How do I loop this using jQuery

Question

I am not lying I am a beginner. So here is my issue. I am receiving a json object in an ajax request and would like to loop its content. I need to create the loop and gather data within the variable "pins."

function(data) {

                   var pins = [
                                   {
                                        "lat":      data.lat,
                                        "lon":      data.lon,
                                        "imageURL": data.imageURL,
                                        "title":    data.title,
                                        "subTitle": data.subTitle,
                                        "pinColor": data.pinColor,
                                        "selected": data.selected,
                                        "pinImage": data.pinImage,
                                        "index":    data.index
                                    }
                               ];
         }

Data is an object and I would like to loop every single element using jQuery.

Thanks

I am data from a json object in an Ajax request. ... success: function(data) {} etc. — user2415277
– user2415277, Commented May 27, 2013 at 19:41
Are you looking to loop over data to generate pins or looking to loop over pins? — Kyle
– Kyle, Commented May 27, 2013 at 19:42
If so, is data an array of objects that each have these attributes? — Kyle
– Kyle, Commented May 27, 2013 at 19:45
Yes data is the json object with all the pins. And I need to loop within my object. I need the brakets otherwise it doesn't work. — user2415277
– user2415277, Commented May 27, 2013 at 19:53
What are you talking about? { and } means it contains an object and [ and ] means it contains an array. Hence pins is an array of objects. — h2ooooooo
– h2ooooooo, Commented May 27, 2013 at 20:17

h2ooooooo · Accepted Answer · 2013-05-27 19:41:01Z

1

This is a javascript array with an object inside, and has nothing to do with jQuery (nor should you use jQuery to loop through it).

for (var i = 0; i < pins.length; i++) {
    console.log(pins[i].lat);
    console.log(pins[i].lon);
    console.log(pins[i].imageURL);
    //etc
}

answered May 27, 2013 at 19:41

h2ooooooo

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3 Comments

DontVoteMeDown Over a year ago

He can use jQuery, if he wants.

Blender Over a year ago

jQuery provides convenience methods to loop through objects, so there's no reason not to use them aside from the performance impact.

user2415277 Over a year ago

Your answer does not work. I would like a full answer where the brakets are kept since I need them to create a pin. Something like var pins = [ $.each etc...

Tomas Kirda · Accepted Answer · 2013-05-27 19:42:56Z

1

$.each({ a: 1, b: 2}, function(key, value){
    console.log(key, value);
});

answered May 27, 2013 at 19:42

Tomas Kirda

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Comments

Kyle · Accepted Answer · 2013-05-27 20:17:41Z

0

The following should work to iterate over data to collect the desired attributes from each object in the array and construct pins.

function(data) {
    var pins = [];

    for (var i=0; i < data.length; i++) {
        var new_pin = {
            "lat":      data[i].lat,
            "lon":      data[i].lon,
            "imageURL": data[i].imageURL,
            "title":    data[i].title,
            "subTitle": data[i].subTitle,
            "pinColor": data[i].pinColor,
            "selected": data[i].selected,
            "pinImage": data[i].pinImage,
            "index":    data[i].index
        };

        pins.push(new_pin);
    }
}

answered May 27, 2013 at 20:17

Kyle

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3 Comments

h2ooooooo Over a year ago

If this is the case, what exactly would be the difference between this and var pins = data;?

Kyle Over a year ago

There would only be a difference if the OP doesn't want to collect all of the attributes... I don't exactly know what the OP is desiring here. Perhaps data isn't structured quite this simply.

Kyle Over a year ago

If that is the case, then it may very well be functionally equivalent to setting var pins = data as h2ooooooo said. @user2415277, if you believe this answer is the solution, please accept it as such. Thank you.

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