visit this link:
FindItGlobally.com - Business Directory
And scroll to the very bottom you will see Resource id #59
I made a Query using this code:
[HIGHLIGHT="PHP"]$queryzz = "SELECT url FROM {$Stb_name}_LINK";
$theurloutput = mysql_query($queryzz) or die('Error, query failed');
echo $theurloutput;[/HIGHLIGHT]
and the Resource id #59 is outputted by the echo.
Thanks for helping. This is my first time using PHP and MySQL.
Resource id #59
Started by TcM, Jan 11 2008 09:11 AM
5 replies to this topic
#1
Posted 11 January 2008 - 09:11 AM
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#2
Posted 11 January 2008 - 09:46 AM
You're not supposed to echo the resource given by mysql_query() The resource contains the result of the query, so you shall use it to get the data. There's a lot of PHP/MySQL-functions. You can find a complete list right here: PHP: MySQL - Manual
#3
Posted 11 January 2008 - 11:01 AM
Thanks man! I finally got it working, now the headache will start :( because I gotta make something working.. let's hope it does!
#4
Guest_Jaan_*
Posted 11 January 2008 - 03:10 PM
Guest_Jaan_*
well you can try this:
it will work 100%
i guess... :D
$query = mysql_query("SELECT * FROM {$Stb_name}_LINK");
$row = mysql_fetch_rows($query);
$theurloutput = $row['url'];
echo $theurloutput;
it will work 100%
i guess... :D
#5
Posted 13 January 2008 - 04:31 AM
Yes I did something similar.. and works fine :)
Thanks
Thanks
#6
Guest_Jaan_*
Posted 13 January 2008 - 08:45 AM
Guest_Jaan_*
Np :)


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