Mysql_fetch_array();

[djcontact]

ok some im getting these error:


[b]Warning:[/b] mysql_fetch_array(): 20 is not a valid MySQL result resource in [b]/home/nelsona/public_html/index.php[/b] on line 9

[b]Warning:[/b] mysql_free_result(): 20 is not a valid MySQL result resource in [b]/home/nelsona/public_html/index.php[/b] on line 17

any idea why im perhaps getting this error...

i first did this site on my server and had no problems,
then transfered it over to the clients server and now im getting this error.

any help would be appricated

thanks
djcontact

[wildteen88]

You usually get this error message when you have an error in your sql query. Please read [a href=\"http://www.phpfreaks.com/forums/index.php?showtopic=95376\" target=\"_blank\"]this thread[/a]. Also seardch the forum too as this question has been answered many times over

[djcontact]

so i read that thread you suggested [a href=\"http://www.phpfreaks.com/forums/index.php?showtopic=95376\" target=\"_blank\"]Invalid Resource, MySQL[/a] and it doesnt really answer my question...
either that or im just not understanding it...

i also went ahead and did a search for mysql_fetch_array and i did get a hand full of search results but nothing that i can relate too...

well with the amount of php knowledege that i know which is not much, i would love to get some assistants on this.

thanks again
djcontact

[thepip3r]

and how are your mysql queries set up for error checking?? here is a simple and reliable way to do error trapping with MySQL:

[code]$result = mysql_query("SELECT * FROM TABLE WHERE blah='$blah'") or die("MySQL Query failed:  ".mysql_error());[/code]

the mysql_error() function will allow you to get the error reported by MySQL

[djcontact]

this is what i have on my connection.php
[code]
#database access info
define ('DB_USER', 'myuser');
define ('DB_PASS', 'mypass');
define ('DB_HOST', 'localhost');
define ('DB_NAME', 'mydbname');

#database connection
$dbc = mysql_connect (DB_HOST, DB_USER, DB_PASS) OR die ('Could not connect to mysql: ' .mysql_error() );
mysql_select_db (DB_NAME) OR die ('Could not select database: ' .mysql_error() );
[/code]


and this is my index.php
[code]include ('_include/connect.php');
//make query
$query = "SELECT * FROM layout";
//run the query
$result = @mysql_query ($query);

if ($result) {
    //fetch
    while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        if($row[0] == 1) {
            include ('layout_01.php');
        }
        if($row[0] == 2) {
            include ('layout_02.php');
        }
    }
    mysql_free_result ($result); //free up the resources
} else { //if it did not run ok
    echo '<p>The content could not be displayed due to system error!</p>';
}
[/code]

[thepip3r]

so you need to add some error checking for your MySQL statement like i already posted. change your code to look like this:

[code]include ('_include/connect.php');
//make query
$query = "SELECT * FROM layout";
//run the query
$result = @mysql_query ($query) or die("MySQL query failed:  ".mysql_error());

if ($result) {
    //fetch
    while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
        if($row[0] == 1) {
            include ('layout_01.php');
        }
        if($row[0] == 2) {
            include ('layout_02.php');
        }
    }
    mysql_free_result ($result); //free up the resources
} else { //if it did not run ok
    echo '<p>The content could not be displayed due to system error!</p>';
}[/code]

[djcontact]

ok so i added the error checking for the mysql statement but im still getting the same error,

[b]Warning:[/b] mysql_fetch_array(): 20 is not a valid MySQL result resource in [b]/home/nelsona/public_html/index.php[/b] on line [b]9[/b]

[b]Warning:[/b] mysql_free_result(): 20 is not a valid MySQL result resource in [b]/home/nelsona/public_html/index.php[/b] on line [b]17[/b]



i really appricate the help
thanks