php search

[pinehead18]

MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");
@mysql_select_db( "$dbName") or die( "Unable to select database");
?>
<?
//error message (not found message)
$XX = "No Record Found";
$query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 ");
while ($row = mysql_fetch_array($query))
   {
        $name = $row["name"];
        $age = $row["age"];
        $location = $row["location"];
print ("this is for $name, and this print the variable2 end so on...");
   }

//below this is the function for no record!!
if (!$name)
{
print ("$XX");
}
//end

}


?>



<form method="post"  action="<?PHP $PHP_SELF ?>">
<select name="metode" size="1">
<option value="row_name1">age</option>
<option value="row_name2">location</option>
</select>
<input type="text" name="search" size="25">
<input type="submit" name=submit value="Begin Searching!!">
</form>

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource on line 16

that is my error.

Any ideas?
Thanks
Anthony

[stolzyboy]

change this

$query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 ");

to this

$query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 ")
               or die(mysql_error());

that will tell you what is wrong with the query, and that is why you are getting an error, because your query isn't returning results

also try splitting the query with the result so you can echo the query alone to ensure you are getting the $variables into the query correctly, i write my queries like

$query = "SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30;";
//here you can do echo $query; to see the actual query you are running
$result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result)) //blah...