$arr = eval("array('foo'=>'bar');");
// returns null
var_dump($arr);
Can someone please explain why did I get null instead of an array?
4 Answers
You need to return the array.
From the docs:
eval()returnsNULLunlessreturnis called in the evaluated code, in which case the value passed toreturnis returned.
So you need to do:
$arr = eval("return array('foo'=>'bar');");
4 Comments
user1643156
got it, thanks. I didn't pay much attention to the
Return Value, just looked at the example which doesn't use a return.
gen_Eric
@user1643156: Why are you using
eval in the first place? It's slow and insecure. There is probably a better way to do what you want.
Swikrit
@gen_Eric I don't know about user16... but I've got two variables and a selector in between. So
'return $rule["value"]' . $selector . '$target;' will give me true==true but that can change depending on the variables. Is there a better way for this case?gen_Eric
I don't know if it's the best way, but I might do
switch($selector){}, instead of using eval(). Then something like case '==': $ret = $rule['value'] == $target;.Did you mean
eval("\$arr = array('foo'=>'bar');");
var_dump($arr);
1 Comment
gen_Eric
You need to escape the
$. You are in double quotes, so PHP is trying to interpolate $arr, since it doesn't exist you wind up evaling " = array('foo'=>'bar');".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');");
Comments
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
1 Comment
user1643156
thanks, I'm aware of the evil part of eval(). the function is only used by me (admin).
evaling that string? There has to be a better way to do this, such asjson_decodeorunserialize.