nuk3 — Sat, 04 Feb 2006 06:09:56 +0000

Technicially the ampersand just surpresses the error message being generated by the function. You are much better off fixing the problem instead of hiding it.

The usual reason for the error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in" is that the sql query is malformed (contains an error, etc). You can find the error in your query by appending an "or clause" to the mysql_query function to display the error and terminate the execution of the script.

$result = mysql_query("SELECT * FROM samples_tbl WHERE id='$category' ",$connect) or die( mysql_error() );

Give that a whirl.

thanks to both of you guys for helping out

pastajon66@juno — Sat, 04 Feb 2006 03:30:41 +0000

i applied the @ to the code and whoala, it is fixed!
thanks again for helping me out!

Dux0r — Fri, 03 Feb 2006 19:39:06 +0000

I hear ya, I remember switching to JavaScript from PHP and spending hours wondering why my <? tags weren't working.

timjpriebe — Fri, 03 Feb 2006 18:14:18 +0000

Actually, I suggested what I meant to suggest, but was thinking of Perl, not PHP. One of the pitfalls of working simultaneously in too many languages.

Dux0r — Fri, 03 Feb 2006 18:10:15 +0000

The error is because this query:

mysql_query("SELECT * FROM img_tbl WHERE category='$section' ORDER BY name",$connect);

..is failing. It can't be due to the connection in config.php failing, otherwise the previous mysql_query on line 11 would also fail.

Therefor it's a problem with your mysql database. The three possible problems are:

There is no table called img_tbl
There is no row called category
There is no row called name

P.S. what Tim suggested is incorrect:

while($myrow = mysql_fetch_array($result))

Will fetch a row from the result and put it in $myrow whereas..

while(@myrow = mysql_fetch_array($result))

Will do nothing, it needs to be assigned to a variable. What he meant to suggest was:

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

Which will do exactly the same thing as it already does, but it wont issue an error message.

timjpriebe — Fri, 03 Feb 2006 15:12:46 +0000

What info is in config.php? It should have $connect defined.

@ wasn't the issue

pastajon66@juno — Fri, 03 Feb 2006 14:58:13 +0000

tim,
thanks for the suggestion
i tried it without success
here is the code for that page:

<?php
// load the configuration file.
include("config.php");
    if (@$_REQUEST['id'] == ""){
    $category = 1;
    } else {
    $category = $_REQUEST['id'];
    }
        $result = mysql_query("SELECT * FROM samples_tbl WHERE id='$category' ",$connect);
        //lets make a loop and get all news from the database
        while($myrow = mysql_fetch_array($result))
             {//begin of loop
               //now print the results:
               $section = $myrow['category'];
               echo "<div id=\"copy\">";
               echo $myrow['text'];
               echo "</div>";
             }//end of loop
        $result = mysql_query("SELECT * FROM img_tbl WHERE category='$section' ORDER BY name",$connect);
        //lets make a loop and get all news from the database
        echo "<div class=\"container\">";
        while($myrow = mysql_fetch_array($result))
             {//begin of loop
               //now print the results:
               echo "<div class=\"float\">";
               echo "<a href=\"display.php?id=".$myrow['name']."\"><img src=\"images/mid_tree_thumbs/".preg_replace('/\s+/', '', $myrow['name']).".jpg\" alt=\"".$myrow['name']."\" height=\"100\" width=\"100\"><br>";
               echo "<p>".$myrow['name']."</p></a>";
               echo "</div>";
             }//end of loop
        echo "</div>";
        mysql_close();
?>

timjpriebe — Fri, 03 Feb 2006 13:26:11 +0000

I'm not sure without seeing the code, but I'm pretty sure that line should be fixed like this:

while(@myrow = mysql_fetch_array($result))

Again, it's nearly impossible to tell if that's the only thing wrong. Often when errors like that are thrown, the mistake may not be on the actual line that PHP thinks it is on. A classic example is where the line of code before is missing a semicolon at the end.

help with a warning message...

pastajon66@juno — Thu, 02 Feb 2006 21:21:35 +0000

i get this warning message on the bottom of a page that i created within the MySQL database:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site306/fst/var/www/html/samples_body.php on line 22

here is line 22:
while($myrow = mysql_fetch_array($result))

what could be wrong?

timjpriebe — Tue, 31 Jan 2006 15:02:35 +0000

Absolutely. All the PHP code should be in sections starting with

<?php
 and ending with 
?>

As long as you stay out of those sections, you won't mess up the PHP at all. Of course, at some point, you may have to adjust something in those sections. There's no real way to tell ahead of time.