I can almost guarantee that it's not the second line that is "failing",
the problem here is that $result is not containing naything, and that is
normally due to the fact that you are not connecting to the db, or the
table "tablename" is not there.
I use the following format as my "standard" MySQL connect and query snippet:
$link = @mysql_connect("localhost",$username,$password) or die ('Could
not connect!'); //@ suppresses the default error message generated by
this function and the "or die()" bit kills the script right then and
there should it not be able to connect.
mysql_select_db("YOUR_DB_NAME",$link);
$sql = "select * from your_table_name";
if ( $result = mysql_query($sql)) { // checks to see if $result
contains anything before it even tries to fetch an associative array
from it.
$row = mysql_fetch_assoc($result);
} else {
echo "Empty result set!";
Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9
out of 10 times, you don't need the array element id's that is returned
by mysql_fetch_array.
Matthew Bielecki wrote:
>I have a couple of scripts that fail with the error of:
>Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
>resource in...
>
>I'm new to both SQL and PHP and I'm wondering if I have some setting
>turned off or what.
>
>Here's the piece of code that is failing (the second line fails):
>
>$result = mysql_db_query($dbname, "SELECT * FROM tablename ORDER BY id");
> $row = mysql_fetch_array($result);
>
>
>Thanks for your help in advance!!
>
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