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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in

I receive the following error when trying to run my script,

"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in html/geturl.php on line 20"


It seems to be this line "while($row = mysql_fetch_array( $result )) {"
<?php
 
$URL = $_GET["URL"];
 
$con = mysql_connect("****","****","****");
 
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("urlshortener1986", $con);
 
 
$result = mysql_query("SELECT * FROM `urls` WHERE `ShortURL` = ".$URL." LIMIT 1");
 
 
 
 
while($row = mysql_fetch_array( $result )) {
 
echo '<meta http-equiv="refresh" content="5;url='.$row['URL'].'">';
 
 
} 
 
 
 
 
mysql_close($con);
 
 
 
 
?>

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JPERKS1985
Asked:
JPERKS1985
  • 2
1 Solution
 
kszurekCommented:

$result = mysql_query("SELECT * FROM `urls` WHERE `ShortURL` = \"".$URL."\" LIMIT 1");

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BarthaxCommented:
In all likelihood your mysql_query has returned an error & not a MySQL resource.  Check what the content of the $result is prior to attempting to use it - the $result will contain false on an error & you can retrieve the error details through mysql_errno() and mysql_error().
0
 
BarthaxCommented:
Or spot the obvious, like kszurek. ;)
0
 
Bernard S.CTOCommented:
Line 16: check if you have results. if there is no anwers, then mysql_fetch_arry cannt work and issues the diagnostice you have got

if (!$result) {
    die('Invalid query: ' . mysql_error()."<br>No answer when looking at [$URL]" );
}
0
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