PHP eval(array_as_string) returns null

问题

$arr = eval("array('foo'=>'bar');");

// returns null
var_dump($arr);

Can someone please explain why did I get null instead of an array?

回答1:

You need to return the array.

From the docs:

eval() returns NULL unless return is called in the evaluated code, in which case the value passed to return is returned.

So you need to do:

$arr = eval("return array('foo'=>'bar');");

回答2:

Did you mean

eval("\$arr = array('foo'=>'bar');"); 

var_dump($arr);

回答3:

The eval function executes the php code given to it. As your code returns nothing, it gives null. You need to return the array and store it in a variable like,

$arr = eval("return array('foo'=>'bar');");

回答4:

First of all, eval is highly discouraged as explained in the manual.

Also, you should be doing something like $arr = eval("return array('foo'=>'bar');"); ie. initialising $arr with the eval function. See it in action here

来源：https://stackoverflow.com/questions/12976254/php-evalarray-as-string-returns-null