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>
?>

Community · Accepted Answer · 2017-05-23 10:29:46Z

9

Never use global.
Pass $connection into your function as an argument.
Logic and representation should be separated.
Read about MVC: here or here.
Global variables are evil, never use it. If someone suggests it - ignore all their answers.

edited May 23, 2017 at 10:29

CommunityBot

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answered May 24, 2011 at 0:06

user680786

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1 Comment

zac1987 Over a year ago

never knew i can do that. I have follow your suggestion, i am using function parameter instead of global :)

Alex Howansky · Accepted Answer · 2011-05-23 23:52:33Z

6

You probably need:

global $connection;

(Inside the function, that is.)

See Variable Scope

answered May 23, 2011 at 23:52

Alex Howansky

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1 Comment

Jim Ford Over a year ago

I voted down for recommending globals. globals very often lead to sloppy code with unexpected results.

Tyrick · Accepted Answer · 2011-05-24 00:13:52Z

4

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 :)

edited May 24, 2011 at 0:13

answered May 24, 2011 at 0:08

Tyrick

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2 Comments

zac1987 Over a year ago

OMG, I have very less knowledge about Class and Object things. May I know any benefits if I use class instead of passing $connection to function parameters?

Kenny Cason Over a year ago

@Zac1987 It can take a bit of time to learn OOP techniques, but a quick google search on OOP + PHP should bring back something interesting. Trust me learning OOP will benefit you so much in the long run as well as make your code more manageable and clean.

Tadeck · Accepted Answer · 2011-05-23 23:53:42Z

2

$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),

answered May 23, 2011 at 23:53

Tadeck

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2 Comments

zac1987 Over a year ago

I was wondering where should i put global $connection; Thanks for pointing me :)

Tadeck Over a year ago

No problem :) It should be placed before you use $connection variable - if you put it at the beginning of the function, this should meet this requirement ;)

gnxtech3 · Accepted Answer · 2011-05-23 23:53:13Z

0

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

global $connection;

answered May 23, 2011 at 23:53

gnxtech3

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Comments

Yoshiyahu · Accepted Answer · 2011-05-23 23:58:43Z

-2

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.

answered May 23, 2011 at 23:58

Yoshiyahu

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