Php function not working ,undefined variable

Question

I getting a php error

undefined variable $fname

 $fname="John";
 function getThis()
 { 
  $complete_name= $fname."Kerry";
echo $complete_name;}
    getThis();

Any help in the right direction is much appriciated! Thanks

yes, it's about scoping, I admit it. otherwise, you have other related codes with it? — Herman Nz
– Herman Nz, Commented Mar 13, 2015 at 7:29

Amit Verma · Accepted Answer · 2015-03-13 07:30:27Z

2

you have a variable scoping issue

Variables defined outside the function have Global scope and they can not be used inside a function, If you want to use them inside a function You must write the global keyword before the variable name.
 global $fname;

And in your case

I noticed that Variable $fname is defined outside the function, and it has a global scope, if you use this inside the function ,then it will throw error.

Try this

  $fname="John";
  function getThis()
 { global $fname;
  $complete_name= $fname."Kerry";
echo $complete_name;}
    getThis();

Source :

http://php.net/manual/en/language.variables.scope.php

edited Mar 13, 2015 at 7:30

answered Mar 13, 2015 at 7:23

Amit Verma

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Comments

Herman Nz · Accepted Answer · 2015-03-13 07:22:51Z

1

Try to place the $fname inside the getThis() function, as in:

<?php

 function getThis() { 
 $fname="John";
 $complete_name= $fname."Kerry";
 echo $complete_name;
 }
 getThis();

?>

answered Mar 13, 2015 at 7:22

Herman Nz

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Comments

Harpartap Singh Permar · Accepted Answer · 2015-03-13 07:24:01Z

1

Use This :-

$fname="John";

    function getThis()
     { 
    global $fname;
    $complete_name= $fname."Kerry";
    echo $complete_name;
    }
    getThis();

For more information click here

answered Mar 13, 2015 at 7:24

Harpartap Singh Permar

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Comments

hugorut · Accepted Answer · 2015-03-13 07:24:09Z

1

You could pass the variable into the function.

Like so:

 $fname="John";

 function getThis($f_name)
 { 
   $complete_name= $f_name."Kerry";

   echo $complete_name;
 }

 getThis($fname);

answered Mar 13, 2015 at 7:24

hugorut

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Comments

user1844933 · Accepted Answer · 2015-03-13 07:20:09Z

0

You can not access variables inside function which declared out side, you have to use global keyword to use....

 $fname="John";
 function getThis()
 { 
global $fname;
$complete_name= $fname."Kerry";
echo $complete_name;}
    getThis();

answered Mar 13, 2015 at 7:20

user1844933

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1 Comment

Anthony Horne Over a year ago

Maybe just add comment as to what was changed and why.

kguest · Accepted Answer · 2015-03-13 08:54:27Z

-1

You have to declare $fname as global. That's it...

edited Mar 13, 2015 at 8:54

kguest

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answered Mar 13, 2015 at 7:41

Arun

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