Get name of key in key/value pair in JSON using jQuery?

Question

Say I have this JSON:

[
    {
        \"ID\": \"1\",
        \"title\": \"Title 1\",
    },
    {
        \"ID\": \"2\",
        \"title\": \"Title 2\",
             


        

Answer 1

This will give you an array of all the string properties that match across an array of objects. Is that what you are looking for?

$.getJSON('testing.json', function(data) {
    var propertiesThatExistInAll = getPropertiesThatExistInAll(data);
});


var getPropertiesThatExistInAll = function(arr) {
    var properties = $.map(data[0], function (prop, value) {
        return prop;
    });

    var propertiesThatExistInAll = [];

    $.each(properties, function (index, property) {
        var keyExistsInAll = true;

        // skip the first one since we know it has all the properties
        for (var i = 1, len = data.length; i < len; i++) {
            if (!data[i].hasOwnProperty(property)) {
                keyExistsInAll = false;
                break;
            }
        }

        if (keyExistsInAll) {
            propertiesThatExistInAll.push(property);
        }
    });

    return propertiesThatExistInAll;
};

Answer 2

This can probably be made more efficient/concise, but the function below will do it.

var testJson = [ {'oi' : 1, 'arf': 2, 'foo' : 0}, {'oi': 5, 'arf': 7}];

function commonKeys(j)
{

    var fillUp = [];
    for(var i in j[0])
       fillUp.push(i);

    for(var i = 1; i < j.length; i++)
    {
       var cur = j[i]; var curArr = [];
       for (var i in cur) {curArr.push(i)};
       fillUp = fillUp.filter(function(x) {return (curArr.indexOf(x) != -1);});
    }

    return fillUp;
}

alert(commonKeys(testJson)); //oi,arf (not foo)

Answer 3

Something like this, perhaps?

items = [];
for (key in jsonobj) {
    if (!itemExists(items, key)) {
        items[items.length] = key
    }
}

function itemExists(items, value) {
    for (i = 0; i < items.length; i++) {
        if (items[i] == value) {
            return true
        }
    }
    return false;
}

Of course, that will return items that exist in any one of the objects, not that exist in all. It's not entirely clear from your question if this is the solution you want.