Problem with accessing a global variable in PHP

Question

Following is code snippet :

function something() {
$include_file = 'test.php';
if ( file_exists($include_file) ) {
    require_once ($include_file);
//      global $flag;
//        echo 'in main global scope flag='.$flag;
    test();
   }
}

something();

exit;

 //in test.php

$flag = 4;
function test() {
   global $flag;

   echo '<br/>in test flag="'.$flag.'"';
   if ($flag) {
       echo 'flag works';
     //do something
   }
}

The above code snippet, echoes 'global scope' $flag value properly but doesnt recognises the $flag with value 4, assumes null value for $flag . Please point out what is wrong in accessing that $flag global variable.

Answer 1

Since your original question has already been answered let me suggest another approach that avoids global variables all together.
You seemingly have a function + a default value that you set as a global variable. You can also define default values for function parameters.

<?php // test.php
function test($flags=4) {
  echo '<br/>in test flag="'.$flags.'"';
  if ($flags!=0) {
    echo ' flag works';
  }
}

No global variable, but you can still call test() without a parameter and the function will use the default value - which you can overwrite by passing the parameter.

function something() {
  $include_file = 'test.php';
  if ( file_exists($include_file) ) {
    require_once ($include_file);
    echo "\nwithout parameter: "; test();
    echo "\nwith a parameter: "; test(16);
  }
}
something();
exit;

prints

without parameter: <br/>in test flag="4" flag works
with a parameter: <br/>in test flag="16" flag works

That's not exactly the same behavior but it may be what you wanted in the first place....

edit: Another point could be the

if ( file_exists($include_file) ) {
  require_once ($include_file);

construct you're using. I don't have a strong opinion (or any strong evidence ;-)) on that but you might want to consider using

$something = include $include_file;
if ( false!==$something ) { /* error handling here... */ }

instead. include "returns" false if php can't include the file. All that's required for this to work is that the included file doesn't explicitly return FALSE. This might be cache friendlier (e.g. when using something like APC). But again: I have no proof for this at hand, just a hunch. At least it gives your code a chance to handle an error "gracefully".

Answer 2

Just do something like this:

$myFlag = getFlag(); 
echo $myFlag; // output: whatever


// test.php 

$flag = 'whatever'; 

function getFlag(){
    global $flag;   
    return $flag; 
}

I would never set a variable in one script and call it from another file. this will cause extreme confusion down the road. Even my solution isn't the best. I would do this as a full blown class w/ ideally.

Answer 3

You are running in a problem here simply because the way PHP is currently interpreting your file. You have nested functions.

Here is how your code currently executes:

function something() {
    $include_file = 'test.php';
    if ( file_exists($include_file) ) {
        //require_once ($include_file);

        $flag = 4;
        function test() {
           global $flag;

           echo '<br/>in test flag="'.$flag.'"';
           if ($flag) {
               echo 'flag works';
             //do something
           }
        }

        //end require_once ($include_file);

        test();
    }
}

something();

exit;

As you can see, when you assign the value of 4 to $flag ($flag = 4), you are within the scope of the function something(), not within the global scope.

In test(), since you declare $flag as global within that function, $flag is a totally different variable, global to the whole script.

In order to avoid this problem, use the superglobal $GLOBALS. It's faster than using global anyway, and that way you don't get mingled in scoping issues as the ones above:

function something() {
    $include_file = 'test.php';
    if ( file_exists($include_file) ) {
        //require_once ($include_file);

        $GLOBALS['flag'] = 4;
        function test() {
           global $flag;

           echo '<br/>in test flag="'.$GLOBALS['flag'].'"';
           if ($GLOBALS['flag']) {
               echo 'flag works';
             //do something
           }
        }

        //end require_once ($include_file);

        test();
    }
}

something();

echo $flag; //echos 4

exit;

Answer 4

$flag = 4; is not in the global scope.

If the include occurs inside a function within the calling file, then all of the code contained in the called file will behave as though it had been defined inside that function.

-- PHP Manual page for include, which also applies for include_once, require, and require_once

I'm going to make a guess that the error you're getting is on the if ($flag) line, because at that point, $flag is uninitialized, because the global $flag variable has never been assigned a value.

Incidentally, echo 'global scope flag='.$flag; isn't displaying the global flag either, as you need a global $flag; in that function to display the global copy, which also has the side effect of making $flag = 4; affect the global copy in the included file.

Answer 5

The line likely giving you problems is this one: echo 'global scope flag'=$flag;

Change it to: echo 'global scope flag = '.$flag;

Edit: Now that you have edited you code snippet to by wrapping your code in the something() function, the problem becomes more clear. I think what is happening here is that the when you initialize $flag with $flag = 4;, it's scope is within the something() method.

Try placing $flag = 4; above/before the line function something() {. Then all will be right with the world.

Answer 6

Shouldn't echo 'global scope flag'=$flag; be echo 'global scope flag =' . $flag;

I received a different error. Which helped me fix the issue quickly.

Parse error: parse error, unexpected '=', expecting ',' or ';' in D:\Websites (Apache)\Test Site1\test.php on line 6