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PHP MySQL Not Equal Comparison between Two Tables

Posted on 2007-11-16
7
2,916 Views
Last Modified: 2010-04-21
I have two tables.  MySQL version 4.1.22-standard -  PHP 5.2.4.  

I'm trying to get the data from the Review Table that do not have entries in the Comment Table.  

Here's what I have..

First Table   Review - Fields ->  (item_id / date / user_id)  

Second Table Comments - Fields -> (comm_id, item_id, user_id,date)

Code Below.. Thanks..
SELECT Review.item_id, Review.date, Review.user_id, Comments.item_id
FROM Review, Comments
WHERE Review.item_id <> Comments.item_id

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0
Comment
Question by:usiff
  • 4
  • 2
7 Comments
 
LVL 48

Accepted Solution

by:
hernst42 earned 500 total points
ID: 20296803
You can try the SQL below
SELECT Review.item_id, Review.date, Review.user_id
FROM Review LEFT OUTER JOIN Comments on (Review.item_id = Comments.item_id)
WHERE Comments.item_id IS NULL

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0
 
LVL 7

Expert Comment

by:dansoto
ID: 20297975

SELECT Review.item_id, Review.date, Review.user_id, Comments.item_id
FROM Review
WHERE Review.item_id NOT IN 
(SELECT Comments.item_id from Comments); 

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0
 

Author Comment

by:usiff
ID: 20299640
Hello hernst42:

Here is another quick issue..

When I add this code below is doesn't work...
WHERE Comments.item_id IS NULL AND Review.user_id = $_SESSION['user_login_id']

But this will work?
WHERE Comments.item_id IS NULL AND Review.user_id = '444'

What the heck I'm I missing?

Thanks

usiff
0
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LVL 48

Expert Comment

by:hernst42
ID: 20299795
maybe $_SESSION['user_login_id'] is empty and does not contain anything. What do ypu get when you print out var_dump($_SESSION['user_login_id']) before you create the SQL?
0
 

Author Comment

by:usiff
ID: 20300288
Hello hernst42,

It's printing our string(1) "444"

It's crazy..
0
 

Author Comment

by:usiff
ID: 20300316
It's working now..  Wow...  I was going crazy!!  

I love typosss

Thanks again!

0
 

Author Closing Comment

by:usiff
ID: 31409512
Thanks!
0

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