mysql error:

[illushinz]

is there a general description or method of troubleshooting for these errors?? :

Warning: Supplied argument is not a valid MySQL result resource in /httpdocs/results.php on line 44

Here is a snippet of the code that is causing this error.
// if the first name has value
if ($HTTP_POST_VARS['first']){
$query = $query." first LIKE ='$form_first%'";
}

// if the last name has value
if ($HTTP_POST_VARS['last']){
$query = $query. " and last LIKE = '$form_last%'";
}

// if city has a value
if ($HTTP_POST_VARS['city']){
$query = $query. " and city LIKE = '$form_city%'";
}

// if state has a value
if ($HTTP_POST_VARS['state']){
$query = $query. " and state LIKE = '$form_state%'";
}

$result = mysql_query($query);
$row = mysql_fetch_array($result); // THIS IS LINE #44
echo $row[First];
?>

Thanks

[sarahk]

here's some debugging advice

// if state has a value 
if ($HTTP_POST_VARS['state'])
{ 
  //note the alternative syntax here
  $query .= " and state LIKE = '$form_state%'"; 
} 

echo "$query<br>";
exit;
$result = mysql_query($query); 
$row = mysql_fetch_array($result); // THIS IS LINE #44 
echo $row[First]; 

now, the query won't run because we exit, but that's ok because we really want to have a look at what's held in $query. There's probably some obvious mistake. Now, if there isn't you need to copy the query from the browser into your mysql front end (phpMyAdmin?) and see if it runs and what the error MySql throws is, it should make sense.

All you error is telling you really is that the query didn't work.

gl

[qbennett]

The other thing to ALWAYS do is to check the $result of running the query before 'use'ing it. If it is whatever mysql_query returns in the event of an error, then use the mysql_error_string type routines to establish what happened.

Happy MySQL'ing

Quentin