如何在嵌套的对象数组中获取对象
发布时间:2022-07-25 23:32:52 326
相关标签: # typescript
在Angular-14应用程序中,我有一个JSON响应:
{
"data": {
"pageItems": [
{
"id": "3fa85f64-5717-4562-b3fc-2c963f66afa6",
"merchantId": "3fa85f64-5717-4562-b3fc-2c963f66afa6",
"userId": "string",
"merchant": {
"id": "3fa85f64-5717-4562-b3fc-2c963f66afa6",
"merchantName": "string",
"accountNumber": "string",
"userLists": [
{
"id": "string",
"firstName": "string",
"lastName": "string",
"email": "string"
}
]
}
}
],
},
"successful": true,
"message": "string",
"statusCode": 0
}
然后,我将it模型界面表示为如下所示:
merchant-user-list.model:
export interface IUserList {
id: string;
firstName: string;
lastName: string;
email: string;
}
export interface IMerchant {
id: string;
merchantName: string;
accountNumber: string;
}
export interface IPageItem {
id: string;
merchantId: string;
userId: string;
merchant: IMerchant;
}
export interface IData {
pageItems: IPageItem[];
}
export interface IMerchantUserList {
data: IData;
successful: boolean;
message: string;
statusCode: number;
}
但是,我想使用firstName、lastName和email搜索DB表。所以我在typescript组件中做了这个。
merchant-users-list.component:
import { MerchantUserService } from 'src/app/services/merchant-user.service';
import { IData, IPageItem, IMerchantUserList } from 'src/app/models/merchant/merchant-user-list.model';
export class MerchantUsersComponent implements OnInit {
allMerchantUserList: any[] = [];
dataBk: IPageItem[] = this.allMerchantUserList;
constructor(
private merchantService: MerchantUserService
) {
}
onMerchantUserSearch() {
this.allMerchantUserList = this.dataBk.filter(
(row) =>
row.merchant.userLists[0].firstName
?.toLowerCase()
.includes(this.selectedName?.toLowerCase()) && row.merchant.userLists[0].lastName
?.toLowerCase()
.includes(this.selectedName?.toLowerCase()) && row.merchant.userLists[0].email
?.toLowerCase()
.includes(this.selectedName?.toLowerCase())
);
}
}
我想在文本框中使用firstName、lastName和email来过滤(搜索)嵌套数组中的所有对象。
我上面的只会得到长度为0的对象。当我做row.merchant.userLists时。firstName,我得到一个错误,firstName,lastName和email在用户列表中不可用。
如何重写onMerchantUserSearch()来实现这一点?
谢谢
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