Sorry, last message was a little vague.
Your SQL will fail because of the ',' after the SET clause.
Therefore affected_rows shows 0, and inserts a new row.
Then when you match with the LIKE clause, it's going to match
multiple times.
Miles, you were SO close with #2. =)
-Szii
----- Original Message -----
From: "Miles Thompson" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, May 16, 2001 3:50 PM
Subject: Re: [PHP-DB] mySQL Addition Problem
>
> Dammit, that code looks right.
>
> So ....
>
> 1. Have you tried it at the mysql console? What results?
> 2. How accurately does day match $day? Have you tried "=" instead of LIKE?
> Maybe it's doing it twice?
> 3. Is day distinct? How do you know you have only one?
>
> When you tested with 2, assuming i had a value of 1, was the result 5 or
6?
>
> Strange - Miles Thompson
>
> At 05:43 PM 5/16/01 -0400, SubvertTRL wrote:
> >I am making a simple script that counts votes per day for a site and uses
a
> >simple database, All i want to do is increment the field by one. the
first
> >time i run the script it works, after that it doubles it. so instead of
1,
> >it adds 2. i know its doubling it because i tried adding 2, 3, and 4 and
it
> >doubled all those... the field is a BIGINT UNSIGNED, i've tried it with
> >just INT, nothing seems to work, i used to have PHP get the field value,
> >increment it, then update the row in the database with the new number, it
> >doubleed it that way. now i use straight mySQL code to do it all in one
> >shot, still doubles it.. can any one help me???
> >
> >heres my code... less database info :)
> >
> >$results = mysql_query("UPDATE trlvotes SET i=i+1, WHERE day LIKE
'$day'");
> >if (mysql_affected_rows($results) == 0){
> > mysql_query("INSERT INTO trlvotes (id, day, i) VALUES (null, '$day',
1)");
> >}
> >mysql_close();
> >
> >
> >you can email me at [EMAIL PROTECTED]
> >
> >thanks!
> >
> >
> >
> >--
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