You could do:
mysql_query("update url set nrviews=nrviews+1 where url='$url'");
if(mysql_affected_rows()==0)
mysql_query("insert into url (urn, nrviews) values ('$url',1)");
Lars
> -----Ursprüngliche Nachricht-----
> Von: Claudiu Bandac [mailto:[EMAIL PROTECTED]
> Gesendet: Montag, 13. Oktober 2003 05:40
> An: [EMAIL PROTECTED]
> Betreff: [PHP-DB] selecting and updating fields
>
>
> Hi !
>
> I've got a form that passes a variable to a script.
> The script connects to a database and I need to check if a
> field containing that variable already exists in the table,
> and if it does, to select the "nr_of_views" field, increment
> it, and update the table and if it doesn't, to create a new
> field with that variable and set nr_of_views to 1
>
> the table in my database that looks like this :
> url nrviews
> www.google.com 1
> www.yahoo.com 3
>
>
> And I've tried something like this:
>
>
> $query = mysql_query("select * from url") or die (mysql_error());
> while ($row = mysql_fetch_row($query))
> {
> if ($url === $row["url"])
> {
> $nrviews = $row["nrviews"];
> $nrviews++
> mysql_query("update url set
> nrviews='$nrviews' where url='$url'") or
> die (mysql_error());
> }
> else
> {
> $nrviews=1;
> mysql_query("insert into url
> (url, nrviews) values ('$url','$nrviews')") or
> die (mysql_error());
> }
> }
>
>
> AND STILL DOESN'T WORK !!!
> When I get a variable that is not in the database, the whole
> thing goes crazy ! (it inserts lots of fields with that
> variable and nrviews=1
>
>
> Please HELP !
>
> Thank You !
>
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