php $GLOBALS variables

Question

I'm writing a bit of the code and I have parent php script that does include() and includes second script, here is snippet from my second code:

echo ($GLOBALS['key($_REQUEST)']);

I'm trying to grab a key($_REQUEST) from the parent and use it in child, but that doesn't work..

this is when I run script using command line:

 mbp:digaweb alexus$ php findItemsByKeywords.php test PHP Notice: Undefined index: key($_REQUEST) in /Users/alexus/workspace/digaweb/findItemsByKeywords.php on line 3 PHP Stack trace: PHP 1. {main}() /Users/alexus/workspace/digaweb/findItemsByKeywords.php:0 mbp:digaweb alexus$

i heard that globals isn't recommended way also, but i don't know maybe it's ok...

The alternative would probably be to define a function/class in the included script and call it/create an instance in the including script. — outis
– outis, Commented Sep 30, 2009 at 5:09
can you maybe answer this question with some examples, i'm not really following what you saying.. — alexus
– alexus, Commented Sep 30, 2009 at 5:13

user153275 · Accepted Answer · 2009-10-02 03:53:43Z

$_REQUEST is a superglobal and will be directly available inside of any function or script, so you don't need to worry about passing it to the child script. However, PHP won't populate $_REQUEST when used from the command line, unless you're using a configuration option I'm unfamiliar with. You'll need to use the $_SERVER['argv'] array.

Globals are indeed not recommended. You'll have an easier time long-term if you go with what outis suggested. Here's an example:

script1.php: <?php $file = $_SERVER['argv'][1]; // 0 is the script's name require_once ('script2.php'); $result = doSomething ($file); echo $result; ?> script2.php: <?php function doSomething ($inputfile) { $buf = file_get_contents($inputfile); $buf = strtolower($buf); // counts as something! return $buf; } ?>

This example doesn't make use of the key($_REQUEST), but I'm not sure what the purpose of that is so I just went with $_SERVER['argv'].

James Skidmore · Accepted Answer · 2009-09-30 16:51:01Z

Based on your comment to my other answer, I think I understand what you're trying to do. You're just trying to pass a variable from one script into another script that's included.

As long as you define a variable before you include the script, it can be used in the included script. For instance:

// script1.php $foo = 'bar'; include_once('script2.php'); // script2.php echo $foo; // prints "bar"

i'll give it a shot, but for now and I dont know if this is best practice, i did through making my script2 as a function.

James Skidmore · Accepted Answer · 2009-09-30 05:04:38Z

echo $_GLOBALS[key($_REQUEST)];

You just need to remove the single quotation marks. It was looking for the literal 'key($_REQUEST)' key, which obviously doesn't exist.

It all depends on what you are trying to do though... what are you trying to do?

when I tried your example, I'm getting: PHP Notice: Undefined variable: _GLOBALS, what I'm trying to do is to pass that key($_REQUEST) to my child script that I called through include(), I need to pass something to that script form parent script

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