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 8981341
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 15, 20262026-06-15T20:18:44+00:00 2026-06-15T20:18:44+00:00

i use a sql query like this to get some results i need: SELECT

  • 0

i use a sql query like this to get some results i need:

SELECT
    *
FROM
    pictures p
WHERE
    p.id NOT IN 
        (
            SELECT
                picture_id
            FROM
                guesses g
            WHERE 
                g.user_id = XXX 
        )
    AND
        p.user_id != XXX
;

Relation is as follows: A user has many pictures and a picture belongs to one user. A user has many guesses and a guess belongs to one picture. The tricky part is that a user is only allowed one guess for the same picture.

XXX = $user_id

I guess that there is a way to rewrite this sub-select using a left join but i can’t get it working.

Can anyone help?

Anja

  • 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
    2026-06-15T20:18:45+00:00Added an answer on June 15, 2026 at 8:18 pm

    Because it is a NOT IN condition you should use a LEFT OUTER JOIN. This is the direct translation to left outer join of your query:

    SELECT
        distinct p.*
    FROM
        pictures p
    LEFT OUTER JOIN
       guesses g      ON g.picture_id = p.id and g.user_id = XXX
    WHERE
       p.user_id != XXX
       and g.user_id is null
    ;
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I am trying to use some parameters in my sql query like this; Answer_table
I want to use a query like: sql(select name where id > 10 and
I want to use this SQL query to get only the records between 8
The answer to another SO question was to use this SQL query: SELECT o.Id,
My Sql function getLive_projects has this query - $query = SELECT p.* FROM projects
How to use conditional select query in SQL Server ? I have a column
I got this: SET @databasename = (SELECT DATABASE()); USE @databasename; ...SQL Querys... But this
I query-ing this from entity framework using WCF, and i want to use the
I get this syntax on the internet and modified some parts. I would like
I want to make one question about oracle/sql query. I have some data like

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.