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Most people will think that global and $GLOBALS[] are only different in the way they are written, but in fact this is not the case. Let's take a look at the differences between them.
According to the official explanation,
$GLOBALS['var'] is the external global variable $var itself.
global $var is a reference or pointer of the same name as external $var. (Error: It is an alias reference, not a pointer!!!)
Give an example:
Usage of php $GAOBAL[]:
01 <?php
02 $var1 = 1;
03 $var2 = 2;
04 function test() {
05 $GLOBALS['var2'] = &$GLOBALS['var1'];
06 }
07
08 test();
09 echo $var2;
10 ?>The normal print result is 1
Use of php global:
01 <?php
02 $var1 = 1;
03 $var2 = 2;
04
05 function test(){
06 global $var1, $var2;
07 $var2 = &$var1;
08 echo $var2;
09 $var2 = 'qianyunlai.com';
10 }
11
12 test(); // 输出 1
13 echo $var2; // 输出 2
14 echo $var1; // 输出 qianyunlai.com
15 ?>$var1 and $va2 in the test() function are both local variables. They just add the global keyword and refer to the global variable $ respectively. var1, $va2. When $var2 = &$var1;, the local variable $var2 no longer points to the global variable $val2, but redirects to the global variable $var1. In other words, the change of the local variable $var2 does not change. It will then affect the global variable $val2, which will affect the redirected global variable $val1.
Let’s look at another example.
1 <?php
2 $var1 = 1;
3 function test(){
4 unset($GLOBALS['var1']);
5 }
6 test();
7 echo $var1;
8 ?>Because $var1 was deleted, nothing was printed.
01 <?php
02 $var1 = 1;
03
04 function test(){
05 global $var1;
06 unset($var1);
07 }
08
09 test();
10 echo $var1;
11 ?>Accidentally printed 1.
It proves that only the alias is deleted, and the reference of $GLOBALS['var'] has not been changed in any way.
Understand?
In other words, global $var is actually $var = &$GLOBALS['var']. It's just an alias for calling an external variable.
global and $GLOBALS in PHP are not only written differently, but the difference between the two is still very big. You need to pay attention to it in practical applications!
Look at the following example first:
1 <?php
2 $id = 1;
3 function test() {
4 global $id;
5 unset($id);
6 }
7 test();
8 echo($id); // 输出 1
9 ?>Reference positioning
Many PHP syntax structures are implemented through the reference mechanism, so everything above about reference binding also applies to these structures. Some constructs, such as pass-by-reference and return-by-reference, have already been mentioned above. Other structures that use references are:
When you declare a variable with global $var, you actually create a reference to the global variable. That is the same as doing:
01 <?php 02 $GLOBALS["var1"] = 1; 03 $var = &$GLOBALS["var1"]; 04 unset($var); 05 echo $GLOBALS['var1']; //输出1 06 //############################################ 07 $GLOBALS["var1"] = 1; 08 $var = &$GLOBALS["var1"]; 09 unset($GLOBALS['var1']); 10 echo $var; //输出1 11 //############################################ 12 //如果写成如下，则会出错 13 $GLOBALS["var"] = 1; 14 $var = &$GLOBALS["var"]; 15 unset($GLOBALS['var']); 16 echo $var; //脚本没法执行 17 //########################################### 18 ?>
This means that, for example, unset $var will not unset a global variable.
unset just breaks the binding between the variable name and the variable content. This does not mean that the variable contents are destroyed.
Return false when using isset($var). $this In a method of an object, $this is always a reference to the object that calls it.
If a reference is assigned to a variable declared as global inside a function, the reference is only visible inside the function.
This can be avoided by using the $GLOBALS array.
Example to reference global variables within a function:
01 <?php
02 $var1 = "Example variable";
03 $var2 = "";
04
05 function global_references($use_globals) {
06 global $var1, $var2;
07 if (!$use_globals) {
08 $var2 = &$var1; // visible only inside the function
09 } else {
10 $GLOBALS["var2"] = &$var1; // visible also in global context
11 }
12 }
13
14 global_references(false);
15 echo "var2 is set to '$var2'\n"; // var2 is set to ''
16 global_references(true);
17 echo "var2 is set to '$var2'\n"; // var2 is set to 'Example variable'
18 ?>Treat global $var; as the abbreviation of $var = &$GLOBALS['var'];. So if you assign another reference to $var, you only change the reference to the local variable.
As mentioned before, references are not pointers. This means that the following construct will not have the expected effect:
1 <?php
2 $bar = 3;
3 function foo(&$var) {
4 $GLOBALS["baz"] = 5;
5 $var = &$GLOBALS["baz"];
6 }
7 foo($bar);
8 echo $bar;//输出3
9 ?>This will cause the $var variable in the foo function to be bound to $bar when the function is called, but then re-bound to $bar $GLOBALS["baz"] above.
It is not possible to bind $bar to other variables in the function call scope through the reference mechanism, because there is no variable $bar in function foo (it is represented as $var, but $var only has variables content without calling the name-to-value binding in the symbol table). You can use reference returns to reference variables selected by the function.
Quoting the explanation of $GLOBALS in the PHP manual:
Global variable: $GLOBALS, note: $GLOBALS is applicable in PHP 3.0.0 and later versions.
An array consisting of all defined global variables. The variable name is the index into the array. This is a "superglobal", or can be described as an automatic global variable.
That is to say, $var1 and $GLOBALS['var1'] in the above code refer to the same variable, not 2 different variables!
If a reference is assigned to a variable declared as global inside a function, the reference is only visible inside the function. This can be avoided by using the $GLOBALS array.
We all know that the variables generated by functions in PHP are private variables of the function, so the variables generated by the global keyword certainly cannot escape this rule. global generates an alias in the function that points to the external variable of the function. variables, rather than real variables external to the function. Once the pointing address of the alias variable is changed, some unexpected situations will occur. $GLOBALS[] is indeed called an external variable, and it will always remain consistent inside and outside the function.
01 <?php
02 $a = 1;
03 $b = 2;
04 function Sum() {
05 global $a, $b;
06 $b = $a + $b;
07 }
08 Sum();
09 echo $b;
10 ?>The output 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.
Why is it not 2? Doesn’t it have no effect outside the function? Please note that $b is not modified by reference in the function, but the modified $b points to the value of physical memory, so the external input is 3.
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PHP Global and $GLOBALS variable scope and differences
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