Topic awaiting preservation: Beating a dead horse: ZIP code radius search (Page 1 of 1)

I stumbled into these forum via a Google search for a project I'm working on. The idea is that someone can type in their zip code and get a list of restaurants within X number of miles. There's a thread on it http://www.ozoneasylum.com/21687 here and a GREAT FAQ here http://www.ozoneasylum.com/21781 for reference.

The thread ends with Pugzly saying there was a problem with the posted code but doesn't explain what it is. I'm a PHP/MySQL newbie and the syntax looks okay to me.

Can anyone explain what the problem is with the code provided in the FAQ?

LansingFood.com

I think the issue was the while loop if there are no results.

Hmm... perhaps Pugz could post the code he ended up using.



.:[ Never resist a perfect moment ]:.

Hi there, I'm the guilty one to the code in the FAQ...

What Pugzly ran into was not getting a valid result from the query ( $result = mysql_query($sql,$connection); ) as he entered the while loop after it:

function doQueryResult($sql){



Global $connection;



    hookUpDb();



    if($sql != ""){



    $result = mysql_query($sql,$connection);



        $item = array();



        while($item = mysql_fetch_array($result)){



        $items[] = $item;



        }



    return $items;



    mysql_free_result($result);



    } else {



    print("Need correctly formed SQL that returns a resultset.");



    }



}

That can be corrected by adding an if($result){ do the loop } around the loop.
It shouldn't be a problem though if you have a valid sql and connection.
Other than that it should work ok.

You could add some checks to see if you are getting an answer if you are testing for a zip that doesn't exist. The example doesn't do that.
/Dan

{cell 260} {Blog}
-{ ?Computer games don?t affect kids; I mean if Pac-Man affected us as kids, we?d all be running around in darkened rooms, munching magic pills and listening to repetitive electronic music.? (Kristian Wilson, Nintendo, Inc, 1989.) }-

(Edited by DmS on 11-25-2004 23:00)

DmS: For clarity reasons, you should edit your post, 'coz UBB tags are not allowed inside the CODE UBB tag and you put a B UBB tag around the line:

$result = mysql_query($sql,$connection);

Sorry guys but it doesn't work. The line in question was in the faq just the way it's presented here.

$result = mysql_query($sql,$connection);

Here's the error I'm getting:

quote:

---

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/lfc/public_html/calu.php on line 87

---

Line 87 reads:

while($item = mysql_fetch_array($result)){

That makes me think it's in the query but I'm frustated with ignorance.

Yummy Places to Eat: LansingFood.com

Just take this line

while($item = mysql_fetch_array($result)){

      $items[] = $item;

}

and wrap it like so:

if ($result) {

     while($item = mysql_fetch_array($result)){

           $items[] = $item;

     }

}

The issue is that if there is no result set then mysql_fetch_array will fail.

(hmm... php has try/catch now right? Gonna have to use that more)



.:[ Never resist a perfect moment ]:.

yeah, php 5 has that. Don't ask me how slow it is, though.
(Exception handling in most languages is rather slow compared to the rest of said languages).

Ok, I've edited the post as poi suggested.

I think the error most likely comes from the connection being bad or nonexistent,
try to add this as well together with the part bitdamaged suggested:

$result = mysql_query($sql,$connection) or die(mysql_error());

That should give you an error message that might give you a hint on where the problem is.
(you have changed the user/password/db parts in the hookUpDb() right?)

/Dan

{cell 260} {Blog}
-{ ?Computer games don?t affect kids; I mean if Pac-Man affected us as kids, we?d all be running around in darkened rooms, munching magic pills and listening to repetitive electronic music.? (Kristian Wilson, Nintendo, Inc, 1989.) }-