Analysis of the difference between global and $GLOBALS in php_PHP tutorial

Some friends said that these two are actually the difference in writing. Now I will prove to you the difference between global and $GLOBALS. Friends in need can refer to it for specific reference.

According to the official explanation it is

1.$GLOBALS['var'] is the external global variable itself.

2.global $var is a reference or pointer of the same name as external $var.

Look at the following example first:

Let’s give an example:

代码如下	复制代码
$var1 = 1;      $var2 = 2;      function test(){           $GLOBALS['var2'] = &$GLOBALS['var1'];      }      test();      echo $var2;      ?>

The normal print result is 1

The code is as follows

Copy code

代码如下	复制代码
$var1 = 1;      $var2 = 2;      function test(){           global $var1,$var2;           $var2 = &$var1;      }      test();      echo $var2;      ?>

$var1 = 1;

$var2 = 2;

Function test(){

global $var1,$var2;
$var2 = &$var1;

}  

test();

echo $var2;

?>

代码如下	复制代码
function test() { global $a; unset($a); } $a = 1; test(); print $a; ?>

Unexpected print result is 2 Why is the printed result 2? In fact, it is because the reference of $var1 points to the reference address of $var2. Resulting in the actual value not changing We all know that the variables generated by functions in PHP are private variables of the function, so the variables generated by global cannot escape this rule. Why do we say this? Look at the following code: PHP code

The code is as follows	Copy code
function test() { global $a; unset($a); } $a = 1; test(); print $a; ?>

The fulfillment effect is:
1

Why is 1 output? Hasn’t $a been unset? Unset failed? PHP bug?
No, in fact, unset works. It unsets $a in the test function. You can add it after the function
print $a;
In other words, global generates the alias variable "$a" for $a outside the test function, in order to differentiate it from the $a outside
Then go back to Example 1 above and look at the code "$var2 =& $var1;" in test_global. The above is a reference assignment operation, that is, $var2 will point to the physical memory address pointed to by var1
So we come to the conclusion that the difference between global and $GLOBALS[] in the function is:
global generates an alias variable in the function that points to the external variable of the function, instead of the actual external variable of the function. Once the address pointed to by the alias variable is changed, some unexpected situations will occur, such as Example 1.
$GLOBALS[] is indeed called an external variable, and it will always remain consistent inside and outside the function!

You can compare the following two examples:

The code is as follows

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global:

代码如下	复制代码
global： function myfunction(){ global $bar; unset($bar); } $bar="someting"; myfunction(); echo $bar; ?> 输出：someting $global[]: function foo() { unset($GLOBALS['bar']); } $bar = "something"; foo(); echo $bar; ?> 输出：空

function myfunction(){

global $bar;
unset($bar);

}

代码如下	复制代码
$a = 1; $b = 2; function Sum() { global $a, $b; $b = $a + $b; } Sum(); echo $b; ?>

$bar="someting";

myfunction();

echo $bar;

?>

Output: something

代码如下	复制代码
$var1 = 1;      function test(){           unset($GLOBALS['var1']);      }      test();      echo $var1;      ?>

$global[]:

function foo()

{

代码如下	复制代码
$var1 = 1;      function test(){          global  $var1;           unset($var1);      }      test();      echo $var1;      ?>

unset($GLOBALS['bar']); } $bar = "something"; foo(); echo $bar; ?> Output: empty

PHP’s global variables are a little different from C language. In C language, global variables take active effect in functions unless they are covered by local variables. This can cause problems, as someone might carelessly mutate a global variable. Global variables in PHP must be declared global when used in functions. Examples of applying global

The code is as follows	Copy code
$a = 1; $b = 2; function Sum() { global $a, $b; $b = $a + $b; } Sum(); echo $b; ?>

The output of the above script will be "3". Global variables $a and $b are declared in the function, and all reference variables of any variable will point to the global variables. Let’s look at another example.

The code is as follows	Copy code
$var1 = 1; Function test(){           unset($GLOBALS['var1']); }   test(); echo $var1; ?>

Because $var1 was deleted, nothing is printed.

The code is as follows	Copy code
$var1 = 1; Function test(){ global $var1;             unset($var1);                                }   test(); echo $var1; ?>

Accidentally printed 1. It proves that only the alias | reference is deleted, and the value itself is not changed in any way.

Global problem analysis:
Question: I have defined some variables ($a) in config.inc.php, and include ("config.inc.php") outside the function in other files. These variables $a need to be used inside the function. If Without a statement, echo $a will not print anything. Therefore, global $a is declared, but there are many functions and many variables. You can't declare it like this repeatedly, right? If there is any good solution, please give me some advice.
answer1: First define the constants in config.inc.php: define (constant name, constant value)
Then require 'config.inc.php' in other places where it is needed,
Then you can use this constant directly in this file.
answer2: I also have a way, which is to define an array, such as $x[a], $x, so that you only need to declare global $x.
answer3: I tried your method, but it didn’t work.
answer4: Change your php.ini file.

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