[SOLVED] Error:

[ryeman98]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/4617/domains/sunnyneo.com/html/admin/customisation/index.php on line 355

 

Line 355:

while($row4 = mysql_fetch_array($editclothing)) {

 

 

Help?

[.Stealth]

Well whats in $editclothing?

Gonna need more info than that!

[ryeman98]

True. Sorry.

 

$editclothing = mysql_query("SELECT * FROM CustomisationClothing WHERE name='$name'");

 

[Pastulio]

all I can think of is a syntax error

 

$editclothing = mysql_query("SELECT * FROM CustomisationClothing WHERE name='$name'");

 

should be:

 

$editclothing = mysql_query("SELECT * FROM `CustomisationClothing` WHERE `name`='$name'");

[ryeman98]

all I can think of is a syntax error

 

$editclothing = mysql_query("SELECT * FROM CustomisationClothing WHERE name='$name'");

 

should be:

 

$editclothing = mysql_query("SELECT * FROM `CustomisationClothing` WHERE `name`='$name'");

 

Nope, didn't work.

[ryeman98]

When I take out the mysql_fetch_array() it works fine...

[Pastulio]

Then I don't know but what I do know is, that you shouldn't be using a while loop because you'll have only one result anyway (well that's what your code looks like), so you could just make it like this

 

$query = "SELECT * FROM `CustomisationClothing` WHERE `name`='$name'";
$result = mysql_query($query);

$row = mysql_fetch_array ($result);

 

I don't know why but when I code it like that it always works for me, unless the table you are looking for does not exist.

[ryeman98]

It's showing the page now but still has that error... :-\

[Pastulio]

Maybe you are not connected to the database, or the right one or something? Because it should always find a valid result if the table exists, even if there are no rows in it.

 

So do a routine check for any typo's or other bugs. Because, when PHP shows an error on a line, your problem could also lie in a different portion of your code