I have read through the FAQ sticky regarding this topic (and have also searched/read here further on this topic) but I am still receiving the following error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a4542527/public_html/index.php on line 8

The scripting for this is basic at this point (I am building something in steps and this is just a test section: The script will call data from the table and display said data.)

The following is the code (two, separate files located on my server)- I am hoping that someone may see something that I am missing.

connection.php

<?php
    $dbhost = 'mysql7.000webhost.com';
    $dbuser = 'a4542527_root';
    $dbpass = '**********';
    
    $db = 'a4542527_test1';
    $conn = mysql_connect($dbhost,$dbuser,$dbpass);
    mysql_select_db($db);

?>

index.php

<?php
    include 'connection.php';
    
    $query = "SELECT * FROM people";
    
    $result = "mysql_query($query)";
    
    while ($person = mysql_fetch_array($result)){
    echo  $person ['name'];
    echo  $person ['descrip'];
    }
?>

Thank-you for any help, suggestions or a point in the right direction.
Matty

Recommended Answers

All 4 Replies

change:
$result = "mysql_query($query)"; to:
$result = mysql_query($query) or die( mysql_error() );

your connection MAY not be proper, or else the sql query is returning some error for sure.

or else the sql query is returning some error for sure

which query? He did NOT query the db at all! He simply assigned a string to $result. He needs to remove the red quotation marks I pointed out from line 6 on his second code block.

Its coding, very little character counts.

$result = "mysql_query($query)";

no need for the " " sign.
;)

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