PHP - die function - Printable Version
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PHP - die function by absorbation on 07-03-2006 at 04:50 PM
Can someone tell me what is wrong with this code:
code:
mysql_query("update referals2 set amount=(amount+1) where website='$referal'") or die (mysql_query("insert into referals2 values('','$referal','')"));
It is simple really, update the feild, but if there is no feild to update add a new one. However, never using the die function before is it designed to echo text or act as an else statement for mysql queries? Thanks
If you can also give me some right code or a better funtion to use when it comes to mysql queries, that would be great as well.
RE: PHP - die function by Plik on 07-03-2006 at 05:10 PM
die echos the parameter you give it, then ends the current scripts execution.
As ever, if you're in doubt about a php function, look it up on php.net http://www.php.net/die
RE: PHP - die function by absorbation on 07-03-2006 at 05:11 PM
quote:
Originally posted by Plik
die echos the parameter you give it, then ends the current scripts execution.
So it could not be used to sumbit another query? Then what could I use
?
RE: PHP - die function by absorbation on 07-03-2006 at 05:18 PM
quote:
Originally posted by thekid
mysql_query will return false if there's an error.
that's perfect for me then, I'll test it now, thanks
.
RE: PHP - die function by WDZ on 07-04-2006 at 03:42 AM
mysql_query() will not return false just because no rows were updated!
Try mysql_affected_rows()
RE: PHP - die function by andrewdodd13 on 07-04-2006 at 08:50 AM
This is what I'd use (off the top of my head):
code:You'll probably also want to query query1 after query2, if I get the way your thing works
$query1 = "update referals2 set amount=(amount+1) where website='$referal'";
$query2 = "insert into referals2 values('','$referal','')";
mysql_query($query1) or die ("Failed to update the database, this is a connection / database not exist problem.");
if (mysql_affected_rows() == 0) {
mysql_query($query2) or die ("Failed to insert new record, this is a connection / database problem.");
}