Simple php program displaying no output or error msg

Question

This is my first program in oop php. Its very simple where i would like to add a numerical value to a variable. And program must output 2.

 <?php
    class MyClass  
    {  
        public $a = 1;  

        public function abc()  
        {  
            if ($a=1){
                $a+1;   
            }  
        }   
    }  

    $obj = new MyClass;  

    echo $obj->abc;  
    ?>

No it must not. But you should get a lot of warnings and notices and even errors. This needs basic debugging first, enable error reporting to the highest level possible, log errors and track the error log. Refer to the PHP reference of your choice to learn about error reporting and logging. — hakre
– hakre, Commented Jul 12, 2012 at 17:47

andrewsi · Accepted Answer · 2012-07-12 17:41:29Z

3

In addition to gview's answer:

if ($a=1){
    $a+1;   
}

Should be:

if ($a == 1){
    $a = $a + 1;   
}

The = operator is for assignment, not for comparisons.

answered Jul 12, 2012 at 17:41

andrewsi

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1 Comment

JoeCortopassi Over a year ago

might want to add something about returning the value for $a

gview · Accepted Answer · 2012-07-12 17:40:54Z

2

The abc() function does not return anything. Thus you get no output. If you add:

return $a;

You'll get something in the echo.

answered Jul 12, 2012 at 17:40

gview

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Comments

JoeCortopassi · Accepted Answer · 2012-07-12 18:05:31Z

1

You aren't returning your results

public function abc()  
{  
   if ($a==1){
      $a++;   
   }  

   return $a;
 }

edited Jul 12, 2012 at 18:05

answered Jul 12, 2012 at 17:41

JoeCortopassi

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Comments

Yash Dayal · Accepted Answer · 2012-07-12 17:42:27Z

0

I think you forgot to return the value from abc()

 public function abc()  
        {  
            if ($a=1){
                $a+1;   
            }  
            return $a;
        }

answered Jul 12, 2012 at 17:42

Yash Dayal

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