PHP - Calling function inside another class -> function

Question

I'm trying to do this:

class database {    
function editProvider($post)
{   
    $sql = "UPDATE tbl SET ";
        foreach($post as $key => $val):
            if($key != "providerId")
            {
                $val = formValidate($val);
                $sqlE[] = "`$key`='$val'"; 
            }
        endforeach;

    $sqlE = implode(",", $sqlE);

    $where = ' WHERE `id` = \''.$post['id'].'\'';
    $sql = $sql . $sqlE . $where;

    $query = mysql_query($sql);

    if($query){
        return true;
    }
}
//
}//end class

And then use this function * INSIDE of another class *:

function formValidate($string){
  $string = trim($string);
  $string = mysql_real_escape_string($string);
  return $string;
}
//

.. on $val. Why doesn't this work? if I write in a field of the form, it's not escaping anything at all. How can that be?

* UPDATE * handler.php:

if(isset($_GET['do'])){

if($_GET['do'] == "addLogin")
{
    $addLogin = $db->addLogin($_POST);
}

if($_GET['do'] == "addProvider")
{
    $addProvider = $db->addProvider($_POST);
}

if($_GET['do'] == "editProfile")
{
    $editProfile = $db->editProfile($_POST);
}

if($_GET['do'] == "editProvider")
{
    $editProvider = $db->editProvider($_POST);
}
}
//end if isset get do

** The editProvider function works fine except for this :-) **

Where have you declared the two functions (in different classes)? — ComFreek
– ComFreek, Commented Nov 18, 2011 at 17:12
Its hard to see what the problem is because you said it is calling a function inside a class but you provide only two functions and no class. Its hard to see the scope and also can you try putting an echo at the beginning and end of the formValidate function to see if it is actually calling it. Please try and provide this debug info. Thanks! — jakx
– jakx, Commented Nov 18, 2011 at 17:15
I've added the different classes, where the functions is inside. editProvider is inside class database, and formValidate is inside class validate. Does this help you? — skolind
– skolind, Commented Nov 18, 2011 at 17:22
Yes please see Jose Vega's comment. Hope that helps! But just a sidenote -- if you had kept the validate as a simple function instead of putting it into a class you would have been able to do what you had originally, but instead putting it into a class forced you to instantiate that class, hopefully that's what you want. Also see calling functions statically:php.net/manual/en/language.oop5.static.php — jakx
– jakx, Commented Nov 18, 2011 at 17:28
But I am trying to use the $this->.. I just thought it had to be inside of a class, since you started talking about it. Now it's NOT inside of a class, but still it doesn't work with $this-> — skolind
– skolind, Commented Nov 18, 2011 at 17:35

Jose Vega · Accepted Answer · 2011-11-18 17:21:23Z

2

You need to instantiate that validate class and than once instantiated you will need to call that function in that class with your value parameters.

Inside your editProvider you can have:

$validator = new validate();
$val = $validator->formValidate($val);

If the above doesn't work, try the following:

$val = mysql_real_escape_string(trim($val));

and see if it works, if it does it has to do with the correct function not being called.

edited Nov 18, 2011 at 17:21

answered Nov 18, 2011 at 17:10

Jose Vega

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3 Comments

Jose Vega Over a year ago

Can you verify that formValidate is actually firing? Add an echo $string and make suer its receiving the correct parameter.

skolind Over a year ago

I am trying to use this without using the functions inside a class. And tried $this-> but it seems it doesn't work.

skolind Over a year ago

It seems like it doesn't even work if I'm doing this: mysql_real_escape_string($val); so it might be something before all this? :-s I'm having a handler.php where im sending the $_POST to the function.. I've edited my question for the handler.php to see.

Mike Purcell · Accepted Answer · 2011-11-18 18:50:35Z

1

Not sure why you are so bent on using $this vs a static implementation. IMO, a static call makes this code much easier. If you really want access to $this->formValidatString() from your database class, you will have to do class database extends MyOtherClass.

Here is how easy it would be to do a static call:

class database {

    public function editProvider($post)
    {   
        $sql = "UPDATE tbl SET ";
        foreach($post as $key => $val):
            if($key != "providerId")
            {
                $val = MyOtherClass::formValidate($val);
                $sqlE[] = "`$key`='$val'"; 
            }
        endforeach;

        $sqlE = implode(",", $sqlE);

        $where = ' WHERE `id` = \''.$post['id'].'\'';
        $sql = $sql . $sqlE . $where;

        $query = mysql_query($sql);

        if($query){
            return true;
        }
    }        
}//end class

class MyOtherClass
{
    public static function formValidate($string) {

        if (strlen($string) < 1) {
            throw new Exception('Invalid $string ' . $string . ');
        }

        $string = trim($string);
        $string = mysql_real_escape_string($string);

        return $string;
    }
}

edited Nov 18, 2011 at 18:50

answered Nov 18, 2011 at 17:48

Mike Purcell

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2 Comments

skolind Over a year ago

This still doesn't work. Please read the other comments. I'm writting that not even mysql_real_escape_string($val) works.. :-S I don't know what's going on here.

Mike Purcell Over a year ago

@Kolind: Updated code, the exception will let you know if a val you pass in is legitimate.

v42 · Accepted Answer · 2011-11-18 17:37:10Z

0

You don't need to have an instance for this purpose. Just do validate::formValidate($val);.

answered Nov 18, 2011 at 17:37

v42

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