MySQL and strange error messages

[dappleyard]

Hi There All

I'm having a problem with the following code. It actually does it's job perfectly, and works, but still produces the good ol error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site35/fst/var/www/html/add.php on line 14". I'd be grateful if someone could have a quick look to see if they can discover what's wrong :-)

<?

$db = mysql_connect("localhost", "xxxxxx", "xxxxxx") or die ("Couldn't connect.");
mysql_select_db("kateskrafts_co_uk", $db) or die ("Couldn't select db.");

// create SQL statement
$sql = "SELECT IGNORE `id`, `sessid`, `itemid`, `quantity`, `type`
        FROM basket WHERE `sessid`='$sessid' AND `itemid`='$itemid' AND `type`='$type' ORDER BY id";

// execute SQL query and get result
$sql_result = mysql_query($sql,$db);

// format results by row
while ($row = mysql_fetch_array($sql_result)) {
$id2 = $row[id];
$sessid2 = $row[sessid];
$itemid2 = $row[itemid];
$quantity2 = $row[quantity];
$type2 = $row[type];

if ($quantity2=="") {

// create SQL statement
 $sql = "

 insert into basket values ('','$sessid','$itemid','1','$type') ";

// execute SQL query and get result
 $sql_result = mysql_query($sql,$db);



} else {

$quantity3 = $quantity2 + 1 ;

$sql = "UPDATE `basket` SET `quantity`='$quantity3' WHERE `sessid`='$sessid' AND `itemid`='$itemid' AND `type`='$type'";

// execute SQL query and get result
$sql_result = mysql_query($sql,$db);

}




}
?>

[bad76]

Hi,
try this to debug out other info:

// execute SQL query and get result
$sql_result = mysql_query($sql,$db) 
       or die ("Error on query:".mysql_error()."<br><i>".$sql."</I>");

[dappleyard]

OK - I added that, my code now looks like this at the beginning:

<?

$db = mysql_connect("localhost", "xxxxxx", "xxxxxx") or die ("Couldn't connect.");
mysql_select_db("kateskrafts_co_uk", $db) or die ("Couldn't select db.");

// create SQL statement
$sql = "SELECT `id`, `sessid`, `itemid`, `quantity`, `type`
        FROM basket WHERE `sessid`='$sessid' AND `itemid`='$itemid' AND `type`='$type' ORDER BY id";

// execute SQL query and get result
$sql_result = mysql_query($sql,$db) or die ("Error on query:".mysql_error()."<br><i>".$sql."</I>");

// format results by row
while ($row = mysql_fetch_array($sql_result)) {
$id2 = $row[id];
$sessid2 = $row[sessid];
$itemid2 = $row[itemid];
$quantity2 = $row[quantity];
$type2 = $row[type];

However, it does not actually echo anything different, just the error i outlined in my initial post. The script does it's job fine. Is there any way to just stop the error from being printed?

[sdjensen]

Shouldn't the code

$id2 = $row[id];
$sessid2 = $row[sessid];
$itemid2 = $row[itemid];
$quantity2 = $row[quantity];
$type2 = $row[type];

be changed to

$id2 = $row['id'];
$sessid2 = $row['sessid'];
$itemid2 = $row['itemid'];
$quantity2 = $row['quantity'];
$type2 = $row['type'];

[dappleyard]

That may have been a problem, but the error is referring to the sql code before i try to format the output.

[bad76]

😕
Yes, just putting @ before fetch_array...

But I don't understand... :eek:

Just moment !

in the if..else statement in the innerloop you overwrite previous $sql_result. Try to change this seconds with $sql_result2 or some like this...

see you

[dappleyard]

Thanks very much mate

Problem solved

Can't believe it was so simple really!

Cheers Guys