Parsing a value using regular expression in java

Question

I am trying to read a line and parse a value using regular expression in java. The line that contains the value looks something like this,

...... TESTYY912345 .......
...... TESTXX967890 ........

Basically, it contains 4 letters, then any two ASCII values followed by numeric 9 then (any) digits. And, i want to get the value, 912345 and 967890.

This is what I have so far in regular expression,

... TEST[\x00-\xff]{2}[9]{1} ...

But, this skips the 9 and parse 12345 and 67890. (I want to include 9 as well).

Thanks for your help.

João Silva · Accepted Answer · 2012-09-12 22:17:43Z

2

You are pretty close. Capture the entire group (9\\d*) after matching TEST\\p{ASCII}{2}. This way, you'll capture the 9 and the following digits:

String s  = "...... TESTYY912345 ......";
Pattern p = Pattern.compile("TEST\\p{ASCII}{2}(9\\d+)");
Matcher m = p.matcher(s);
if (m.find()) {
  System.out.println(m.group(1)); // 912345
}

edited Sep 12, 2012 at 22:17

answered Sep 12, 2012 at 22:12

João Silva

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Community · Accepted Answer · 2020-06-20 09:12:55Z

2

See my comment for a working expression, "TEST.{2}(9\\d*)".

final Pattern pattern = Pattern.compile("TEST.{2}(9\\d*)");
for (final String str : Arrays.asList("...... TESTYY912345 .......",
         "...... TESTXX967890 ........")) {
  final Matcher matcher = pattern.matcher(str);
  if (matcher.find()) {
    final int value = Integer.valueOf(matcher.group(1));
    System.out.println(value);
  }
}

See the result on ideone:

912345

967890

This will match any two characters (except a line terminator) for what is XX and YY in your example, and will take any digits after the 9.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 12, 2012 at 22:14

obataku

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