Home/ Questions/Q 7064267
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-28T04:48:48+00:00

Pattern pattern = Pattern.compile(<a>([a-zA-Z]+)</a>) Matcher matcher = pattern.matcher(<a>Text</a>); matcher.find() String str = matcher.group(); I

  • 0
Pattern pattern = Pattern.compile("<a>([a-zA-Z]+)</a>")
Matcher matcher = pattern.matcher("<a>Text</a>");
matcher.find()
String str = matcher.group();

I want to get "Text" to str, but I get "<a>Text</a>". Why and how should I do it properly?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    You need to specify the index of the group, 1 in this case:

    Pattern pattern = Pattern.compile("<a>([a-zA-Z]+)</a>")
Matcher matcher = pattern.matcher("<a>Text</a>");
matcher.find()
String str = matcher.group(1);

    Documentation for Matcher.group(int)

    • 0