CODEWITHSUNDEEP
javaregex
user1205079
user1205079

Reputation: 405

java regular expression find and replace

I am trying to find environment variables in input and replace them with values.

The pattern of env variable is ${\\.}

Pattern myPattern = Pattern.compile( "(${\\.})" );
String line ="${env1}sojods${env2}${env3}";

How can I replace env1 with 1 and env2 with 2 and env3 with 3, so that after this I will have a new string 1sojods23?

Upvotes: 18

Views: 63209

Answers (7)

Brian Roach
Brian Roach

Reputation: 76898

Strings in Java are immutable, which makes this somewhat tricky if you are talking about an arbitrary number of things you need to find and replace.

Specifically you need to define your replacements in a Map, use a StringBuilder (before Java 9, less performant StringBuffer should have been used) and the appendReplacements() and appendTail() methods from Matcher. The final result will be stored in your StringBuilder (or StringBuffer).

Map<String, String> replacements = new HashMap<String, String>() {{
    put("${env1}", "1");
    put("${env2}", "2");
    put("${env3}", "3");
}};

String line ="${env1}sojods${env2}${env3}";
String rx = "(\\$\\{[^}]+\\})";

StringBuilder sb = new StringBuilder(); //use StringBuffer before Java 9
Pattern p = Pattern.compile(rx);
Matcher m = p.matcher(line);

while (m.find())
{
    // Avoids throwing a NullPointerException in the case that you
    // Don't have a replacement defined in the map for the match
    String repString = replacements.get(m.group(1));
    if (repString != null)    
        m.appendReplacement(sb, repString);
}
m.appendTail(sb);

System.out.println(sb.toString());

Output:

1sojods23

Upvotes: 68

user11404376
user11404376

Reputation: 11

    Map<String, String> replacements = new HashMap<String, String>() {
        {
            put("env1", "1");
            put("env2", "2");
            put("env3", "3");

        }
    };

    String line = "${env1}sojods${env2}${env3}";

    String rx = "\\$\\{(.*?)\\}";

    StringBuffer sb = new StringBuffer();
    Pattern p = Pattern.compile(rx);
    Matcher m = p.matcher(line);

    while (m.find()) {
        // Avoids throwing a NullPointerException in the case that you
        // Don't have a replacement defined in the map for the match
        String repString = replacements.get(m.group(1));
        if (repString != null)
            m.appendReplacement(sb, repString);
    }
    m.appendTail(sb);

    System.out.println(sb.toString());

In the above example we can use map with just key and values --keys can be env1 ,env2 ..

Upvotes: 1

Steven Spungin
Steven Spungin

Reputation: 29081

You can use a StringBuffer in combination with the Matcher appendReplacement() method, but if the the pattern does not match, there is no point in creating the StringBuffer.

For example, here is a pattern that matches ${...}. Group 1 is the contents between the braces.

static Pattern rxTemplate = Pattern.compile("\\$\\{([^}\\s]+)\\}");

And here is sample function that uses that pattern.

private static String replaceTemplateString(String text) {
    StringBuffer sb = null;
    Matcher m = rxTemplate.matcher(text);
    while (m.find()) {
        String t = m.group(1);
        t = t.toUpperCase(); // LOOKUP YOUR REPLACEMENT HERE

        if (sb == null) {
            sb = new StringBuffer(text.length());
        }
        m.appendReplacement(sb, t);
    }
    if (sb == null) {
        return text;
    } else {
        m.appendTail(sb);
        return sb.toString();
    }
}

Upvotes: 0

Kent
Kent

Reputation: 195039

This gives you 1sojods23:

String s = "${env1}sojods${env2}${env3}";
final Pattern myPattern = Pattern.compile("\\$\\{[^\\}]*\\}");
Matcher m = myPattern.matcher(s);
int i = 0;
while (m.find()) {
    s = m.replaceFirst(String.valueOf(++i));
    m = myPattern.matcher(s);
}

System.out.println(s);

and this works too:

final String re = "\\$\\{[^\\}]*\\}";
String s = "${env1}sojods${env2}${env3}";
int i = 0;
String t;
while (true) {
    t = s.replaceFirst(re, String.valueOf(++i));
    if (s.equals(t)) {
        break;
    } else {
        s = t;
    }
}

System.out.println(s);

Upvotes: 5

aretai
aretai

Reputation: 1641

Use groups once it is matched ${env1} will be your first group and then you use regex to replace what is in each group.

Pattern p = Pattern.compile("(${\\.})");
Matcher m = p.matcher(line);
while (m.find())
  for (int j = 0; j <= m.groupCount(); j++)
    //here you do replacement - check on the net how to do it;)

Upvotes: -2

Enissay
Enissay

Reputation: 4953

I know this is old, I was myself looking for a, appendReplacement/appendTail example when I found it; However, the OP's question doesn't need those complicated multi-line solutions I saw here.

In this exact case, when the string to replace holds itself the value we want to replace with, then this could be done easily with replaceAll:

String line ="${env1}sojods${env2}${env3}";

System.out.println( line.replaceAll("\\$\\{env([0-9]+)\\}", "$1") );

// Output => 1sojods23

DEMO

When the replacement is random based on some conditions or logic on each match, then you can use appendReplacement/appendTail for example

Upvotes: 13

Boris Strandjev
Boris Strandjev

Reputation: 46943

Hopefully you would find this code useful:

    Pattern phone = Pattern.compile("\\$\\{env([0-9]+)\\}");
    String line ="${env1}sojods${env2}${env3}";
    Matcher action = phone.matcher(line);
    StringBuffer sb = new StringBuffer(line.length());
    while (action.find()) {
      String text = action.group(1);
      action.appendReplacement(sb, Matcher.quoteReplacement(text));
    }
    action.appendTail(sb);
    System.out.println(sb.toString());

The output is the expected: 1sojods23.

Upvotes: 6

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