Error but it works

Mitch · June 3rd, 2006, 11:28 AM

I have a page that is generating the following errror and I can't figure out why, especially when I use the same code on another page (but different database) and it does not generate the error, but when I change the database reference i get it:

Code:

Warning: mysql_list_fields(): Unable to save MySQL query result in /usr/www/users/xyz/abc/123/list.php on line 9

But it does work, the only thing it does is populate a dropdown list and that works.

Here is the code:

Code:

<?
 require_once("bookmark_fns.php");
 do_html_header("");



$connection = mysql_pconnect("x", "y", "z"); 

$fields = mysql_list_fields("A", "B", $connection);

?> 


<form method="post" action="show_list.php">
<p align="center">
<select name="searchteam">

<?php
$query = mysql_query('select username from onmylist_user order by username') or die
(mysql_error());
while($row = mysql_fetch_array($query)) {
           echo("<option value=\"${row['username']}\">${row['username']}\n");
}
?>

</select>

<input type=submit value="Show A Persons People On My List">

</form>


<?
 do_html_footer();
?>

Mitch

richard.york · June 3rd, 2006, 03:09 PM

Validate the arguments that you are passing to mysql_list_fields and call mysql_error / mysql_errno after calling mysql_list_fields.

For example,

Code:

if ($connection && is_resource($connection))
{
    $fields = mysql_list_fields('A', 'B', $connection);

    if (!$fields)
    {
        // $fields evaluates to a false value
        echo mysql_error();
    }
    else if (is_resource($fields)
    {
        // $fields has a value that doesn't evaluate to false
        // and it is a resource, do something with it here.
    }
    else
    {
        // You should never get here, but I'm not one to say never
        echo "Unable to retrieve database fields.";
    }
}
else
{
    // something's botched with the connection.
    echo "No connection to the database!";
}

Alot of what you do in programming is track down bugs, so to build a solid application, you'll need to include that part of it in your planning. That means validating the data that you pass as arguments to other functions, verify that the $connection is a resource handle and that it doesn't have a false value. Then, put in as much error reporting as is possible. Check returned values. Call built-in error reporting functions like mysql_error(), after every function where it's possible an error could be generated. Put in error reporting for every possible scenario. I see that you've at least started doing that from your script here, but, you need more!

The alternative is, if you're in a rush and just want a quick and dirty solution to the problem and aren't interested in the possible academic reasons why you're seeing errors, then just supress the error, in PHP that's done with the '@' operator.

Code:

$fields = @mysql_list_fields('A', 'B', $connection);

Of course, I recommend the former route.

In about six years of PHP programming, I've never seen this particular error, so I can't comment on what is likely to be wrong. Looking at the manual for this function, however, it is marked as deprecated. You should use a straight query instead. Either way, follow my advice and you'll spend less time wondering what went wrong and more time fixing what went wrong and working on other projects.

Quote:

quote:From: http://www.php.net/mysql_list_fields

This function is deprecated. It is preferable to use mysql_query() to issue a SQL SHOW COLUMNS FROM table [LIKE 'name'] statement instead.

The concept would be the same using the updated method since mysql_query also returns a resource handle. Just use the suggested SQL above and you'll be good to go.

For your reading pleasure:
http://www.php.net/is_resource
http://www.php.net/mysql_list_fields
http://www.php.net/mysql_error
http://www.php.net/mysql_errno
http://www.php.net/errorfunc
http://www.php.net/manual/en/languag...rorcontrol.php

HTH!

Regards,
Rich

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Mitch · June 4th, 2006, 09:25 AM

Thanks very much1

Little by little I will get there. :)

Mitch