Error condition not triggering correct response

[mmckimson1]

First of all, I apologize for my "newbieness" in advance.  I've got a form processing script that is working fine except for one section of code, snippets of which is shown below:

          //default values
          $formResponse = "How are you really doing?  Use this to find out!";
          $imgSource = "blue.png";

				If ($country == "United States") {
						//US zip code query
						$sql = "SELECT location, revParData FROM zipData WHERE zipCode = '$zipCode' LIMIT 1";
						mysql_select_db('db');
						$retval = mysql_query( $sql, $conn );
						if(! $retval ) {
							$formResponse = "It looks like you entered an invalid zip code. Please try again!";
							$imgSource = "yellow.png";
						} else {
									$sql = "SELECT location, revParData FROM zipData WHERE zipCode = '$zipCode' LIMIT 1";
									mysql_select_db('db');
									$retval = mysql_query( $sql, $conn );

									while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
										$location = $row['location'];
										$revParData = $row['revParData'];
									}
						}

What I thought would happen is this:

 

1. If the zip code entered by the end user does not exist in the table, the $formResponse and $imgSource values would be the "... invalid zip code" and "yellow.png" values.
2. If a valid zip code is entered, it completes the second query.

In fact, it works fine if a zip code that exists in the table is entered, but shows the default values for $formResponse and $imgSource if a zip code that doesn't exist in the database is entered.

 

I can't figure out why it is doing this... if anyone can point out my coding error, it would be appreciated.

 

Cheers!

Mike

 

 

[ginerjm]

Horribly formatted code post.  Let me look at it for a bit.

[Barand]

You are working on the incorrect assumption that returning no results will result in query failure. No results is a perfectly valid result. Check the number of rows returned instead to see if any records were found.

[ginerjm]

 Wow!  This is a pretty incomplete section of code!

 

Here it is cleaned up a bit.

 

//default values
$formResponse = "How are you really doing?  Use this to find out!";
$imgSource = "blue.png";
If ($country == "United States")
{
    //US zip code query
    $sql = "SELECT location, revParData FROM zipData WHERE zipCode = '$zipCode' LIMIT 1";
    mysql_select_db('db');
    $retval = mysql_query( $sql, $conn );
    if(!$retval )
    {
        $formResponse = "It looks like you entered an invalid zip code. Please try again!";
        $imgSource = "yellow.png";
    }
    else
    {
        $sql = "SELECT location, revParData FROM zipData WHERE zipCode = '$zipCode' LIMIT 1";
        mysql_select_db('db');
       $retval = mysql_query( $sql, $conn );
       while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
       {
            $location = $row['location'];
            $revParData = $row['revParData'];
       }
   }
 

 

So incomplete.

 

As Barand said - no results is not the same as "!$results)".

 

Going thru this code I see that  you check to see if you are in the US and you do a query for a single zip code record.  Incorrectly you then output the 'invalid code' message if the query fails.  Skipping that you then re-run the exact same query if the query succeeded and once again output what you think is going to be a result.  Actually you need to not only check if the query ran (which you do) but also if it returned a result/row (which you don't.  Why you run the query twice - who know?

 

Furthermore - if you use the php manual you must have noticed that the MySQL_* functions are deprecated.  Use At Your Own Risk.  But since you are set in your ways here is how I would re-write it:

 

$formResponse = "How are you really doing?  Use this to find out!";
$imgSource = "blue.png";
mysql_select_db('db');
If ($country == "United States")
{
   $sql = "SELECT location, revParData FROM zipData WHERE zipCode = '$zipCode'";
   $retval = mysql_query( $sql, $conn );
   if(!$retval )
   {
       $formResponse = "Error running query - ".mysql_error();
       $imgSource = "yellow.png";
   }
   if ( mysql_num_rows() == 0)
   {
       $formResponse = "It looks like you entered an invalid zip code. Please try again!";
       $imgSource = "yellow.png";
   }
   else
   {
       $row = mysql_fetch_array($retval, MYSQL_ASSOC))
       $location = $row['location'];
       $revParData = $row['revParData'];
   }
...

 

No dupe query, no dupe select db, proper handling of all results.

 

Now stop using MySQL_*