MySQL PHP4 problem

Anon

Hi,

I am migrating from PHP3 to PHP4, and I
am receiving the following error message:

Warning: Supplied argument is not a valid MySQL-Link resource in test.php3 on line 7

Here's the code:

$mysql_link = mysql_connect("localhost", --, --);
mysql_select_db("test");
$query = "INSERT INTO testtable (createdate) values ('testdata')";
$mysql_result = mysql_query($query, $mysql_link);
$cid=mysql_insert_id($mysql_result);

The amazing thing to me is that when I
look into TESTTABLE, I find that Insert
was successful -- a new row with
'testdata' was inserted.

How can I fix this?

Anon

Yeah,

this is funky but try doing mysql_insert_id(); with not result index - or add an @ sign before the mysql_insert_id()

Ie.

$id = mysql_insert_id(); // or
$id = @mysql_insert_it($result);

NOTE: the @ sign suppresses error messages produced by the function after it - I use it religiously on include() mysql_affected_rows() and mysql_num_rows()

--Joe
http://www.miester.org

Anon

Joe, thanks much, I got it to work with;

$id = mysql_insert_id();

Now, is this considereda a 'bug', and should
I report it or just le it slide?

Lee

Anon

From the manual:

int mysql_insert_id ([int link_identifier])

You are using a result pointer as a link_identifier.
That's not good.