未定义函数eval（） - PHP

Question

我想用那样的eval（）函数：

$foo = 'eval'; 
$bar = 'echo 1;'; 
$foo($bar); 

但我发现了一个错误：致命错误：调用未定义的函数eval（）

，由于下面的代码工作它

$foo = 'base64_encode'; 
$bar = 'foobar'; 
echo $foo($bar); 

谁能帮助奇怪？

Answer 1

从eval documentation：

Note: Because this is a language construct and not a function, it cannot be called using variable functions .

继在注意链接，你还会发现：

Variable functions won't work with language constructs such as echo , print , unset() , isset() , empty() , include , require and the like. Utilize wrapper functions to make use of any of these constructs as variable functions.