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Filter_var what are you doing?


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  #1  
Old March 13th, 2013, 03:20 AM
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Filter_var what are you doing?
Why does it print Not valid?
PHP Code:
<?php 
class Validator
    {
        public function validate($string, $type)
            {
                if($type=='email')
                    {
                        if(!filter_var($string, FILTER_VALIDATE_EMAIL))    
                            {
                                return FALSE;    
                            }
                    }
            }
    
    }

$object = new Validator();
$string = 'ben@yahoo.com';
$type = 'email';
if($object->validate($string, $type))
    {
        echo "Valid";    
    }
else
    {
        echo "Not Valid";    
    }
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  #2  
Old March 13th, 2013, 03:49 AM
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Hi,

because that piece of code isn't finished. What if filter_var() does succeed like it's supposed to do in your case? You don't seem to have considered that case in your if statement.

That's why it's usually a good idea to complete the code before starting to test it.

You should actually get rid of those unnecessary if statements and use the return value of filter_var() directly:

PHP Code:
<?php  

class Validator 
    { 
        public function validate($string, $type) 
            { 
                if($type == 'email') 
                    return filter_var($string, FILTER_VALIDATE_EMAIL) !== false;
            } 
     
    }
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  #3  
Old March 13th, 2013, 08:56 AM
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filter_var changes the data and returns the changed data. It does not return a boolean.

PHP Code:
if(filter_var($string, FILTER_VALIDATE_EMAIL) != $string) 
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  #4  
Old March 13th, 2013, 09:21 AM
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Quote:
Originally Posted by ManiacDan
filter_var changes the data and returns the changed data. It does not return a boolean.


filter_var() with FILTER_VALIDATE_EMAIL returns false if the validation failed and the original string if it succeeded.

You seem to be confusing this with the FILTER_SANITIZE constants. Those do in fact make the function change the input.
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