[英]Adding the values from an object to an array of objects without overriding existing key values
你怎么做的:
var array = [{key: [3]}, {key1: [3]}, {key1: [3]}]
var object = {key1: [3], key2: [3]};
對此:
{key: [3], key1: [9], key2: [3]}
所有“key”都是像“LQVjUacPgK”這樣的userIds,如下面的obj示例所示。
[N] =是N個對象的數組,每個對象在其中具有大約10個鍵值對。
N = {obj, obj, obj};
obj = {_account: "JDQEPxoy3ktZRP9VEzAMtXLa7rXXedhQ4bARq"
_id: "oQER3vznDwikxm1wdLzJFdVjKL6XomcORMxDL"
amount: 170
category: Array[2]
category_id: "21003000"
date: "2015-06-09"Object
type: Object
userId: "LQVjUacPgK"}
現在我這樣做:
var test = _.reduce(_.flatten(array.concat([object])),function(a,b){
return _.extend(a, b);
});
}
};
而得到這個結果:
console.log(test)//{key: [3], key1: [3], key2: [3]}
要清楚,問題是key1在所有對象之間具有不同的值。 我想保留兩者的值,以便key1:[9] 。
這不是一個下划線的答案,但基本上我不會為此使用reduce操作,而是做一個簡單的for-each:
var array = [{key: [3]}, {key1: [3]}, {key1: [3]}] var object = {key1: [3], key2: [3]}; array.forEach(function(current) { Object.keys(current).forEach(function(name) { // object[name] = [((object[name] || [])[0] || 0) + current[name][0]]; object[name] = (object[name] || []).concat(current[name]); }); }); console.log(JSON.stringify(object)); // {"key1":[3,3,3],"key2":[3],"key":[3]}
類似傑克的答案(也非下划線),但它創建了一個新的對象,它不修改現有對象的對象:
var array = [{key: [3]}, {key1: [3]}, {key1: [3]}]
var object = {key1: [3], key2: [3]};
var x = array.concat([object]).reduce(function(prev, curr) {
Object.keys(curr).forEach(function(key){
if (prev.hasOwnProperty(key)) {
prev[key][0] += curr[key][0];
} else {
prev[key] = curr[key];
}
});
return prev;
},{});
console.log(JSON.stringify(x)); // {"key":[3],"key1":[9],"key2":[3]}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.