> mysql_insert_id() is a PHP function, not a MYSQL function. You are also
> missing a close quote on your insert string. So you have to do it like
> this:
Correct, but LAST_INSERT_ID() is a MySQL function.... which is what he's
using...
> $sql2 = "insert into acl (adminId,transportId,securityId) values
('$userid',".last_insert_id().",'1')";
So, no, you're trying to use a MySQL function here in PHP code. The original
syntax was correct.
> $lastid = mysql_insert_id();
> $sql2 = "insert into acl (adminId,transportId,securityId) values
('$userid',".$lastid.",'1')";
>
> which is the same as:
>
> $lastid = mysql_insert_id();
> $sql2 = "insert into acl (adminId,transportId,securityId) values
('$userid',$lastid,'1')";
Both of those are correct, too, if you want to use the PHP function.
> > $sql1 = "insert into transport (domain,transport) values
> > ('$domain','$transport')";
> > $sql2 = "insert into acl (adminId,transportId,securityId) values
> > ('$userid',last_insert_id(),'1'";
> >
> > mysql_query($sql1);
> > mysql_query($sql2);
You're missing a closing paranthesis on the end of the second SQL statement.
That is probably what's causing your error. In the future, use mysql_error()
to find out why it's failing...
mysql_query($query) or die("Error in $query: " . mysql_error());
---John Holmes...
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