CODEWITHSUNDEEP
javaregex
Abhishek
Abhishek

Reputation: 1068

Remove double quotes from output Java

I am trying to extract a url from the string. But I am unable to skip the double quotes in the output.

import java.util.regex.Matcher;
import java.util.regex.Pattern;
class Main {
  public static void main(String[] args) {
    String s1 = "<a id=\"BUTTON_LINK\" style=\"%%BUTTON_LINK%%\" target=\"_blank\" href=\"https://||domainName||/basketReviewPageLoadAction.do\">%%CHECKOUT%%</a>";
    //System.out.println(s1);
    Pattern pattern = Pattern.compile("\\s*(?i)href\\s*=\\s*(\"([^\"]*\")|'[^']*'|([^'\">\\s]+))");
    Matcher matcher = pattern.matcher(s1);

    if(matcher.find()){
      String url = matcher.group(1);
      System.out.println(url);
    }

  }
}

My Output is:

"https://||domainName||/basketReviewPageLoadAction.do"

Expected Output is:

https://||domainName||/basketReviewPageLoadAction.do

I cannot do string replace. I have add few get param in this output and attach back it to original string.

Upvotes: 0

Views: 306

Answers (5)

Srdjan M.
Srdjan M.

Reputation: 3405

Regex: (?<=href=")([^\"]*) Substitution: $1?params...

Details:

  • (?<=) Positive Lookbehind
  • () Capturing group
  • [^] Match a single character not present in the list
  • * Matches between zero and unlimited times
  • $1 Group 1.

Java code:

By using function replaceAll you can add your params ?abc=12 to the end of the capturing group $1 in this case href.

String text = "<a id=\"BUTTON_LINK\" style=\"%%BUTTON_LINK%%\" target=\"_blank\" href=\"https://||domainName||/basketReviewPageLoadAction.do\">%%CHECKOUT%%</a>";
text = text.replaceAll("(?<=href=\")([^\"]*)", String.format("$1%s", "?abc=12"));
System.out.print(text);

Output:

<a id="BUTTON_LINK" style="%%BUTTON_LINK%%" target="_blank" href="https://||domainName||/basketReviewPageLoadAction.do?abc=12">%%CHECKOUT%%</a>

Code demo

Upvotes: 1

Vimal
Vimal

Reputation: 73

You will try this out,

s1 = s1.Replace("\"", "");

Upvotes: -1

Abhishek
Abhishek

Reputation: 1068

This solution worked for now.

Pattern pattern = Pattern.compile("\\s*(?i)href\\s*=\\s*\"([^\"]*)");

Upvotes: 0

lin
lin

Reputation: 1589

ugly, seems works.Hope this help.

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.stream.Collectors;
import java.util.stream.Stream;

class Main {
    public static void main(String[] args) {
        String s1 = "<a id=\"BUTTON_LINK\" style=\"%%BUTTON_LINK%%\" target=\"_blank\" href= \"https://||domainName||/basketReviewPageLoadAction.do\">%%CHECKOUT%%</a>";
        //System.out.println(s1);
        Pattern pattern = Pattern.compile("\\s*(?i)href\\s*=\\s*(\"([^\"]*)\"|'([^']*)'|([^'\">\\s]+))");
        Matcher matcher = pattern.matcher(s1);

        if (matcher.find()) {
            String url = Stream.of(matcher.group(2), matcher.group(3),
                    matcher.group(4)).filter(s -> s != null).collect(Collectors.joining());
            System.out.print(url);

        }

    }
}

Upvotes: 0

Eugene S
Eugene S

Reputation: 6910

You can try one of these options:

System.out.println(url.replaceAll("^\"|\"$", ""));
System.out.println(url.substring(1, url.length()-1));

Upvotes: 1

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