函数名称必须是一个字符串的PHP？

Question

经过几个小时的谷歌搜索和搜索答案，我放弃了，并非常沮丧与PHP，没有支持在那里的基本错误。无论如何，这是我的功能，我做错了什么？

$accessToken = function(){ 
     $array = array(
      "foo" => "bar", 
      "bar" => "foo" 
     ); 

     $obj = json_encode($array); 
     $obj = json_decode($obj); 

     return $obj->foo; 
}; 

$getInfo = function(){ 

     $code = $accessToken(); 
     return $code; 


}; 

$getInfo(); 

我得到错误

Notice: Undefined variable: accessToken in C:\inetpub\wwwroot\mysite\lab\cfhttp.php on line 43

Fatal error: Function name must be a string in C:\inetpub\wwwroot\mysite\lab\cfhttp.php on line 43

Answer 1

use()可以带可变本域内的范围

$getInfo = function() use ($accessToken) { 
    $code = $accessToken(); 
    return $code; 
}; 

https://3v4l.org/BOtnf

Answer 2

$accessToken心不是内部$getInfo()

$accessToken = function(){ 
     $array = array(
      "foo" => "bar", 
      "bar" => "foo" 
     ); 

     $obj = json_encode($array); 
     $obj = json_decode($obj); 

     return $obj->foo; 
}; 

$getInfo = function($accessTokenFunction){ 

     $code = $accessTokenFunction(); 
     return $code; 


}; 

$getInfo($accessToken); 

Answer 3

在PHP中，您通常以另一种方式定义函数。请尝试以下操作：

function accessToken() { 
    $array = array(
      "foo" => "bar", 
      "bar" => "foo" 
    ); 

    $obj = json_encode($array); 
    $obj = json_decode($obj); 

    return $obj->foo; 
} 

function getInfo() { 
    $code = accessToken(); 
    return $code; 
} 

getInfo(); 

您也可以到这里看看http://php.net/manual/en/functions.variable-functions.php和http://php.net/manual/en/functions.user-defined.php