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 894289
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 15, 20262026-05-15T14:21:35+00:00 2026-05-15T14:21:35+00:00

I have a MySql table having the following structure: ontology_term pathway_id pathway_name I want

  • 0

I have a MySql table having the following structure:

ontology_term
pathway_id
pathway_name

I want to write a query using which we can get mapping between various pathways (having unique id’s -> pathway_id) based on the number of common ontology terms.

So the output should be,

Pathway_id1, Pathway_id2, No. of common terms

I know, it can be easily done using a server side language, will it be faster to use MySql instead?

  • 1 1 Answer
  • 3 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
    2026-05-15T14:21:36+00:00Added an answer on May 15, 2026 at 2:21 pm

    If I understood you right, that is

    select a.pathway_id, b.pathway_id, count(*)
    from t a
    inner join t b on a.ontology_term = b.ontology_term
    group by a.pathway_id, b.pathway_id
    

    There is no record if two pathways do not have common terms

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

Sidebar

Related Questions

I have created a table in MSSQL having following structure - **Table Name -
I have a column called CODE in a MySQL table which can be NULL.
I have a mysql InnoDB table with the following structure: id artist_id There are
I am using MySQL and I am having the following scenario: Table nodes: node_id,
So im having a problem (obviously). I have the following MySQL table data 7
I have a rather large mysql table that has the following structure: fieldid |
Hi I have a mysql table having the following fields A INT, B INT,
I have the following two mySQL tables with the 'child' table having a DELETE
I'm having the following problem using Django with MySQL 5.5.22. Given a table with
I have a MySQL table with an auto-incremented integer primary key. I want to

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.