MySQL Rank, Update, Select Order By Help

Posted on 2011-02-18
Medium Priority
Last Modified: 2012-06-21
I would like to update a database ranking in 1 swoop instead of ranking and then updating.

Here is my rank implementation:
SET @rank=0;
SELECT @rank:=@rank+1 AS rank, id, fd1, fd2, fd3 FROM statsmsmobsters ORDER BY fd2 DESC, fd3 DESC;

I tried to do this and I get 0 results:
SET @rank=0;
UPDATE dbTable SET rank_fd= @rank:=@rank+1 ORDER BY fd2 DESC, fd3 DESC;

Thanks for any help!
Question by:PiZzL3
LVL 21

Accepted Solution

K V earned 2000 total points
ID: 34931724
I'm not sure what do you mean by " I get 0 results: " as the query should work proper and will update all the data under dbTable.
LVL 41

Expert Comment

ID: 34932566
try this.
update statsmsmobsters 
join (select id,fd1,fd2,fd3,@curRank := @curRank + 1 AS rank 
        from statsmsmobsters 
        JOIN      (SELECT @curRank := 0) r 
     order by fd2 desc,fd3 desc) ranks 
on (ranks.id = statsmsmobsters.id) 
set statsmsmobsters.rank_fd = ranks.rank ;

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LVL 111

Expert Comment

by:Ray Paseur
ID: 34933192
Regarding this, "I get 0 results" -- why do you think that you get zero results?  How does your code test for the results?  Please post a complete code segment that shows the PHP you used, thanks.

Author Comment

ID: 34933281
I get this back from phpMyAdmin when I run the "update" query (the select works perfectly fine).

"Your SQL query has been executed successfully
# MySQL returned an empty result set (i.e. zero rows).".

You did just make me think of something though... I ran the following query:
"UPDATE dbTable SET rankfd= 0;"

then ran:
"SET @rank=0;
UPDATE dbTable SET rankfd= @rank:=@rank+1 ORDER BY fd2 DESC, fd3 DESC;"

and it worked as you said it would. I suppose it returned 0 results because none of the ranks were changing.

I'm currently having difficultly following that example.. I don't have a clue how it works. However my original update statement does work, so I don't need to use it anymore, but ty.

I wasn't using a script to test this, I was just in phpMyAdmin. I find it easier to test statements before I use them in my code that way. I got 0 results due to the reason stated above.
LVL 111

Expert Comment

by:Ray Paseur
ID: 34933625
Yep - the mysql_affected_rows() function does not care how many rows are matched, only how many were actually updated.  If you tell MySQL to SET myvalue="X" WHERE key < 10000, the resulting count will only represent the number of rows where myvalue was not already "X" and had to be changed.

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