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"PHP Error (Array)"

Have a look at this PHP. It takes in data from 2 forms, and later on uses a switch statement to allow or deny access to a page.

$query = "SELECT memberPass FROM member WHERE memberName = '$userentry'";
$result = mysql_query($query)
or die("SYerror 03 : Could Not Execute Query");
$row = mysql_fetch_array($result,MYSQL_ASSOC);
extract($row);
switch ($memberPass)

The problem is, when a log in fails, a warning appears :

Warning: extract(): First argument should be an array in...

Need to get rid of it, somehow. Any ideas?

phuzzy wrote:
> What is the is_array thing? What does it do exactly?

Its just a built in php function that checks whether the variable you pass to it is an array or not. Check http://www.php.net/is_array if you want to know more.

Thanks matey :D Works a dream.

What is the is_array thing? What does it do exactly?

Your problem is when the login fails, then your mysql result contains 0 rows, so when you call mysql_fetch_array it returns false to $row rather than an array which then gets passed to extract which expects an array as its first argument, hence the error.
Several solutions I guess, something like:

if (is_array($row)) {
extract($row);
} else {
// if row isn't an array then init the vars to something so they make
// some sense to later code.
$memberPass="";
}

That should fix your error. Its not the best solution, but it works hopefully.

Have a look at this PHP. It takes in data from 2 forms, and later on uses a switch statement to allow or deny access to a page.

$query = "SELECT memberPass FROM member WHERE memberName = '$userentry'";
$result = mysql_query($query)
or die("SYerror 03 : Could Not Execute Query");
$row = mysql_fetch_array($result,MYSQL_ASSOC);
extract($row);
switch ($memberPass)

The problem is, when a log in fails, a warning appears :

Warning: extract(): First argument should be an array in...

Need to get rid of it, somehow. Any ideas?