\"?if_exists}");
System.out.println("1"+tempprop+"2");

Output:

1(kfsdk"?if_exists}2

Explanation:

• (?<!\\s) Negative lookbehind which asserts what preceeds is not a space character.
• \\)# Matches the literal )# symbols.

Answer 2

To exclude [^\\s] from your capture, use the regex (?<![^\\s])\\)\\# . Another option is to enforce letters using (?<=\\w)\\)\\# or even forcing an opening bracket by using

out = str.replaceAll("(\\(\\w+)\\)\\#", "$1\"?if_exists}");

See here for more on lookaround (look-behind and look-ahead).