A Query In the String

Question

I didn't know what to write about the title of the problem:) I want to add a string to the query but I get internal 500 error

$arr = json_decode($json, true);
$name = "Peter";
echo $arr[$name];  // Output: 65

It's correct but,

$obj = json_decode($json);
echo $obj->$name;

not working.

Of course this is not normal code but caybe this could be a method?

A 500 error means that there's an internal server error. Check your servers error log (or turn display_errors on) to see what the actual error message is. Have you even defined `$name` in the second code? — M. Eriksson, Sep 18 '18 at 09:44
I don't understand the question. Your first code is correct. Your second code isn't. So don't use the second code. — Quentin, Sep 18 '18 at 09:44

score 0 · Answer 1 · answered Sep 18 '18 at 10:32

Given a json string {"name":"foo"} You can display name by writing

$json = '{"name":"foo"}';
$data = json_decode($json);
echo $data->name ;

Since name is a property of object $data .

if you'd write $data->$name , it would mean you have a $name variable somewhere you wanted to refer.

1 Answers1