Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8685869
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 12, 20262026-06-12T22:41:13+00:00 2026-06-12T22:41:13+00:00

I have: Pattern pat = Pattern.compile((\\d+) (\\d+) (1$)); Matcher mat = pat.matcher(line); with matches

  • 0

I have:

Pattern pat = Pattern.compile("(\\d+) (\\d+) (1$)");
Matcher mat = pat.matcher(line);

with matches for:

1 2 1

but not for:

1     2     1

How can I achieve, that pattern matching is insenitive according to spaces between numbers?

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-12T22:41:14+00:00Added an answer on June 12, 2026 at 10:41 pm

    Use \s for one space and add a + that means one ore more spaces.

    "(\\d+)\\s+(\\d+)\\s+(1$)"
    

    If you want zero or more spaces you have to use a * instead of the +.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Let's say I don't have photoshop, but I want to make pattern files (.pat)
I have a pattern like this that matches multiple sets of values: (((\w+) (\d+))+)
I have this pattern: [0-9]*\.?[0-9] It matches numbers but it also matches 3.5.4 as:
I have this pattern ^(?:http://)?(?:www.)?(.*?)/?(.*?)$ but it's still not perfect. Let's say we have
I have the following pattern matching case in a scala function: def someFunction(sequences: Iterable[Seq[Int]]):Seq[Int]
I have this module pattern that stores a bunch of vars. I want to
I currently have a: @Pattern(regexp=\p{Alpha}+, message=Only Alphabetic chars allowed) That restricts the user to
I have multiple pages that have this pattern: <iframe frameborder =0 src=[someURL] width=100% height=900>
In perl, if I have a string $str and a pattern saved in $pat
I need help with re module. I have pattern: pattern = re.compile('''first_condition\((.*)\) extra_condition\((.*)\) testing\((.*)\)

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.