On Wed,  3 Oct 2001 17:35, Daniel AlsÚn wrote:
> Hi,
>
> i need help with a, seemingly, simple problem...
>
> I have a form that is supposed to update a MySql table with only two
> fields (id and show_id). I use show_id to determine what is showing on
> my first page and never store more than one value in the table.
>
> Now - i cant seem to get the REPLACE to work. First it only added more
> rows in the table with a new value each time. I solved that by
> inserting I inserted a hidden value into the form to make sure that the
> correct (and only) value will be replaced:
>
> <input type="hidden" name="fid" value="1">
>
> But, that┤s when it stopped working.
> Show_id is determined by radio buttons in the form and that part is
> working. In the actual query i REPLACE the value of show_id:
>
>     $query = "REPLACE INTO puff";
>     $query .= "(id, show_id)";
>     $query .= " values('$fid''$show_id')";
>
> I am sure it┤s a simple solution. But please point it out for me.

A quick look at the REPLACE syntax tells me that 

"REPLACE works exactly like INSERT, except that if an old record in the 
table has the same value as a new record on a unique index, the old 
record is deleted before the new record is inserted."

So if id isn't a unique index, you could expect the values to be added.

-- 
David Robley      Techno-JoaT, Web Maintainer, Mail List Admin, etc
CENTRE FOR INJURY STUDIES      Flinders University, SOUTH AUSTRALIA  

   Multitasking = 3 PCs and a chair with wheels!

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