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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Re: [PHP-DB] mySQL Addition Problem
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!
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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PHP Database Mailing List (http://www.php.net/)
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To contact the list administrators, e-mail: [EMAIL PROTECTED]
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Re: [PHP-DB] mySQL Addition Problem
WOW - sharp eyes!!
Miles
At 04:02 PM 5/16/01 -0700, Szii wrote:
Remove the comma after the SET clause?
-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!
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
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