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 4067328
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 20, 20262026-05-20T16:14:08+00:00 2026-05-20T16:14:08+00:00

I’m new to SQLite and I’m trying to create a view of joined tables,

  • 0

I’m new to SQLite and I’m trying to create a view of joined tables, where table_results references table_names. The user can add, edit and remove items from table_names, but they can not change the references within table_results. What I’m trying to accomplish is that if an entry in table_names is removed, during a JOIN table_names ON (table_results.name_id=table_names._id) will return all rows in table_results, but where a name entry is removed will display “NO NAME”

Example:

table_names:
_id name
1   John
2   Bill
3   Sally
4   Nancy

table_results:
_id  name_id  score_1  score_2
1    1        50       75
2    4        80       60
3    2        83       88
4    3        75       75
5    2        93       95

where:
Select table_results._id table_names.name, table_results.score_1, table_results.score_2
FROM table_results
JOIN table_names ON (table_reults.name_id = table_names._id);

produces:
1 John   50   75
2 Nancy  80   60
3 Bill   83   88
4 Sally  75   75
5 Bill   93   95

Now if the user was to remove Bill from table_names, the same query string would produce:

1 John   50   75
2 Nancy  80   60
4 Sally  75   75

My Question:
Is there a way to have the query that substitutes values when the join doesn’t find a match? I’d like the above example to produce the following output, but I’m not sure how to write the query string for SQL.

1 John     50   75
2 Nancy    80   60
3 NO NAME  83   88
4 Sally    75   75
5 NO NAME  93   95

Thanks for your help.

  • 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
    2026-05-20T16:14:09+00:00Added an answer on May 20, 2026 at 4:14 pm

    Sure, you are doing an inner join, just change it up to a left outer join..

    SELECT table_results._id
           ,COALESCE(table_names.name, 'NO NAME') AS [name]
           ,table_results.score_1
           ,table_results.score_2
    FROM table_results
    LEFT OUTER JOIN table_names
        ON (table_reults.name_id = table_names._id);
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm new to using the Perl treebuilder module for HTML parsing and can't figure
I'm trying to create an if statement in PHP that prevents a single post
I have a jquery bug and I've been looking for hours now, I can't
link Im having trouble converting the html entites into html characters, (&# 8217;) i
I am trying to understand how to use SyndicationItem to display feed which is
I have a string like this: La Torre Eiffel paragonata all’Everest What PHP function
I want use html5's new tag to play a wav file (currently only supported
I am trying to render a haml file in a javascript response like so:
I'm parsing an RSS feed that has an ’ in it. SimpleXML turns this
I need to clean up various Word 'smart' characters in user input, including but

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.