add an... or die(mysql_error()) to the $resultm = mysql_query($sqlm);
statement and you'll have a lot more info for yourself, and the list if
you still need assistance.
-Original Message-
From: Klaus Haberkorn [mailto:[EMAIL PROTECTED]]
Sent: Saturday, April 21, 2001 6:18 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Supplied argument is not a valid MySQL result
resource...
Looks like many reasons cause this error-message,
but I cannot find
Looks like many reasons cause this error-message,
but I cannot find a problem in the Query:
ie.:
$db_handle = mysql_connect("localhost", "mysql", "mysql");
if($db_handle)
{
$result = mysql_db_query("mysql", "select * from user", $db_handle);
if($result = TRUE)
{
$row =
Hi,
you usually get the error if your query is wrong and
you try to call mysql_fetch_array in a while loop.
It's easier if you post the relevant lines of code but
I'd say there is something wrong with your query.
Johannes
""Klaus Haberkorn"" [EMAIL PROTECTED] schrieb im Newsbeitrag
thank You, Johannes,
but I think we can skip this,
we reduced the query statement to the most simple one:
"SELECT * FROM {tablename}"
if the tablename is wrong, then another error message is issued.
the result is not treated by a while-loop but just by fetch_num_rows,
which at least reports the
Hi,
i don't get that.
Examlpe: tablename: test , entry: id 001; name: John Doe
$result = mysql_query("SELECT * FROM test");
$row = mysql_fetch_whatever($result); // everything ok
$result = mysql_query("SELECT * FROM tet");
$row = mysql_fetch_wahtever($result); // Warning: Supplied arggument
