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How do I stop duplicates in Mysql results?

Posted on 2010-11-23
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Last Modified: 2013-12-12
Experts,
How do I prevent dupicate results in my select statement?  I am attempting to assign a "category id" and a "parent id" to the category list.  There are hundreds of products with the same category and parent id.  I need to limit to 1 result for each.

Thanks!

Randy
mysql_query("TRUNCATE TABLE `OCcategory`");


$result5 = mysql_query("SELECT * FROM `temp3` ORDER BY `cat_id` LIMIT 1");

while($row5 = mysql_fetch_assoc($result5))
{
if (trim($row5['category'])=="") { continue;}
mysql_query("INSERT INTO OCcategory (category_id, parent_id) VALUES ('".$row5['cat_id']."','$zero')");

}

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Comment
Question by:rlb1
7 Comments
 
LVL 143

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 800 total points
ID: 34198698
I have written an article for this kind of problems ...
http://www.experts-exchange.com/A_3203.html

however, with mysql, the solutions are a bit trickier...
you might try this:
$result5 = mysql_query("SELECT * FROM `temp3` GROUP BY cat_id, parent_id ORDER BY `cat_id` ");

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Author Comment

by:rlb1
ID: 34198971
OK, I have tried GROUP BY, DISTINCT, INSERT IGNORE, etc.  and I am still getting duplicates.  Any other thought as how to solve?

Thanks!
0
 

Author Comment

by:rlb1
ID: 34198995
Using
$result5 = mysql_query("SELECT * FROM `temp3` GROUP BY cat_id, parent_id ORDER BY `cat_id` ");
I got the following error.

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
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LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 34199152
ok.
can you please show a data sample, and show the requested output from there?
0
 
LVL 2

Assisted Solution

by:benschwartz
benschwartz earned 800 total points
ID: 34199998
your query above (SELECT * FROM `temp3` GROUP BY cat_id, parent_id ORDER BY `cat_id`) looks right. Are you sure that 'temp3' exists and has those fields?
0
 
LVL 3

Assisted Solution

by:roynaufal
roynaufal earned 400 total points
ID: 34203350
I usually use select distinct....from
would be helpful if u can show us some of the data though
0
 

Author Comment

by:rlb1
ID: 34206885
Thanks for your help!  I got it to work!
$result20 = mysql_query("SELECT * FROM `ctgstorefeed` GROUP BY picturesku ORDER BY `picturesku`") or die( mysql_error() );

while($row20 = mysql_fetch_assoc($result20))
{

mysql_query("INSERT IGNORE INTO images (image) VALUES ('".$row20['picturesku']."')");

}

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