使用可变变量填充数组导致错误

Question

我有以下代码：

function filterUsers(array $setOfAllUsers) {
   if (empty($setOfAllUsers)) {
      return array(array(), array());
   }

   $activeUsers   = array();
   $inactiveUsers = array();
   foreach($setOfAllUsers as $userRow) {
       $var = ($userRow['IsActive'] ? '' : 'in') . 'activeUsers';

       $$var[$userRow['CID']]['Label'] = $userRow['UserLabel'];
       // Error happens here ---^

       $$var[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
       );
   }

   return array($activeUsers, $inactiveUsers);
}

我收到以下错误：Warning: Illegal string offset 'Label' in ...

我该如何解决这个问题？我尝试首先像这样定义标签部分：$$var[$userRow['CID']] = array(); $$var[$userRow['CID']]['Label'] = ''; 但没有用。

为了弄清楚我想要达到的目标是：

if ($userRow['IsActive']) {
   $activeUsers[$userRow['CID']]['Label'] = $userRow['UserLabel'];

   $activeUsers[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
   );
} else {
   $inactiveUsers[$userRow['CID']]['Label'] = $userRow['UserLabel'];

   $inactiveUsers[$userRow['CID']]['UserList'][$userRow['UID']] = array(
           'FirstName' => $userRow['FName'],
           'LastName'  => $userRow['LName'],
           ... More data
   );
}

我不想在 if/else 中重复上面的内容，而是想使用 $$

来实现它

Answer 1

尝试使用 ${$var} 而不是 $$var。

编辑

来自PHP手册（http://php.net/manual/en/language.variables.variable.php）：

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

Answer 2

尝试以下操作：

$myUsers = $$var;
$myUsers[...][...] = ...;

使用 $$var[...][...] 的问题是首先计算 $var[...][...]，然后它试图找到名为 $$var[...][...].

的变量

Answer 3

function filterUsers(array $setOfAllUsers) {
   if (empty($setOfAllUsers)) {
      return array(array(), array());
   }

   $users = array(
     'inactiveUsers' => array(), 
     'activeUsers'   => array()
   );
   foreach($setOfAllUsers as $userRow) {
       $status = ($userRow['IsActive'] ? '' : 'in') . 'activeUsers';
       $users[$status][$userRow['CID']] = array();
       $users[$status][$userRow['CID']]['Label'] = $userRow['UserLabel'];

       $users[$status][$userRow['CID']]['UserList'] = array(
           $userRow['UID'] => array(
             'FirstName' => $userRow['FName'],
             'LastName'  => $userRow['LName'],
           )
       );
   }

   return $users;
}

Answer 4

不要像这样使用 var-vars。您运行陷入 PHP 语法错误。即使没有语法错误，您也不应该首先使用 var-vars。它们生成几乎不可能调试的代码。

$x = 'foo';
$foo = array();
$$x[0] = 1;
var_dump($x);  // string(3) "foo"
var_dump($foo); // array(0) { }
$$x[1][2] = 3;
PHP Notice:  Uninitialized string offset: 2 in php shell code on line 1

注意数组如何没有被 $$x[0] 赋值修改，以及 2+ 维赋值如何导致 "undefined string offset"。您的 var-var 没有被视为一个数组 - 它被视为一个字符串，并且失败了，因为它是一个语法故障。您可以将字符串视为数组，但不能使用 var vars:

$bar = 'baz';
$bar[1] = 'q';
echo $bar; // output: bqz

似乎没有办法将 var-var 用作数组，尤其是多维数组。

使用可变变量填充数组导致错误

Using variable variables to populate array results in error

php

variable-variables