How to put mysql inside a php function?

Question

I have problem about putting mysql into a function showMsg(). The mysql is working fine if it is not wrapped by function showMsg(), but when I wrap it with function showMsg(), it gives error message "Warning: mysql_query(): supplied argument is not a valid". How to put mysql inside a php function? Below is my codes:

<?php    
function showMsg(){   
        $query2 = "SELECT id, message, username, datetime FROM messageslive ORDER BY id DESC LIMIT 20";
        $result2 = mysql_query($query2,$connection) or die (mysql_error());
        confirm_query($result2);
        $num = mysql_num_rows($result2); 
        while($msginfo = mysql_fetch_array($result2)){
            echo $msginfo['message'];
            echo $msginfo['username'];
        }
}

<div>
   <?php showMsg(); ?>
</div>
?>

Answer 1

You probably need:

global $connection;

(Inside the function, that is.)

See Variable Scope

Answer 2

1. Never use global .
   Pass $connection into your function as an argument.

2. Logic and representation should be separated.
   Read about MVC: here or here .

3. Global variables are evil, never use it. If someone suggests it - ignore all their answers.

Answer 3

As everyone mentioned, the issue has to do with variable scoping. Instead of add global $connection; you could consider a more OOP approach and consider:

A: passing the $connection variable into the function.

B: placing related functions in a class and pass the DB connection into the Class constructor. for example:

class YourClass  {

   private $connection;

   public function __construct($connection) {
       $this->connection = $connection;
   }

   public function showMsg(){   
        $query2 = "SELECT id, message, username, datetime FROM messageslive ORDER BY id DESC LIMIT 20";
        $result2 = mysql_query($query2,$this->connection) or die (mysql_error());
        confirm_query($result2);
        $num = mysql_num_rows($result2); 
        while($msginfo = mysql_fetch_array($result2)){
            echo $msginfo['message'];
            echo $msginfo['username'];
        }
   }

}

I don't have enough rep to comment. But I also like OZ_'s answer:)

Answer 4

$connection variable has no value assigned. The following code should solve your problem (add it at the beginning of the function):

global $connection;

But you should be aware of the fact, that using globals is not a good idea and you may want to:

• (preferably) pass $connection variable within the parameter of the function, or
• move $connection declaration from outside the function just into the function (if it does not cause additional problems), or
• redeclare $connection variable within the function (again: if it will not cause additional problems),

Answer 5

Because variables are local to functions. You need to add this inside your function:

global $connection;

Answer 6

Simply put, functions ignore outside variables due to variable scope . You must let the declare the variable as being from the outside by using global or you can send $connection through a parameter.