PHP，两个外部脚本上的全局变量

Question

How would I pass a global var in two external scripts?

<div>
<!-- INCLUDEPHP 1.php -->
</div>
<div>
<!-- INCLUDEPHP 2.php -->
</div>

I have tried creating global variables on 1.php and `2.phpp but it didn't work.

1.php:

<?php
global $someVar;
$sql = ...;
$someVar= $db -> sql_query($sql);
?>

2.php:

<?php
global $someVar;
echo "$someVar";
?>

Am I doing something wrong?

Answer 1

I would try including the scripts via PHP:

<div>
<?php require "1.php" ?>
</div>
<div>
<?php require "2.php" ?>
</div>

Answer 2

If both includes are loaded into the same page, and the variables exist already in the global scope, all your functions can access them with the global statement. Since everything is already global, the statement is not required in the global scope, only inside functions. This also permits functions to share variables by casting them onto the global scope.

There are many dangers to this, though, which I'll not pretend to be fully aware of, so one is best advised to make prudent use of the global scope in large complex applications as they can become very volatile if naming conventions are relaxed.

Basically, we're looking at,

function arrow() { global $a; $a = "arrow"; return $a; }
function sky() { global $b; $b = "sky"; return $b; }
echo "I shot an " . arrow() . " into the " . sky() . ".";
echo "I shot an $a into the $b.";

which is child's play, it demonstrates how exposed the variables are, sitting out there with no protection. Now another function can come along and blow the whole thing apart:

function whammo() { global $a, $b; $c = $a; $a = $b; $b = $c;}
echo "I shot an " . arrow() . " into the " . sky() . ".";
whammo();
echo "I shot an $a into the $b.";

See what I mean?

Your solution probably lies in a closure of some sort, wherein all the functions are contained that need this 'global'. It will be much better protected.