You might want to try this:
mysql_query("INSERT INTO links (count) VALUES ("$count") WHERE
lid=$id");
Kurth Bemis wrote:
>
> At 12:31 PM 1/26/2001, Krznaric Michael wrote:
>
> sorry - here you all go..
>
> // number crunching time
> $count++;
> $time = date("Y-m-d H:i:s");
>
> $result = mysql_query("INSERT INTO links (count) VALUES $count WHERE lid=$id");
>
> echo $result;
>
> > You need to be a little more specific about DB and problem type.
> >You may have to commit the transaction if commit is not implict (ex Oracle).
> >There could be many resons including your SQL statement.
> >
> >Mike
> >
> >-----Original Message-----
> >From: Kurth Bemis [mailto:[EMAIL PROTECTED]]
> >Sent: Friday, January 26, 2001 12:23 AM
> >To: [EMAIL PROTECTED]
> >Subject: [PHP] arg....
> >
> >
> >i'm having a horrible time updating 2 fields in the same db. I don't get
> >an error but the fields aren't updated....can anyone send me a snippet for
> >the to learn from?
> >
> >~kurth
> >
> >
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