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 1070007
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T20:30:08+00:00 2026-05-16T20:30:08+00:00

I don’t understand MySQL very well, here are the table structures I am using.

  • 0

I don’t understand MySQL very well, here are the table structures I am using.

users

id | first_name | last_name | username
| password

categories

id | user_id | name | description

links

id | user_id | category_id | name |
url | description | date_added |
hit_counter

I am trying to return a result set like this, to give information about the category for a user that includes how many links are in it.

id | user_id | name | description | link_count

At the moment I have this query, but it only returns rows for categories that have links. It should return rows for categories that do not have any links (empty categories).

SELECT categories.*, COUNT(links.id)
FROM categories LEFT JOIN links ON
categories.id=links.category_id;

How to do this query? Thanks.

  • 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
    2026-05-16T20:30:08+00:00Added an answer on May 16, 2026 at 8:30 pm

    we can’t do select table dot “star” with an aggregate.

    what you wanna do is something like (pseudocode):

    select
        categories.field1,
        categories.field2,
        {etc.}
        count(links.id)
    from categories
    left join links
        on categories.id = links.category_id
    group by
        categories.field1,
        categories.field2,
        {etc.}
    

    iow: you’re missing the group by code-block to get the right aggregate in your query result set.

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

Sidebar

Related Questions

I don't edit CSS very often, and almost every time I need to go
I don't understand where the extra bits are coming from in this article about
I don’t think I’ve grokked currying yet. I understand what it does, and how
Don't understand why #include <Header.h> is not compiling while #include Header.h is compiling with
I don't have much experience with RegEx so I am using many chained String.Replace()
I don't know when to add to a dataset a tableadapter or a query
I don't want PHP errors to display /html, but I want them to display
I don't remember whether I was dreaming or not but I seem to recall
I don't expect a straightforward silver bullet answer to this, but what are the
I don't want to take the time to learn Obj-C. I've spent 7+ years

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.