我有一个包含多个对象的现有数组。使用间隔我想用另一个数组中的值更新现有数组。请参阅下面的(简化)示例。
我已经成功了:
到目前为止我可以看到,我可以使用一些es6函数:),如:
当前数组:
[{
"fan_count": 1234,
"id": "1234567890",
"picture": {
"data": {
"url": "https://scontent.xx.fbcdn.net/v/photo.png"
}
}
},
{
"fan_count": 4321,
"id": "09876543210",
"picture": {
"data": {
"url": "https://scontent.xx.fbcdn.net/v/photo.png"
}
}
}, ...
]
新阵列:
[{
"fan_count": 1239,
"picture": {
"data": {
"url": "https://scontent.xx.fbcdn.net/v/photo.png"
}
"id": "1234567890"
},
{
"fan_count": 4329,
"picture": {
"data": {
"url": "https://scontent.xx.fbcdn.net/v/photo.png"
}
},
"id": "09876543210"
}, ...
]]
答案 0 :(得分:0)
您可以使用array.filter删除新数组中不存在的元素,并且可以遍历新数组以更新当前数组中的同一对象:
var currArr = [
{
"fan_count": 1234,
"id": "1234567890",
},
{
"fan_count": 4321,
"id": "09876543210",
},
{
"fan_count": 4321,
"id": "09876543215",
}
];
var newArr = [
{
"fan_count": 1234,
"id": "1234567890"
},
{
"fan_count": 5555,
"id": "09876543210"
}
];
currArr = currArr.filter(obj => newArr.some(el => el.id === obj.id));
newArr.forEach(obj => {
var found = currArr.find(o => o.id === obj.id);
if (found) {
found.fan_count_new = obj.fan_count;
}
});
console.log(currArr);
后来我意识到最好转过来,将cur_Arr形成currArr添加到新的。这是因为它更容易处理新对象,您不必处理已删除的对象。所以,任何人如何寻找这样的东西:
newArr.forEach(obj => {
var found = currArr.find(o => o.id === obj.id);
if (found) {
console.log('found: ', found.fan_count, obj.fan_count)
obj.fan_count_prev = found.fan_count;
obj.fan_count_diff = Math.round(obj.fan_count - found.fan_count);
}
if (typeof obj.fan_count_prev === "undefined") {
obj.fan_count_prev = obj.fan_count;
obj.fan_count_diff = 0
}
});