我必须将对象数组作为来自node.js的有效载荷,并将其存储在不同的变量中并进行迭代,但是* ngFor只迭代第一个循环而不是第二个循环,因此如何在相同的div上进行迭代 status.component.html
<div *ngFor="let payload1 of payload1;let payload2 of payload2;">
<h4>{{payload2.username}}</h4>
<h4>{{payload1.date}}</h4>
</div>
status.component.ts
payload1 = [];
payload2 = [];
ngOnInit() {
this.statusService.getStatusDetails(this.statusObj).subscribe(
(data) => {
if(data.status == 26){
this.payload1 = data.payload[0];
this.payload2 = data.payload[1];
}
}
);
}
答案 0 :(得分:4)
我认为,如果您的有效载荷长度相同且有序匹配(您的问题似乎暗示确实如此),这应该可行
<div *ngFor="let p of payload1;let i = index">
<h4>{{payload2[i].username}}</h4>
<h4>{{payload1[i].date}}</h4>
</div>
答案 1 :(得分:1)
为什么不从中制作单个阵列?
payload = [];
ngOnInit() {
this.statusService.getStatusDetails(this.statusObj).subscribe(
(data) => {
if(data.status == 26){
this.payload = data.payload[0]
.map((item, index) => ({ ...item, ...data.payload[1][index] }));
}
}
);
}
片段证明有效:
const d1 = [
{ id: 0 },
{ id: 1 },
];
const d2 = [
{ name: '0' },
{ name: '1' },
];
const d3 = d1.map((item, index) => ({ ...item, ...d2[index] }));
console.log(d3);
如果数组的长度不同,您甚至可以构建此合并的有效负载。
const d1 = [
{ id: 0 },
{ id: 1 },
];
const d2 = [
{ name: '0' },
{ name: '1' },
{ name: '2' },
];
const longest = d1.length > d2.length ? d1 : d2;
const shortest = d1.length <= d2.length ? d1 : d2;
const d3 = longest.map((item, index) => ({ ...item, ...shortest[index] }));
console.log(d3);
角度版本:
payload = [];
ngOnInit() {
this.statusService.getStatusDetails(this.statusObj).subscribe(
(data) => {
if(data.status == 26){
const longest = data.payload[0].length > data.payload[1].length ?
data.payload[0] : data.payload[1];
const shortest = data.payload[0].length <= data.payload[1].length ?
data.payload[0] : data.payload[1];
this.payload = longest
.map((item, index) => ({ ...item, ...shortest[index] }));
}
}
);
}