首先,我是Typescript和Angular 2的新手,这似乎是一个明显的答案,但我似乎无法让它工作.我有以下型号
export interface IBrand {
brandID: number;
name: string;
image: string;
}
然后我有组件类
import { Component,OnInit } from '@angular/core';
import { Router,RouteParams } from '@angular/router-deprecated'
import { bootstrap } from '@angular/platform-browser-dynamic';
import { IBrand } from './brand';
import { BrandService } from './brand.service';
@Component({
selector: 'data-bind',templateUrl: 'app/dashboardApp/brands/brand-list.component.html',styleUrls: ['app/dashboardApp/brands/brand-list.component.css']
})
export class BrandListComponent implements OnInit {
brands: IBrand[];
errorMessage: string;
newBrand: IBrand;
pageTitle: string = 'Brands';
constructor(private _brandService: BrandService,private _router: Router) {
}
ngOnInit(): void {
this._brandService.getBrands()
.subscribe(
brands => this.brands = brands,error => this.errorMessage = <any>error);
}
}
然后我有以下的HTML
<div class="form-group">
<label for="brandName">Brand:</label>
<input type="text" class="form-control" placeholder="Name" id="brandName" [(ngModel)]="newBrand.name" />
</div>
platform-browser.umd.js:962 ORIGINAL EXCEPTION: TypeError: Cannot read property ‘name’ of undefined
但是我可以轻松地将它绑定到像pageTitle这样的东西.任何可以指引我正确方向的想法？
解决方法
我问了这个问题已经有一段时间了,它开始得到一些观点,所以我会添加我的答案.
export class Brand {
constructor() {}
brandID: number;
name: string;
image: string;
}
现在我有一个构造函数,我可以使用new运算符来实例化对象.
export class BrandListComponent implements OnInit {
brands: Brand[];
errorMessage: string;
newBrand: Brand = new Brand();
pageTitle: string = 'Brands';
(...)
}
现在我在没有任何数据的情况下初始化了所需的品牌,我可以将其绑定到模型.希望这可以帮助.
