我正在尝试解决问题，但无法弄清楚如何做到这一点。 我有两个数组：
var arr1 = [
{Week: "WK1", CreatedTickets: 1}, {Week: "WK3", CreatedTickets: 12},
{Week: "WK2", CreatedTickets: 3}, {Week: "WK5", CreatedTickets: 5}
];
var arr2 = [
{Week: "WK1", ClosedTickets: 4}, {Week: "WK6", ClosedTickets: 40},
{Week: "WK3", ClosedTickets: 33}, {Week: "WK2", ClosedTickets: 2}
];
预期的产出将是和周也应该是有序的。
var output = [
{Week: "WK1", CreatedTickets: 1, ClosedTickets: 4},
{Week: "WK2", CreatedTickets: 3, ClosedTickets: 2},
{Week: "WK3", CreatedTickets: 12, ClosedTickets: 33},
{Week: "WK5", CreatedTickets: 5, ClosedTickets: 0},
{Week: "WK6", CreatedTickets: 0, ClosedTickets: 33}
];
请帮我解决这个问题。非常感谢。
杰米
答案 0 :(得分：0)
要达到预期效果，请使用以下选项，
获得准确输出的两个循环
以下面的格式创建对象的第一个循环
{
“WK1”：{
“CreatedTickets”：1，
“ClosedTickets”：4
}，
“WK2”：{
“CreatedTickets”：3，
“ClosedTickets”：2
}，
“WK3”：{
“CreatedTickets”：12，
“ClosedTickets”：33
}，
“WK5”：{
“CreatedTickets”：5，
“ClosedTickets”：0
}，
“WK6”：{
“CreatedTickets”：0，
“ClosedTickets”：33
}
}
用于将上述临时对象转换为输出数组
的第二个循环 [
{Week：“WK1”，CreatedTickets：1，ClosedTickets：4}，
{Week：“WK2”，CreatedTickets：3，ClosedTickets：2}，
{Week：“WK3”，CreatedTickets：12，ClosedTickets：33}，
{Week：“WK5”，CreatedTickets：5，ClosedTickets：0}，
{Week：“WK6”，CreatedTickets：0，ClosedTickets：33}
];
JS
var output =[];//output array
var temp = {};//temp object
var arr1 = [
{Week: "WK1", CreatedTickets: 1}, {Week: "WK3", CreatedTickets: 12},
{Week: "WK2", CreatedTickets: 3}, {Week: "WK5", CreatedTickets: 5}
];
var arr2 = [
{Week: "WK1", ClosedTickets: 4}, {Week: "WK6", ClosedTickets: 40},
{Week: "WK3", ClosedTickets: 33}, {Week: "WK2", ClosedTickets: 2}
];
var concatArr = arr1.concat(arr2)
for(var o of concatArr){
if(Object.keys(temp).indexOf(o.Week)==-1){
temp[o.Week] ={}
o.ClosedTickets?temp[o.Week].ClosedTickets = o.ClosedTickets:temp[o.Week].ClosedTickets = 0
o.CreatedTickets?temp[o.Week].CreatedTickets = o.CreatedTickets:temp[o.Week].CreatedTickets = 0
}else{
o.ClosedTickets?temp[o.Week].ClosedTickets = temp[o.Week].ClosedTickets + o.ClosedTickets:temp[o.Week].ClosedTickets
o.CreatedTickets?temp[o.Week].CreatedTickets = temp[o.Week].CreatedTickets + o.CreatedTickets:temp[o.Week].CreatedTickets
}
}
for( var key of Object.keys(temp)){
output.push({
Week:key,
CreatedTickets:temp[key].CreatedTickets,
ClosedTickets:temp[key].ClosedTickets
})
}
console.log(output)
答案 1 :(得分：-1)
您可以映射（使用reduce）初始数组以更快地访问其键，然后使用forEach执行循环以构建所需的结果。
现在，使用sort值（只是数字）执行Week结果数组。
var arr1 = [{Week: "WK1", CreatedTickets: 1}, {Week: "WK20", CreatedTickets: 7}, {Week: "WK3", CreatedTickets: 12}, {Week: "WK2", CreatedTickets: 3}, {Week: "WK5", CreatedTickets: 5}],
arr2 = [{Week: "WK1", ClosedTickets: 4}, {Week: "WK6", ClosedTickets: 40},{Week: "WK3", ClosedTickets: 33}, {Week: "WK2", ClosedTickets: 2}],
// Convert to something like this:
// {
// "WK1": {
// "Week": "WK1",
// "CreatedTickets": 1
// },
// "WK20": {
// "Week": "WK20",
// "CreatedTickets": 7
// },
// }
//
// This is to speed up the access and comparisons.
