SyntaxError: JSON Parse error: Unexpected identifier "object"

Question

Please help me to understand what's wrong. I want to parse JSON reply as object.

PHP process.php code:

<?php
    $return = array();
        array_push($return['amount']="$amount");
        array_push($return['fee']="$fee");
        array_push($return['total']="$total");
    echo json_encode($return);
?>

Returns JSON string:

{"amount":"350","fee":"0","total":"350"}

JS (jquery) code:

$.getJSON("process.php?amount="+amount, function(data,status) {
   var obj = $.parseJSON(data);
   alert (obj.amount);
});

I receive error:

SyntaxError: JSON Parse error: Unexpected identifier "object"

BUT! When I try to insert result instead data (but insert ' quotes left/right):

var obj = $.parseJSON('{"amount":"350","fee":"0","total":"350"}');

And I see alert = 350. So, it's working good.

I try to make something like that:

var jsonreply = "'"+data+"'";
var obj = $.parseJSON(jsonreply);

But received below error:

SyntaxError: JSON Parse error: Single quotes (') are not allowed in JSON

when datatype is json recieved in jQuery ajax methods you don't need to use $.parseJSON on reponse, it has already been converted internally. To prove this try console.log( $.type(data)) in your success callback — charlietfl
– charlietfl, Commented Oct 28, 2013 at 1:46

icktoofay · Accepted Answer · 2013-10-28 01:48:16Z

7

getJSON parses the JSON for you — calling $.parseJSON will convert the object into the string [object Object] and then try to parse that, giving you an error. Just omit the $.parseJSON call and use data directly.

Furthermore, I should note that the calls to array_push are strange and unnecessary. array_push usually takes an array and a value to push on to it, but (for example) in your first line you're setting $return['amount'] to "$amount" and then passing $return['amount'] to array_push, which does nothing at best and might give you a warning or error at worst. You'd get the exact same behavior if you did this:

$return['amount']="$amount";
$return['fee']="$fee";
$return['total']="$total";

Then you might also realize that the quotes around, say, "$amount" are unnecessary, and you could actually do this:

$return['amount']=$amount;
$return['fee']=$fee;
$return['total']=$total;

Finally, you can actually condense all five lines using some special array syntax very easily:

echo json_encode(array(
    'amount' => $amount,
    'fee' => $fee,
    'total' => $total
));

This is quite a bit nicer if I do say so myself.

edited Oct 28, 2013 at 1:48

answered Oct 28, 2013 at 1:40

icktoofay

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3 Comments

andrewvnice Over a year ago

Also, why are you using array_push in your PHP script? If you would like to add values to an array, you can do so more easily like so: $return['amount']=$amount;

icktoofay Over a year ago

@mazinsights: Your comment might better belong on the question myself, as I wasn't the one who wrote the code in the first place! It's a good suggestion, though, so I've added that (and a bit more, actually) to my answer.

Vasvas Over a year ago

In example above I make a simple code. Really, I used array, but when I array_push one array to another I received not good results for json. Something like ["0",{"key0":"value","key1":"value"},{"key0":"value"}]

Vicky Gonsalves · Accepted Answer · 2013-10-28 01:49:17Z

1

Actually u dont need to parse it. U can directly access it

$.getJSON("process.php?amount="+amount, function(data,status) {
 alert (data.amount); 
});

edited Oct 28, 2013 at 1:49

answered Oct 28, 2013 at 1:41

Vicky Gonsalves

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Comments

lolrs2013 · Accepted Answer · 2013-10-28 01:39:52Z

0

It seems like your error is here:

var jsonreply = "'"+data+"'";

Try to escape those ' with "\". Like

var jsonreply = "\'"+data+"\'";

answered Oct 28, 2013 at 1:39

lolrs2013

518 bronze badges

1 Comment

icktoofay Over a year ago

' is not a special character within double quotes; backslashing it won't do anything. Perhaps you meant "\\'", but then you're still generating invalid JSON.

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