INNER JOIN GOT error for mysql_fetch_array()

[jimjack145]

 

Hi ALL,

 

When i run below code then i got and error as below:

mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/datalist_year.php on line 21

 

I am not getting y i got such an error. Any guidance and assist welcome.

 

Regards,

JIMI

-----------------------------------------------------------

<?PHP

 

$loginname = ''";

 

$password = "";

 

$database = "";

 

$con = mysql_connect("localhost", "$loginname", "$password") or die("could not connect");

mysql_select_db("$database");

 

$query = "SELECT candidate.first_name, candidate.last_name, candidate_foreign.value from candidate, candidate_foreign INNER JOIN candidate_foreign ON candidate.candidate_id = candidate_foreign.assoc_id WHERE candidate_foreign.field_name = 'IITYear' AND value >= '" .$_POST['from']. "' AND '".$_POST['to']."'";

 

$result = mysql_query ("$query");

 

echo "You are seaching between year '" .$_POST['from']. "' to '" .$_POST['to']. "'";

 

echo "<TABLE BORDER=1><TH>First Name</TH><TH>Last Name</TH><TH>Cell No</TH><TH>Email</TH>";

 

 

  while ($in = mysql_fetch_array($result))

      {

        echo "<TR>";

        echo "<TD>" . $in[first_name] . "</TD>";

        echo "<TD>" . $in[last_name] ."</TD>";

        echo "<TD>" . $in[value] ."</TD>";

        echo "</TR>";

      }

 

echo "</TABLE>";

 

mysql_close($con);

 

?>

[onlyican]

try this

 

2 Things

$result = mysql_query($query);

You dont need the "

and also try this under that line

 

if(!$result){

echo "Error.<br />\n".mysql_error();

}

[jimjack145]

hi

 

Thks for your reply i have done from my side and that works pretty good.

 

Once again thks for your reply.

 

Jimi