json返回字符串而不是对象

Question

i have

 $age = implode(',', $wage);   // which is object return:  [1,4],[7,11],[15,11]
  $ww = json_encode($age);

and then i retrieve it here

    var age = JSON.parse(<?php echo json_encode($ww); ?>); 

so if i make

   alert(typeof(<?php echo $age; ?>))   // object
   alert(typeof(age))                   //string

in my case JSON.parse retuned as string.

how can i let json return as object?

EDIT:

 var age = JSON.parse(<?php echo $ww; ?>); // didnt work , its something syntax error

Answer 1

implode returns a string, so it is only natural that json_encode encodes it as such. It does not recognize already JSON-like data passed as a string.

If you want to get an object, you have to pass an associative array to json_encode :

$foo = array(
    1 => 4,
    7 => 11,
    15 => 11
);

echo json_encode($foo); // {1:4,7:11,15:11}

With so little info about what $wage looks like before it's imploded, it's hard to tell exactly what you want to get. How is that structure ( [1,4],[7,11],[15,11] ) an object? Is the first element of each tuple a key? That's what I assumed with my example, but it might be off.

Answer 2

a. You get a syntax error because you need to enclose the string within quotes, like so:

var age = JSON.parse("<?php echo $ww; ?>");

b. Moreover, you don't need JSON.parse . You can simply echo the php var after it was already json_encoded in the server side:

var age = <?php echo $ww; ?>;

JSON.parse is there to convert a JavaScript string to an object. In the case of PHP string, once it is built as JSON, echoing it in the right place is equivalent to coding it yourself.

Answer 3

var age = [<?php echo $age; ?>];