Go Premium for a chance to win a PS4. Enter to Win

x
?
Solved

please help with error supplied argument is not a valid MySQL result resource

Posted on 2004-04-20
6
Medium Priority
?
289 Views
Last Modified: 2008-03-06
Hi guys can you help me out please, i am not sure whats up with this error, code is listed below.
Working with php 4.3.6 running Apache 2.0.49 on XP

if (!$id) {

    // print the list if there is not editing

    $result = mysql_query("SELECT * FROM Athletes",$db);

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

      printf("<a href=\"%s?id=%s\">%s %s</a> \n", $PHP_SELF, $myrow["AthleteID"], $myrow["Name"], $myrow["Address"]);

        printf("<a href=\"%s?id=%s&delete=yes\">(DELETE)</a><br>", $PHP_SELF, $myrow["AthleteID"]);

    }

  }
0
Comment
Question by:jess_z0
  • 3
  • 2
6 Comments
 
LVL 6

Expert Comment

by:CosminB
ID: 10866975
you could either try
   $result = mysql_query("SELECT * FROM Athletes");
   or make sure $db is the resource link identifier

also you could do
   $result = mysql_query("SELECT * FROM Athletes");
   if (!result)
   {
     echo 'MYSQL ERROR: ', mysql_error(); //this will apear if you have an error in your sql sintax(wrong table name or stuff like that)
   }



use this script and tell me what is says:
if (!$id) {
    // print the list if there is not editing
    $result = mysql_query("SELECT * FROM Athletes");
    if (!result)
   {
     echo 'MYSQL ERROR: ', mysql_error(); //this will apear if you have an error in your sql sintax(wrong table name or stuff like that)
   }
    while ($myrow = mysql_fetch_array($result))
   {
      printf("<a href=\"%s?id=%s\">%s %s</a> \n", $PHP_SELF, $myrow["AthleteID"], $myrow["Name"], $myrow["Address"]);
       printf("<a href=\"%s?id=%s&delete=yes\">(DELETE)</a><br>", $PHP_SELF, $myrow["AthleteID"]);
    }
}
0
 

Author Comment

by:jess_z0
ID: 10867099
thanks for the script mate! much appriecated your help

from the script at the bottom you asked me to run, this is the response, same error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache2\htdocs\display2.php on line 123
0
 
LVL 6

Accepted Solution

by:
CosminB earned 1000 total points
ID: 10867139
add this:
echo 'MYSQL ERROR: ', mysql_error();
right before the while and see if it says anything
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 

Author Comment

by:jess_z0
ID: 10867190
i have used this code before so i am not sure why this is happening.

why is it not recognising the fuctions, is something screwed up with this installation?

thanks
jess
0
 

Author Comment

by:jess_z0
ID: 10867217
ok good work!
so it is saying no database is selected
0
 
LVL 10

Expert Comment

by:eeBlueShadow
ID: 10867307
you need to add a

mysql_select_db("yourDatabaseName", $db);

just after the mysql_connect line.

But you probably got that by now
0

Featured Post

Free Tool: IP Lookup

Get more info about an IP address or domain name, such as organization, abuse contacts and geolocation.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Author Note: Since this E-E article was originally written, years ago, formal testing has come into common use in the world of PHP.  PHPUnit (http://en.wikipedia.org/wiki/PHPUnit) and similar technologies have enjoyed wide adoption, making it possib…
It’s a season to be thankful, and we’re thankful for users like you who engage on site, solve technology problems, and network with others in the industry. What tech are we most thankful for? Keep reading.
Learn how to match and substitute tagged data using PHP regular expressions. Demonstrated on Windows 7, but also applies to other operating systems. Demonstrated technique applies to PHP (all versions) and Firefox, but very similar techniques will w…
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.
Suggested Courses

886 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question