//
map1 = arr1.reduce((a, o) => ({...a, ...{[o.Week]: o}}), {}),
map2 = arr2.reduce((a, o) => ({...a, ...{[o.Week]: o}}), {});
// Now let's loop the keys of the previously mapped 'map1'.
Object.keys(map1).forEach(k => {
if (map2[k]) {
// This is to avoid the values ("WK1", "WK2", and so on.)
// because map1 already has that value.
delete map2[k].Week;
// Will generate something like this:
// {
// WK1: {
// "Week": "WK1",
// "CreatedTickets": 1,
// "ClosedTickets": 4
// }
// }
map1[k] = { ...map1[k], ...map2[k] };
} else {
// Will generate something like this:
// {
// WK1: {
// "Week": "WK1",
// "CreatedTickets": 7,
// "ClosedTickets": 0 // To initialize qith zero because is an object that doesn't exists in map2
// }
// }
map1[k] = { ...map1[k], ...{CreatedTickets: map1[k].CreatedTickets, ClosedTickets: 0} };
}
});
// Now the same forEach for the map2 to complete with the missing objects (weeks) in map1
Object.keys(map2).forEach(k => {
if (!map1[k]) {
map1[k] = { ...map1[k], ...{Week: map2[k].Week, CreatedTickets: 0, ClosedTickets: map2[k].ClosedTickets} };
}
});
// Justa a sort
var sorted = Object.values(map1).sort((a,b) => a.Week.replace(/[a-z]/gi, '') - b.Week.replace(/[a-z]/gi, ''));
console.log(sorted);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 2 :(得分：-1)
如果使用ESNext功能，您可以使用一些快捷方式，但由于您使用的是var而不是let / const，我只假设是ES5。
以下是一个简单的解决方案。
var arr1 = [
{Week: "WK1", CreatedTickets: 1},
{Week: "WK3", CreatedTickets: 12},
{Week: "WK2", CreatedTickets: 3},
{Week: "WK20", CreatedTickets: 7},
{Week: "WK5", CreatedTickets: 5}
];
var arr2 = [
{Week: "WK1", ClosedTickets: 4},
{Week: "WK6", ClosedTickets: 40},
{Week: "WK3", ClosedTickets: 33},
{Week: "WK2", ClosedTickets: 2}
];
//first lets make a copy of arr1
var output = arr1.slice();
//now loop arr2, and see if exists in ouput or not
arr2.forEach(function (item) {
//lets find in arr1
var f = output.find(function(i) { return item.Week === i.Week; });
if (f) {
//ok found lets update
f.ClosedTickets = item.ClosedTickets;
} else {
//not found lets add
output.push(item);
}
});
//ok if we want 0, for OpenTickets and ClosedTickets if none found
output.forEach(function (item) {
item.ClosedTickets = item.ClosedTickets | 0;
item.CreatedTickets = item.CreatedTickets | 0;
});
//finally lets's sort
//one gotcha, sorting by WK1,WK2,WK10 etc in string sort would give
//WK1, WK10, WK2.. so we will use regex to extract just the wk part
//convert to integer and then sort
var regexnum = /\d+/;
function numWk(s) {
return parseInt(regexnum.exec(s.Week)[0], 10);
}
output.sort(function (a,b) { return numWk(a) - numWk(b); });
console.log(output);