Why string variable in JSON does not work?

Question

I have got the code:

$(document).ready(function() {

  var adress = [];
  var locations = ' [';
  $('.office p').each(function(el) {
    adress[el] = $(this).text();
    locations = locations + '{address:\''+adress[el]+'\', data: \'0\', options:{icon: "http://selectner.com/img/bullet.png"}},';
  });

  locations = locations + ']';
        console.log(locations);


     $('#tablink').click(function (e) {

     $('#map').gmap3({
      map:{
         options:{
           center:[51.4675954,0.048876],
           zoom: 2,
           scrollwheel: true,
          draggable: true, 
          }
      },
     marker:{
         values : locations,
         options:{
          draggable: true
        },
        events:{
                    }
    }
   });
  });
});

If I will copy console.log results and paste it into JSON (in gmap3 function - locations place) - all will work, but now it is not working.

I think, that I must to use JSON.parse(), but it write error to me:

SyntaxError: JSON.parse: expected property name or '}' at line 1 column 3 of the JSON data

How I can solve this problem?

Stop trying to cobble together JSON yourself – create proper data structures out of arrays/objects, and then have those encoded as JSON when needed using the browser’s native methods (or a framework/polyfill if necessary). — C3roe
– C3roe, Commented Mar 14, 2015 at 19:59
@LShetty: I rebuild with double squares, but it still don't work. Now code console.log is [{"address":'Toronto, SCARBOROUGH JUNCTIO', "data": '0', "options":{"icon": 'http://selectner.com/img/bullet.png'}},{"address":'Moscow, tverskaya street, 18', "data": '0', "options":{"icon": 'http://selectner.com/img/bullet.png'}},{"address":'Leipzig, SÜDVORSTADT', "data": '0', "options":{"icon": 'http://selectner.com/img/bullet.png'}},] — Сергей Илларионов
– Сергей Илларионов, Commented Mar 14, 2015 at 20:16
You still have invalid JSON over there. That's why it doesn't work. Maybe you can try simply using a JSON serializer instead of concatenating those strings. — Darin Dimitrov
– Darin Dimitrov, Commented Mar 14, 2015 at 20:18

Darin Dimitrov · Accepted Answer · 2015-03-14 20:00:42Z

1

Don't build JSON using string concatenation operations. Just don't. Never, ever.

You may try using a simple JSON serializer which is designed for this purpose so that you never get invalid JSON:

var addresses = [];
$('.office p').each(function() {
    addresses.push({
        address: $(this).text(),
        data: 0,
        options: {
            icon: 'http://selectner.com/img/bullet.png'
        }
    });
});

var locations = JSON.stringify(addresses);
console.log(locations);
// At this stage it is guaranteed that the addresses variable will
// contain valid JSON string

answered Mar 14, 2015 at 20:00

Darin Dimitrov

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5 Comments

Michelangelo Over a year ago

I agree with you, but OP is using a plugin and this is how the documentation says how to do it :/

Darin Dimitrov Over a year ago

Oh that's really weird. Could you point to this documentation? Maybe the authors of it should be notified about the mistake they did. I mean if a documentation is showing an example where JSON is being built with string concatenation operations, then, well, OMG.

Darin Dimitrov Over a year ago

I am not sure where in this example a JSON is being built using string concatenations. They seem to be building some DOM elements using string concatenations (which is also bad) but not JSON.

Darin Dimitrov Over a year ago

The OP's problem is that he is building JSON using string concatenation operations. Chances of getting this right are close to 0.

Michelangelo Over a year ago

Yep I see now, sorry about the confusion. Well also trying to build your own JSON, we all have been there :)

Joseph · Accepted Answer · 2015-03-14 19:59:31Z

0

That's because your "string" isn't valid JSON. Read more about the structure of JSON here: http://json.org/

Also, you're already in JavaScript, why are you building JSON when you can already use objects?

answered Mar 14, 2015 at 19:59

Joseph

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