Displaying Stuff With MySQL - Please Help

Swefx · 05-14-2006, 04:56 AM

Ok so here's the code:

Code:

<?php
// The MySQL info, edit these
$host = "myhost"; // the database location. if you dont know it, leave it as it is
$dbuser = "myuser"; // the database username
$dbpass = "mypass"; // the databases user's pass
$dbname = "voting"; // the name of the database

// The actual connection. DO NOT EDIT THESE
mysql_connect("$host","$dbuser","$dbpass"); // Using the variables stated above

$display = ("SELECT challenger FROM battle");
echo $display;
?>

The problem is that instead of displaying the data it display "SELECT challenger FROM battle".
????: NamePros.com http://www.namepros.com/programming/197174-displaying-stuff-with-mysql-please-help.html

I'm quite new to MySQL so I dont know what to do.

Shorty · 05-14-2006, 06:16 AM

Read my article on database queries:
http://www.goarticles.com/cgi-bin/showa.cgi?C=188575

NonProphet · 05-14-2006, 06:33 AM

theres more efficient ways to do it, but you can try this.

Code:

mysql_connect("$host","$dbuser","$dbpass");
mysql_select_db($dbname);
$display = "SELECT challenger FROM battle";
$result = mysql_query($display) or die(mysql_error());
while($row = mysql_fetch_assoc($result))
{
    $stuff = $row['challenger'];
     echo $stuff;
}

BillyConnite · 05-14-2006, 07:44 AM

Furthering what NonProphet said:

Your code Swefx, simply echos a MySQL query, but first you need to run that query by MySQL, then MySQL will send back data from that query, in the form of an array (or a few other choices).

NonProphet's code should work fine

PoorDoggie · 05-14-2006, 01:54 PM

Originally Posted by BillyConnite

Furthering what NonProphet said:

Your code Swefx, simply echos a MySQL query, but first you need to run that query by MySQL, then MySQL will send back data from that query, in the form of an array (or a few other choices).
????: NamePros.com http://www.namepros.com/showthread.php?t=197174

NonProphet's code should work fine

In your code $display will be either "true" or "false" depending on whether or not the mysql query worked or not.

You could do this with what you have:

PHP Code:

  if($display){

  // the mysql query worked:

  while($row = mysql_fetch_array($display)){

    echo $row['name_of_field_you_want_to_display'];

  }

}

else{

  // the mysql query didn't work

  echo "Something Wrong!<br />".mysql_error();

}

and there you have a simple error check as well!

05-14-2006, 07:44 AM	#4 (permalink)
BillyConnite Join Date: Jul 2005 Location: Coffs H, Australia Posts: 3,456	Furthering what NonProphet said: Your code Swefx, simply echos a MySQL query, but first you need to run that query by MySQL, then MySQL will send back data from that query, in the form of an array (or a few other choices). NonProphet's code should work fine __________________ Free Forums / GoDaddy Coupon Codes (NEW DOMAIN!) / Free Arcade Script / <?='Your computer is '.(1?fine:broken).'.'?>

05-14-2006, 06:16 AM	#2 (permalink)
Shorty Senior Member Join Date: Sep 2005 Location: England Posts: 1,034	Read my article on database queries: http://www.goarticles.com/cgi-bin/showa.cgi?C=188575

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