[SOLVED] Error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL...

[ryeman98]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/4617/domains/somesite.com/html/admin/customisation/index.php on line 515

 

Lines 512-531:

// BACKGROUND=EDIT
$id = $_POST['id'];
$getback = ("SELECT id FROM CustomisationBackgrounds WHERE id='$id'");
$row5 = mysql_fetch_array ($getback);
?>

<form name="editBackground" action="editBackground.php" method="post">
<input type="hidden" name="id" value="<?php echo $row5['id']; ?>" />
<table width="100%" border="0" cellpadding="0" cellspacing="5">
<tr>
	<td><b>Name:</b></td> <td><input type="text" name="name" value="<?php echo $row5['name']; ?>" /></td>
</tr>
<tr>
	<td><b>IMG URL:</b></td> <td><input type="text" name="img_url" value="<?php echo $row5['img_url']; ?>" size="60" /></td>
</tr>
<tr>
	<td> </td> <td><input type="submit" name="submit" value="Edit Background" /></td>
</tr>
</table>
</form>

[teng84]

"SELECT id FROM CustomisationBackgrounds WHERE id='".$id."'"

[pocobueno1388]

I'm assuming this is line 515:

$getback = ("SELECT id FROM CustomisationBackgrounds WHERE id='$id'");

 

If it is, change it to this:

<?php
  $getback = ("SELECT id FROM CustomisationBackgrounds WHERE id='$id'")or die(mysql_error());
?>

 

That will tell you what the problem is.

[ryeman98]

I'm assuming this is line 515:

$getback = ("SELECT id FROM CustomisationBackgrounds WHERE id='$id'");

 

If it is, change it to this:

<?php
  $getback = ("SELECT id FROM CustomisationBackgrounds WHERE id='$id'")or die(mysql_error());
?>

 

That will tell you what the problem is.

 

I still get the same error :-\

 

I checked the table name and it's right so I just don't get it...

[teng84]

"SELECT id FROM CustomisationBackgrounds WHERE id='".$id."'"

[pocobueno1388]

Just realized something, You don't ever use mysql_query()

 

 

<?php

$getback = mysql_query("SELECT id FROM CustomisationBackgrounds WHERE id='$id'")or die(mysql_error());

?>

 

 

[ryeman98]

"SELECT id FROM CustomisationBackgrounds WHERE id='".$id."'"

 

Maybe you should learn more.

That doesn't matter.

[ryeman98]

Just realized something, You don't ever use mysql_query()

 

 

<?php

$getback = mysql_query("SELECT id FROM CustomisationBackgrounds WHERE id='$id'")or die(mysql_error());

?>

 

 

 

Thanks! After a full day of coding, that sort of thing can happen

[pocobueno1388]

Haha, yeah...I hear you

[teng84]

echo this and give me the result then tell me f i have to learn more

"SELECT id FROM CustomisationBackgrounds WHERE id='".$id."'"

 

 

 

[pocobueno1388]

echo this and give me the result then tell me f i have to learn more

"SELECT id FROM CustomisationBackgrounds WHERE id='".$id."'"

 

Conocation doesn't matter when the variable is in between double quotes. Your way and ryeman98's way would both work. ryeman98 was just stating that it doesn't matter, so it was pointless to suggest it =]