Question

如何合并对象中的重复键和一个对象中对象的concat值我有像这样的对象

var object1 = {
    role: "os_and_type", 
    value: "windows"
};
var object2 = {
    role: "os_and_type", 
    value: "Android"
};
var object3 = {
    role: "features", 
    value: "GSM"
};

我怎样才能实现这个目标

new_object = [{
    role: "os_and_type",
    value: ["windows", "android"]         
}, {
    role: "features",
    value: ["GSM"]
}];

Answer 1

你走了：

var object1 = {
    role: "os_and_type", 
    value: "windows"
};
var object2 = {
    role: "os_and_type", 
    value: "Android"
};
var object3 = {
    role: "features", 
    value: "GSM"
};

function convert_objects(){
    var output  = [];
    var temp    = [];
    for(var i = 0; i < arguments.length; i++){  // Loop through all passed arguments (Objects, in this case)
        var obj = arguments[i];                 // Save the current object to a temporary variable.
        if(obj.role && obj.value){              // If the object has a role and a value property
            if(temp.indexOf(obj.role) === -1){  // If the current object's role hasn't been seen before
                temp.push(obj.role);            // Save the index for the current role
                output.push({                   // push a new object to the output,
                    'role':obj.role,
                    'value':[obj.value]         //   but change the value from a string to a array.
                });
            }else{                              // If the current role has been seen before
                output[temp.indexOf(obj.role)].value.push(obj.value); // Save add the value to the array at the proper index
            }
        }
    }
    return output;
}

这样称呼：

convert_objects(object1, object2, object3);

您可以根据需要为函数添加任意数量的对象。

Answer 2

太糟糕了，我们还没有看到任何尝试。

function merge(array) {
    var temp = {},
        groups = [],
        l = array.length,
        i = 0,
        item;
    while (item = array[i++]) {
        if (!temp[item.role]) {
            temp[item.role] = {
                role: item.role,
                value: [item.value]
            };
        } else if (temp[item.role].value.indexOf(item.value) === -1) {
            temp[item.role].value.push(item.value);
        }
    }
    for (var k in temp) {
        groups.push(temp[k]);
    }
    return groups;
}

用法：

var groups = merge([object1, object2, object3]);

Answer 3

这是一个使用地图的版本，以避免反复扫描重复项。还使用一些很酷的方法

它最终也略微变小了。

function merge(objects) {
    var roles = {};
    objects.forEach(function(obj){
        roles[obj.role] = roles[obj.role] || {};
        roles[obj.role][obj.value] = {};
    });
    return Object.keys(roles).map(function(role){
        return {
            role: role, 
            value: Object.keys(roles[role])
        };
    });
}

http://jsfiddle.net/mendesjuan/cD7uu/1/

如何在没有覆盖值的情况下合并javascript中的对象

3 个答案: