how to copy certain objects of an existing json array of objects to a new json array

Question

My JSON array of objects is:

[{
  "stateCd": "IL",
  "value": "Illinois",
  "selected": "false"
},
{
  "stateCd": "CA",
  "value": "California",
  "selected": "true"
},
{
  "stateCd": "NY",
  "value": "New york",
  "selected": "false"
}]

I want to create a new JSON array of objects which should only contain those objects from above array where selected:false.

I've tried concat:

angular.forEach($scope.oldJsonArr, function (value, index) {
    if($scope.oldJsonArr[index].selected=="false"){
        $scope.newJsonArr.concat($scope.oldJsonArr[index]);
    }
});

But this is returning the newJsonArr as undefined. Any help is appreciated!

you can use where in underscore.js if you are used underscore or use simple filter ! — Venkat.R
– Venkat.R, Commented Jan 20, 2016 at 15:07
Added reference as well for your better understanding. checkout ! — Venkat.R
– Venkat.R, Commented Jan 20, 2016 at 15:23

k102 · Accepted Answer · 2016-01-20 14:56:12Z

2

Here is the code:

$scope.oldJsonArr = [{"stateCd": "IL", "value": "Illinois", "selected": "false"}, {
        "stateCd": "CA",
        "value": "California",
        "selected": "true"
    }, {"stateCd": "NY", "value": "New york", "selected": "false"}];

    $scope.newJsonArr = [];

    angular.forEach($scope.oldJsonArr, function (value, index) {
        if ($scope.oldJsonArr[index].selected == "false") {
            $scope.newJsonArr.push($scope.oldJsonArr[index]);
        }
    });

    console.info($scope.newJsonArr);

You have to define variables, or they will remain undefined
You had to use .push() method

answered Jan 20, 2016 at 14:56

k102

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Comments

forevergenin · Accepted Answer · 2016-01-20 14:59:51Z

0

You can use Array.prototype.filter() method.

var falseFilter = function(value) {
    return value.selected == "false";
};

var newJsonArray = oldJsonArray.filter(falseFilter);

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter

answered Jan 20, 2016 at 14:59

forevergenin

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Comments

Venkat.R · Accepted Answer · 2016-01-20 15:21:21Z

Use Filter from JavaScript rather using library utility function.

var people = [
  { firstname:"Micro",  hasSocialNetworkSite: false, lastname:"Soft", site:"http://microsoft.com" },
  { firstname:"Face",   hasSocialNetworkSite: true, lastname:"Book", site:"http://facebook.com" },
  { firstname:"Go",     hasSocialNetworkSite: true, lastname:"ogle", site:"http://google.com" },
  { firstname:"Twit",   hasSocialNetworkSite: true, lastname:"Ter", site:"http://twitter.com" },
  { firstname:"App",    hasSocialNetworkSite: false, lastname:"Le",   site:"http://apple.com" },
  { firstname:"Master", hasSocialNetworkSite: false, lastname:"Card", site:"http://mastercard.com" }
];

var filteredResult = people.filter(function(v) {
  return v.hasSocialNetworkSite == true; // Filter out the appropriate one
});

document.write("Social Networking Site<br>");

for (var i in filteredResult) {
  document.write(filteredResult[i].site + "<br>")
}

Reference to know in detail:
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
https://msdn.microsoft.com/en-us/library/ff679973(v=vs.94).aspx
http://www.w3schools.com/jsref/jsref_filter.asp

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