I am betting the problem is simply your semicolon... never use a semicolon in a php
mysql query. It
doesn't need one. In general, though, you should write your query such that it will
tell you exactly
what went wrong... Re-write this like so:
$sql = 'select * from aannh_towns';
$result = mysql_query($sql) or exit("Query failed. Error: ".mysql_error()."<BR>".$sql);
This will spit out the mysql error as well as the query as mysql tried to execute it.
You could even
make this a function if you like... then all you'd have to do is pass the query into
it and it'll
spit out the result resource. Something like:
function sql_query($sql) {
$result = mysql_query($sql) or exit("Query failed. Error:
".mysql_error()."<BR>".$sql);
return $result;
}
This will save you a lot of time searching for what broke! :-)
Cheers,
# Nathan
----- Original Message -----
From: "Dan McCullough" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Sunday, April 28, 2002 4:14 PM
Subject: [PHP] Supplied argument is not a valid MySQL result resource
What does that error mean?
$result = mysql_query('select * from aannh_towns;');
echo 'info stored';
while ($query_data = mysql_fetch_array($result)) {
echo "", $query_data['town'], "", $query_data['town_id'], ""; }
?>
=====
dan mccullough
--------------------------------------------------------
"Theres no such thing as a problem unless the servers are on fire!"
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