Hello again,

I got the fetch_array problems fixed.  I was using the actual server name, 
when I switched back to localhost everything worked!!

Now I have a question about how to make cookies work on a Windows machine. 
 This is what I have these parameters set to but it's not working.  Do I 
have slashes/backslashes wrong, or is there something else I have to do in 
Apache??:

session.save_path = "C:/Program Files/Apache 
Group/Apache/web/php/dir/files/temp"

; Whether to use cookies.
session.use_cookies = 1


; Name of the session (used as cookie name).
session.name = PHPSESSID

; Initialize session on request startup.
session.auto_start = 0

; Lifetime in seconds of cookie or, if 0, until browser is restarted.
session.cookie_lifetime = 0

; The path for which the cookie is valid.
session.cookie_path = "c:/Program Files/Apache 
Group/Apache/web/php/dir/files/temp"

; The domain for which the cookie is valid.
session.cookie_domain = www.mydomain.com



As always, thanks for your help.








PHPCoder <[EMAIL PROTECTED]>
07/24/02 01:50 PM

 
        To:     Matthew Bielecki <[EMAIL PROTECTED]>
        cc:     php-general <[EMAIL PROTECTED]>
        Subject:        Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is "failing", 
the problem here is that $result is not containing naything, and that is 
normally due to the fact that you are not connecting to the db, or the 
table "tablename" is not there.

I use the following format as my "standard" MySQL connect and query 
snippet:

$link = @mysql_connect("localhost",$username,$password) or die ('Could 
not connect!'); //@ suppresses the default error message generated by 
this function and the "or die()" bit kills the script right then and 
there should it not be able to connect.
        mysql_select_db("YOUR_DB_NAME",$link);
        $sql = "select * from your_table_name";
        if ( $result = mysql_query($sql)) {  // checks to see if $result 
contains anything before it even tries to fetch an associative array 
from it.
             $row = mysql_fetch_assoc($result);
        } else {
        echo "Empty result set!";

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 
out of 10 times, you don't need the array element id's that is returned 
by mysql_fetch_array.

Matthew Bielecki wrote:

>I have a couple of scripts that fail with the error of:
>Warning: mysql_fetch_array(): supplied argument is not a valid MySQL 
result 
>resource in...
>
>I'm new to both SQL and PHP and I'm wondering if I have some setting 
>turned off or what.
>
>Here's the piece of code that is failing (the second line fails):
>
>$result = mysql_db_query($dbname, "SELECT * FROM tablename ORDER BY id");
>        $row = mysql_fetch_array($result);
>
>
>Thanks for your help in advance!!
>




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