[PHP][MYSQL]Display dynamic data from database

[Rdogg112]

im making a game and i need to show a users money but i dont know how help?

[MatthewJ]

You should probably learn to program before trying to code a game.

 

[Rdogg112]

true

 

this is wat i got so far

 

$sql = mysql_query("SELECT * FROM `users` WHERE `money` = $money") or die(mysql_error());
$blah = mysql_fetch_array($sql);
echo 'Money: '. $blah['money'];

 

but i dont know what to put in $money

+ i dont know if that code works.

[lonewolf217]

code is basically correct.  where $money comes from is entirely up to you though based on your "game"

[Rdogg112]

what could i acctualy put in $money?

[lonewolf217]

um, how about a value of how much money you are searching for ?

[Rdogg112]

what do you mean how much money im searching for?

[lonewolf217]

i suggest you start with the basics.  like what you want to do with your game, and how to use SQL queries .

 

http://www.tizag.com/phpT/index.php

http://www.tizag.com/sqlTutorial/

 

Once you understand what you are talking about, then you can get specific help

[Muddy_Funster]

You don't want to search WHERE money = $money in the first place!  search where userID = $userID (or simmilar) if you knew the value of $money you wouldn need to ask the database what is would you?

[PaulRyan]

Your code is almost correct, you need to get the money based on the logged in users ID or username.

 

  $sql = mysql_query("SELECT money FROM `users` WHERE id = {$user['id']}");
  if($sql) {
    $blah = mysql_fetch_assoc($sql);
    echo 'Money: '. $blah['money'];
  } else }
    echo 'This Query Failed: '.mysql_error();
  }

 

Try the above, but change the $user['id'] to the method you are using to identify a user.

 

Regards, PaulRyan.

[PFMaBiSmAd]

This is a two year old thread that got bumped by sdfpw to post a link.

[PaulRyan]

Well spotted...I feel such an idiot now