Need to access a variable outside of a jQuery each loop

Question

So I have a json string that Looks like this

var jsonString = '[{"name":"John Smith","uid":1234,{"name":"Susie Something","uid":4567"}]';

I'm making a function that will return all the UIDs seperated by commas. So for example: 1234,4567.

What I do now is this

function getUIDS(json){
var uids = '';
parsed = JSON.parse(json);

$.each(friends, function() {
        uids = uids + this.uid.toString() + ',';
 });
 reuturn uids;

}

However, nothing returns. How can I access a variable that was created by a jquery thread, outside of the jquery thread.

Maybe nothing returns because of reuturn? Furthermore, check the JSONstrings. Third, you should parse the JSON before you can use it as an object. — KooiInc
– KooiInc, Commented Jun 27, 2012 at 7:36
Fourth, you're iterating over friends (which is not defined), not parsed. — penartur
– penartur, Commented Jun 27, 2012 at 7:40

Adil · Accepted Answer · 2012-06-27 07:38:08Z

1

Pass parsed instead of friends, if this is complete code and friends is not defined anywhere

function getUIDS(json){
var uids = '';
parsed = JSON.parse(json);

$.each(parsed, function() {
        uids = uids + this.uid.toString() + ',';
 });
 reuturn uids;

}

answered Jun 27, 2012 at 7:38

Adil

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Comments

Per Salbark · Accepted Answer · 2012-06-27 07:40:37Z

1

Try fixing your json:

var jsonString = '[{"name":"John Smith","uid":1234},{"name":"Susie Something","uid":4567"}]';

And change your code slightly:

function getUIDS(json) {
   var uids = '',
       friends = JSON.parse(json);

   $.each(friends, function() {
       uids += this.uid + ',';
   });

   return uids;
}

The important parts are looping over friends and spelling return correctly.

You should also consider using join as suggested by someone. That will make sure you don't get a trailing "," at the end of your uids string (unless that's what you want).

answered Jun 27, 2012 at 7:40

Per Salbark

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Comments

Blaster · Accepted Answer · 2012-06-27 07:43:55Z

0

Try:

function getUIDS(json){
  var uids = [];
  var friends = JSON.parse(json);

  $.each(friends, function(i) {
     uids.push(friends[i].uid);
  });

  return uids.join(',');
}

Problems in your code were: (See code comments)

function getUIDS(json){
var uids = '';
parsed = JSON.parse(json);

$.each(friends, function() { // using friends instead of parsed
        uids = uids + this.uid.toString() + ','; // using this here
 });
 reuturn uids; // spelling mistake for return keyword

}

edited Jun 27, 2012 at 7:43

answered Jun 27, 2012 at 7:38

Blaster

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