我有一些代码工作得很好，但我遇到了问题。
基本上当它到达一个NULL的记录时，它会向数组添加一个...
在这种情况下，第二条记录为NULL，所以我得到了：
[10,0,20]
我需要它做的是，如果thsub为NULL，则不向数组添加任何内容并继续下一条记录。
因此，在这种情况下，期望的结果是：
[10,20]
这是完整的代码：
var data = {
"cars": [{
"id": "1",
"name": "name 1",
"thsub": [{
"id": "11",
"name": "sub 1",
"stats": {
"items": 5,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 5,
},
"translations": null
}],
"image": null
},
{
"id": "2",
"name": "name 2",
"thsub": null, //this will break the code
"image": null
},
{
"id": "54",
"name": "name something",
"thsub": [{
"id": "65",
"name": "sub 1",
"stats": {
"items": 10,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 10,
},
"translations": null
}],
"image": null
}
]
}
var thCount = [];
for (var l = 0, m = data.cars.length; l < m; l++) {
thCount[l] = 0;
if (data.cars[l].thsub) {
for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
if (data.cars[l].thsub[i].stats) {
thCount[l]+=data.cars[l].thsub[i].stats.items;
}
}
}
}
console.log(thCount);
我该怎么做？
答案 0 :(得分：1)
如果设置了thsub，则只能推送一个值。
var data = { cars: [{ id: "1", name: "name 1", thsub: [{ id: "11", name: "sub 1", stats: { items: 5, }, ions: null }, { id: "22", name: "sub 2", stats: { items: 5, }, translations: null }], image: null }, { id: "2", name: "name 2", thsub: null, image: null }, { id: "54", name: "name something", thsub: [{ id: "65", name: "sub 1", stats: { items: 10, }, ions: null }, { id: "22", name: "sub 2", stats: { items: 10, }, translations: null }], image: null }] },
thCount = [];
for (var l = 0, m = data.cars.length; l < m; l++) {
if (data.cars[l].thsub) {
thCount.push(0);
for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
if (data.cars[l].thsub[i].stats) {
thCount[thCount.length - 1] += data.cars[l].thsub[i].stats.items;
}
}
}
}
console.log(thCount);
答案 1 :(得分：1)
你需要添加变量，然后只有在有东西时才推送到数组。
var data = {
"cars": [{
"id": "1",
"name": "name 1",
"thsub": [{
"id": "11",
"name": "sub 1",
"stats": {
"items": 5,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 5,
},
"translations": null
}],
"image": null
},
{
"id": "2",
"name": "name 2",
"thsub": null, //this will break the code
"image": null
},
{
"id": "54",
"name": "name something",
"thsub": [{
"id": "65",
"name": "sub 1",
"stats": {
"items": 10,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 10,
},
"translations": null
}],
"image": null
}
]
}
var thCount = [];
for (var l = 0, m = data.cars.length; l < m; l++) {
if (data.cars[l].thsub) {
var tmp = 0;
for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
if (data.cars[l].thsub[i].stats) {
tmp+=data.cars[l].thsub[i].stats.items;
}
thCount.push(tmp);
}
}
}
console.log(thCount);
答案 2 :(得分：0)
您可以在for循环内创建新变量，如果不是0，则使用push()。
var data = {"cars":[{"id":"1","name":"name 1","thsub":[{"id":"11","name":"sub 1","stats":{"items":5},"ions":null},{"id":"22","name":"sub 2","stats":{"items":5},"translations":null}],"image":null},{"id":"2","name":"name 2","thsub":null,"image":null},{"id":"54","name":"name something","thsub":[{"id":"65","name":"sub 1","stats":{"items":10},"ions":null},{"id":"22","name":"sub 2","stats":{"items":10},"translations":null}],"image":null}]}
var thCount = [];
for (var l = 0, m = data.cars.length; l < m; l++) {
var count = 0
if (data.cars[l].thsub) {
for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
if (data.cars[l].thsub[i].stats) {
count += data.cars[l].thsub[i].stats.items;
}
}
}
if (count != 0) thCount.push(count)
}
console.log(thCount);
答案 3 :(得分：0)
行thCount[l] = 0;将导致向thCount添加元素，无论您的条件是if (data.cars[l].thsub)
您可以使用另一个变量，只有在我们想要向数组添加内容时才增加它：
var data = {
"cars": [{
"id": "1",
"name": "name 1",
"thsub": [{
"id": "11",
"name": "sub 1",
"stats": {
"items": 5,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 5,
},
"translations": null
}],
"image": null
},
{
"id": "2",
"name": "name 2",
"thsub": null, //this will break the code
"image": null
},
{
"id": "54",
"name": "name something",
"thsub": [{
"id": "65",
"name": "sub 1",
"stats": {
"items": 10,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 10,
},
"translations": null
}],
"image": null
}
]
}
var thCount = [];
for (var l = 0, k = -1, m = data.cars.length; l < m; l++) {
if (data.cars[l].thsub) {
thCount[++k] = 0;
for (var i = 0, j = data.cars[l].thsub.length; i < j; i++) {
if (data.cars[l].thsub[i].stats) {
thCount[k] += data.cars[l].thsub[i].stats.items;
}
}
}
}
console.log(thCount); //alert(thCount);
答案 4 :(得分：0)
关于如何修复循环有一些答案，所以基本问题在于thCount[l] = 0行。您还可以使用更高阶函数来遍历对象，从而产生更好的可读代码：
var data = {
"cars": [{
"id": "1",
"name": "name 1",
"thsub": [{
"id": "11",
"name": "sub 1",
"stats": {
"items": 5,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 5,
},
"translations": null
}],
"image": null
},
{
"id": "2",
"name": "name 2",
"thsub": null, //this will break the code
"image": null
},
{
"id": "54",
"name": "name something",
"thsub": [{
"id": "65",
"name": "sub 1",
"stats": {
"items": 10,
},
"ions": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 10,
},
"translations": null
}],
"image": null
}
]
}
var thCount = data.cars.filter(function(car){
return car.thsub;
}).map(function(car){
return car.thsub.reduce(function(a,b){
return a.stats.items + b.stats.items;
});
});
console.log(thCount)
首先，您filter输出所有没有thsub的对象。然后你map将这些条目转换为新值。您可以使用reduce - 函数获取这些值，该函数汇总items中的所有stats - 值。
答案 5 :(得分：0)
还可以考虑使用嵌套reduce方法，例如如下......
var data = {
"cars": [{
"id": "1",
"name": "name 1",
"thsub": [{
"id": "11",
"name": "sub 1",
"stats": {
"items": 5
},
"translations": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 5
},
"translations": null
}],
"image": null
}, {
"id": "2",
"name": "name 2",
"thsub": null,
"image": null
}, {
"id": "54",
"name": "name something",
"thsub": [{
"id": "65",
"name": "sub 1",
"stats": {
"items": 10
},
"translations": null
}, {
"id": "22",
"name": "sub 2",
"stats": {
"items": 10
},
"translations": null
}],
"image": null
}]
};
var thCount = data.cars.reduce(function (collector, carItem) {
var
thSubs = carItem.thsub;
//if (Array.isArray(thSubs)) { ... }
if ((thSubs != null) && (thSubs.length >= 1) && Number.isFinite(thSubs.length)) {
collector.push(
thSubs.reduce(function (count, subItem) {
return (count + subItem.stats.items);
}, 0)
);
}
return collector;
}, []);
console.log(thCount);