如何创建一个只在某个位置匹配字符串的java正则表达式模式？

Question

I would like to create a regular expression pattern that would succeed in matching only if the pattern string not followed by any other string in the test string or input string ! Here is what i tried :

      Pattern p = Pattern.compile("google.com");//I want to know the right format

      String input1 = "mail.google.com";
      String input2 = "mail.google.com.co.uk";

      Matcher m1 = p.matcher(input1);
      Matcher m2 = p.matcher(input2);

      boolean found1 = m1.find();
      boolean found2 = m2.find();//This should be false because "google.com" is followed by ".co.uk" in input2 string

Any help would be appreciated!

Answer 1

Your pattern should be google\\.com$ . The $ character matches the end of a line. Read about regex boundary matchers for details.

Answer 2

Here is how to match and get the non-matching part as well.

Here is the raw regex pattern as an interactive link to a great regular expression tool

^(.*)google\\.com$

^ - match beginning of string
(.*) - capture everything in a group up to the next match
google - matches google literal
\\. - matches the . literal has to be escaped with \\
com - matches com literal
$ - matches end of string

Note: In Java the \\ in the String literal has to be escaped as well! ^(.*)google\\\\.com$

Answer 3

Pattern p = Pattern.compile("google\.com$");

The dollar sign means it has to occur at the end of the line/string being tested. Note too that your dot will match any character, so if you want it to match a dot only, you need to escape it.

Answer 4

You should use google\\.com$ . $ character matches the end of a line.

Pattern p = Pattern.compile("google\\.com$");//I want to know the right format

String input2 = "mail.google.com.co.uk";

Matcher m2 = p.matcher(input2);


boolean found2 = m2.find();
System.out.println(found2);

Output = false