Regex in Java not working while same regex is working in shell

Question

I want to replace all :variable (word starting with :) with ${variable}$ .

For example,

:aks_num with ${aks_num}$
:brn_num with ${brn_num}$

Following is my code, which does not work:

public static void main(String[] argv) throws Exception 
{
    CharSequence chSeq = "AND ((:aks_num = -1) OR (aks_num = :aks_num AND ((:brn_num = -1) OR (brn_num = :brn_num))))";

    // replaceAll also not working
    //String s = chSeq.replaceAll(":\\([a-z_]*\\)","\\${ $1 \\}$");

    Pattern p = Pattern.compile(":\\([a-z_]*\\)");
    Matcher m = p.matcher(chSeq);

    if (m.find()) {
         System.out.println("Found value: " + m.group(0) );
         System.out.println("Found value: " + m.group(1) );
         System.out.println("Found value: " + m.group(2) );
      } else {
         System.out.println("NO MATCH");
      }
}

While in shell script the following regex works perfectly:

s/:\([a-z_]*\)/${\1}$/g

ReplaceAll works - s.replaceAll(":([a-z_]+)", "\\${$1}\\$") — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 11, 2016 at 11:28
@praxnet: I posted an answer below with explanations and more ideas on the enhancements. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 11, 2016 at 11:38

Arnaud Denoyelle · Accepted Answer · 2016-04-11 11:35:21Z

2

:\$[a-z_]*\$ (with escaped parenthesis) means that you want to match expressions like :(aks_num). Obviously, there are no such expression in the input string. That explains why there are no matches.

Instead, if you want to use parenthesis in order to capture some variables, you should not escape the parenthesis.

Example :

CharSequence chSeq = "AND ((:aks_num = -1) OR (aks_num = :aks_num AND ((:brn_num = -1) OR (brn_num = :brn_num))))";
Pattern p = Pattern.compile(":([a-z_]*)");
Matcher m = p.matcher(chSeq);

while (m.find()) {
  System.out.println("Found value: " + m.group(0)+". Captured : "+m.group(1));
}

Output:

Found value: :aks_num. Captured : aks_num
Found value: :aks_num. Captured : aks_num
Found value: :brn_num. Captured : brn_num
Found value: :brn_num. Captured : brn_num

edited Apr 11, 2016 at 11:35

answered Apr 11, 2016 at 11:33

Arnaud Denoyelle

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1 Comment

Arnaud Denoyelle Over a year ago

@Shafizadeh some golds are easy to get but it is not my goal :)

rock321987 · Accepted Answer · 2016-04-11 11:38:44Z

2

CharSequence chSeq = "AND ((:aks_num = -1) OR (aks_num = :aks_num AND ((:brn_num = -1) OR (brn_num = :brn_num))))";

// replaceAll also not working
//String s = chSeq.replaceAll(":\\([a-z_]*\\)","\\${ $1 \\}$");

Pattern p = Pattern.compile(":(\\w+)");
Matcher m = p.matcher(chSeq);

while (m.find()) {
     System.out.println("Found value: " + m.group(1) );
}

Ideone Demo

Working fine with replaceAll

Pattern p = Pattern.compile("(:\\w+)");
Matcher m = p.matcher(x);
x = m.replaceAll("\\${$1}\\$");

edited Apr 11, 2016 at 11:38

answered Apr 11, 2016 at 11:31

rock321987

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Comments

dejvuth · Accepted Answer · 2016-04-11 11:28:28Z

1

You don't need to escape the parentheses, so

Pattern.compile(":([a-z_]*)");

should work.

answered Apr 11, 2016 at 11:28

dejvuth

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Comments

Wiktor Stribiżew · Accepted Answer · 2016-04-11 11:36:39Z

1

I believe you got confused with the Java's regex syntax that is different from regular sed syntax. You do not need to escape parentheses to make them "special" grouping operators. Vice versa, in Java, when you escape parentheses, they start matching literal ( and ) symbols.

In the replacement pattern, $ must be escaped for the regex engine to replace with literal $ symbols, but you do not need to escape braces there.

So, just use

.replaceAll(":([a-z_]+)", "\\${$1}\\$")

See the IDEONE demo

I suggest the + quantifier because I doubt you need to match a : followed with a space, or digits - any non-letter.

BTW, you do not need any /g flag in Java since replaceAll will replace all matches with the provided replacement pattern.

NOTE: you can further adjust the pattern to match all letters/digits/underscores with ":(\\w+)". Or just alphanumerics/underscore: ":([\\p{Alnum}_]+)".

edited Apr 11, 2016 at 11:36

answered Apr 11, 2016 at 11:33

Wiktor Stribiżew

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2 Comments

Wiktor Stribiżew Over a year ago

@Shafizadeh: You are thinking this up, see s/:$[a-z_]*$/${\1}$/g

Shafizadeh Over a year ago

Yes, maybe just a-z is up for OP. Anyway I think your NOTE is enough.

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