Create an array from object data

Question

I have this type of data in an object in my project.

{
    "name": "Travel",
    "map": [
        {
            "name": "Montreal",
            "lat": "45.498649",
            "lng": "-73.492729"
        },
        {
            "name": "Brossard",
            "lat": "45.466667",
            "lng": "-73.450000"
        }
    ]
}

How can I create this type of structure code to show my Google Map ?

var locations = [
  {
    name: "Montreal",
    latlng: new google.maps.LatLng(45.498649, -73.492729)
  },
  {
    name: "Brossard",
    latlng: new google.maps.LatLng(45.466667, -73.450000)
  }
];

Thanks a lot.

ifm · Accepted Answer · 2014-11-22 14:16:03Z

3

It's more simple if you use 'forEach':

json.map.forEach(function(val){
    locations.push({
        name: val.name, 
        latlng: new google.maps.LatLng(val.lat, val.lng)
    });
});

edited Nov 22, 2014 at 14:16

answered Nov 22, 2014 at 14:01

ifm

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1 Comment

user4158797 Over a year ago

It do not conserves my initial wanted structure with the new google.maps.LatLng part.

Amir Popovich · Accepted Answer · 2014-11-22 15:18:23Z

2

You were close:

for (i = 0; i < dmap.length; i++) {
    locations.push({name: dmap[i].name, lat: dmap[i].lat,lng:dmap[i].lng});
}

Another way is doing it with Array.map:

dmap.map(function(item){
    return {name: item.name, lat: item.lat, lng:item.lng};
});

JSFIDDLE Example.

Since You Editted this is the relevant code:

for (i = 0; i < dmap.length; i++) {
    locations.push({name: val.name, 
        latlng: new google.maps.LatLng(val.lat, val.lng)
    });
}

Or

dmap.map(function(item){
    return {name: val.name, 
        latlng: new google.maps.LatLng(val.lat, val.lng)
    };
});

edited Nov 22, 2014 at 15:18

answered Nov 22, 2014 at 13:52

Amir Popovich

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3 Comments

user4158797 Over a year ago

Good but how can I include the new google.maps.LatLng part ?

Amir Popovich Over a year ago

thats an external class that probally comes with adding the google maps js file.

user4158797 Over a year ago

No I really need this part, if not, it will not work.

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