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

  • Home
  • SEARCH
  • 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 3856170
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 19, 20262026-05-19T17:53:19+00:00 2026-05-19T17:53:19+00:00

SELECT v.*, COUNT(a.*) FROM vacancies AS v, applications AS a WHERE a.vacancy_id = v.id

  • 0
SELECT v.*, COUNT(a.*) 
FROM vacancies AS v, applications AS a 
WHERE a.vacancy_id = v.id

Basically I need a way of counting the number of applications that correspond to each vacancy.

Applications have a vacancy id. So I want to get the vacancies from the vacancies table and iterate through them outputting the amount of applications that have the same vacancy_id.

Is this the right way to do it? It says I have a problem with the COUNT(a.*) bit.

Any ideas?

Thank you.

Edit

Ok, so I tried the group by thing, and now have this:

SELECT v.*, COUNT(a.candidate_id)
FROM vacancies AS v, applications AS a 
WHERE a.vacancy_id=v.id 
GROUP BY v.title

But it only returns one result, because theres only one application. Is there any way to return EVERYTHING from the vacancies table, even if there are no applications associated with the vacancy?

  • 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
    2026-05-19T17:53:20+00:00Added an answer on May 19, 2026 at 5:53 pm

    COUNT is a group operation. You need to use GROUP BY at the end.

    EDIT
    Use left join

    SELECT v.*, COUNT(a.candidate_id)
    FROM vacancies AS v
    LEFT JOIN applications AS a on a.vacancy_id=v.id
    GROUP BY v.title

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

Sidebar

Related Questions

SELECT COUNT(*) FROM BigTable_1 Which way I should to use to get number of
When we execute select count(*) from table_name it returns the number of rows. What
Is there a better way of doing this ? SELECT (SELECT count(*) FROM `tbl`
SELECT count(*) FROM table WHERE column ilike '%/%'; gives me the number of values
What's an efficient way to select count(*) from DB in order to count registered
I have come across articles that state that SELECT COUNT(*) FROM TABLE_NAME will be
I'm trying to do a simple Select Count(*) from PRODUCTS where date > xxx
I have the following string expression in a PowerShell script: select count(*) cnt from
this subquery works in SQL Server: select systemUsers.name, (select count(id) from userIncidences where idUser
I have a query like this: SELECT t1.id, (SELECT COUNT(t2.id) FROM t2 WHERE t2.id

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.