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 752957
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T14:50:30+00:00 2026-05-14T14:50:30+00:00

The following statements give the same result (one is using on , and the

  • 0

The following statements give the same result (one is using on, and the other using where):

mysql> select * from gifts INNER JOIN sentGifts ON gifts.giftID = sentGifts.giftID;
mysql> select * from gifts INNER JOIN sentGifts WHERE gifts.giftID = sentGifts.giftID;

I can only see in a case of a Left Outer Join finding the “unmatched” cases:
(to find out the gifts that were never sent by anybody)

mysql> select name from gifts LEFT OUTER JOIN sentgifts 
           ON gifts.giftID = sentgifts.giftID 
           WHERE sentgifts.giftID IS NULL;

In this case, it is first using on, and then where. Does the on first do the matching, and then where does the “secondary” filtering? Or is there a more general rule of using on versus where? 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
    2026-05-14T14:50:30+00:00Added an answer on May 14, 2026 at 2:50 pm

    WHERE is a part of the SELECT query as a whole, ON is a part of each individual join.

    ON can only refer to the fields of previously used tables.

    When there is no actual match against a record in the left table, LEFT JOIN returns one record from the right table with all fields set to NULLS. WHERE clause then evaluates and filter this.

    In your query, only the records from gifts without match in ‘sentgifts’ are returned.

    Here’s the example

    gifts
    
    1   Teddy bear
    2   Flowers
    
    sentgifts
    
    1   Alice
    1   Bob
    
    ---
    SELECT  *
    FROM    gifts g
    LEFT JOIN
            sentgifts sg
    ON      g.giftID = sg.giftID
    
    ---
    
    1  Teddy bear   1     Alice
    1  Teddy bear   1     Bob
    2  Flowers      NULL  NULL    -- no match in sentgifts
    
    ---
    SELECT  *
    FROM    gifts g
    LEFT JOIN
            sentgifts sg
    ON      g.giftID = sg.giftID
    WHERE   sg.giftID IS NULL
    
    ---
    
    2  Flowers      NULL  NULL    -- no match in sentgifts
    

    As you can see, no actual match can leave a NULL in sentgifts.id, so only the gifts that had not ever been sent are returned.

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

Sidebar

Related Questions

I have tried the following two statements: SELECT col FROM db.tbl WHERE col (LIKE
When using IF statements in Python, you have to do the following to make
I have the following statement SELECT id, descr from mytable which returns 1, 'Test1'
I am doing the following statements in Java, Obj t[] = new Obj[10]; Obj
Frequently I come across the following statements in C/C++ source code: $Id: lzio.c,v 1.24
Given following Ruby statements: (Read input and store each word in array removing spaces
What is the better practice of the following two switch/case statements? Is there an
I use the following statement prepared and bound in ODBC: SELECT (CASE profile WHEN
I tried running the following statement: INSERT INTO VOUCHER (VOUCHER_NUMBER, BOOK_ID, DENOMINATION) SELECT (a.number,
When I am running the following statement: @filtered = map {s/ //g} @outdata; it is

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.