在我的Angular应用中，我有一个组件:
import { MakeService } from './../../services/make.service';
import { Component, OnInit } from '@angular/core';
@Component({
selector: 'app-vehicle-form',
templateUrl: './vehicle-form.component.html',
styleUrls: ['./vehicle-form.component.css']
})
export class VehicleFormComponent implements OnInit {
makes: any[];
vehicle = {};
constructor(private makeService: MakeService) { }
ngOnInit() {
this.makeService.getMakes().subscribe(makes => { this.makes = makes
console.log("MAKES", this.makes);
});
}
onMakeChange(){
console.log("VEHICLE", this.vehicle);
}
}
但是在“制造”属性中我犯了一个错误。
我不知道该怎么办……
在我的情况下，它与不同的声明根据新的typescript严格的特性:
@ViewChild(MatSort, {static: true}) sort!: MatSort;
如果在tsonfig中禁用typescript新的严格特性。json和
"compilerOptions": {
///
,
"strictPropertyInitialization":false
}
旧的Angular指南代码工作得很好
@ViewChild(MatSort) sort: MatSort;
这里有4种方法来解决这个问题
Arunkumar Gudelli (2022) https://www.angularjswiki.com/angular/property-has-no-initializer-and-is-not-definitely-assigned-in-the-constructor/
1)你可以像下面的代码一样应用它。当你这样做的时候，系统不会给出一个错误。
“明确赋值断言”(!)来告诉TypeScript我们知道这个值
详细信息
@Injectable()
export class Contact {
public name !:string;
public address!: Address;
public digital!: Digital[];
public phone!: Phone[];
}
2)第二种方法是在这里创建一个构造函数并定义值。
export class Contact {
public name :string;
public address: Address;
public digital: Digital[];
public phone: Phone[];
constructor( _name :string,
_address: Address,
_digital: Digital[],
_phone: Phone[])
{
this.name=_name;
this.address=_address;
this.digital=_digital;
this.phone=_phone;
}
}
3)第三种选择是创建一个get属性并按如下方式赋值
export class Contact {
public name :string="";
public address: Address=this._address;
get _address(): Address {
return new Address();
}
}
这个已经在Angular Github的https://github.com/angular/angular/issues/24571上讨论过了
我认为这是每个人都会转向的方向
引用自https://github.com/angular/angular/issues/24571#issuecomment-404606595
For angular components, use the following rules in deciding between:
a) adding initializer
b) make the field optional
c) leave the '!'
If the field is annotated with @input - Make the field optional b) or add an initializer a).
If the input is required for the component user - add an assertion in ngOnInit and apply c.
If the field is annotated @ViewChild, @ContentChild - Make the field optional b).
If the field is annotated with @ViewChildren or @ContentChildren - Add back '!' - c).
Fields that have an initializer, but it lives in ngOnInit. - Move the initializer to the constructor.
Fields that have an initializer, but it lives in ngOnInit and cannot be moved because it depends on other @input fields - Add back '!' - c).
转到你的tsconfig。Json文件，并更改属性:
"noImplicitReturns": false
然后加上
"strictPropertyInitialization": false
在"compilerOptions"属性下。
你的tsconfig。Json文件应该是这样的:
{
...
"compilerOptions": {
....
"noImplicitReturns": false,
....
"strictPropertyInitialization": false
},
"angularCompilerOptions": {
......
}
}
希望这能有所帮助!!
祝你好运
