如果我在函数之前更改全局变量，为什么不起作用？

Question

接下来的 PHP 代码将名称的首字母大写，其他字母小写：

<?php
$a1 = "WILLIAM";
$a2 = "henry";
$a3 = "gatES";
fix_names();
echo $a1 . " " . $a2 . " " . $a3;
function fix_names()
{
 global $a1;
 global $a2;
 global $a3;
 $a1 = ucfirst(strtolower($a1));
 $a2 = ucfirst(strtolower($a2));
 $a3 = ucfirst(strtolower($a3));
}
/*Output: William Henry Gates*/
?>

如果我在函数外将变量范围更改为全局，它不起作用：

<?php
$a1 = "WILLIAM";
$a2 = "henry";
$a3 = "gatES";
global $a1;
global $a2;
global $a3;
fix_names();
echo $a1 . " " . $a2 . " " . $a3;
function fix_names()
{
$a1 = ucfirst(strtolower($a1));
$a2 = ucfirst(strtolower($a2));
$a3 = ucfirst(strtolower($a3));
}
/*Output: WILLIAM henry gatES*/
?>

请解释为什么它不起作用！

Answer 1

Any variable used inside a function is by default limited to the local function scope. For example:

<?php
    $a = 1; /* global scope */ 

    function test()
    { 
        echo $a; /* reference to local scope variable */ 
    } 

    test();
?>

This script will not produce any output because the echo statement refers to a local version of the $a variable, and it has not been assigned a value within this scope. You may notice that this is a little bit different from the C language in that global variables in C are automatically available to functions unless specifically overridden by a local definition. This can cause some problems in that people may inadvertently change a global variable.

In PHP global variables must be declared global inside a function if they are going to be used in that function.

参考：http://php.net/variables.scope.

因此，为了从函数外部访问变量，您需要在函数 global 内部声明它们，否则您将尝试执行 ucfirst(strtolower($a1)) 在一个甚至不存在于当前（本地）作用域中的变量 $a1 上。

如果我在函数之前更改全局变量，为什么不起作用？

Why doesn't work if I change variables global, before the function?

php

variables

scope

global

function