提供的参数与调用目标的任何签名都不匹配 - Typescript
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Supplied parameters do not match any signature of call target - Typescript
提问by Herbert Corazao
I have a model called FilterOperation.ts that goes like this:
我有一个名为 FilterOperation.ts 的模型，它是这样的：
export class FilterOperations {
constructor(
public mainFilter : string,
public currencyType : string,
public status : string
) { }
}
Then have my component defined like this:
然后让我的组件像这样定义：
import { Component, OnInit, Input } from "@angular/core";
import { FormsModule } from '@angular/forms';
import 'rxjs/operator/finally';
//models
import { Session } from './src/models/session';
import { Client } from './src/models/client';
import { Operation } from './src/models/operation';
import { FilterOperations } from './src/models/filterOperations';
//services
import { OperationSearchService } from './src/services/operations_admin.service';
//constants
import * as constantsValues from "../../core/constants/constants";
@Component({
templateUrl: './app/views/operations_admin/src/templates/operations_admin.html',
styleUrls: ['./app/views/operations_admin/css/operations_admin.css'],
providers: [ OperationSearchService ]
})
export class OperationsAdminComponent implements OnInit {
@Input() filter: FilterOperations;
errorMessage: string;
clientFound: Client;
sessionList: Session[];
operationCount: number;
mainSearchSelector: string;
constructor (private operationSearchService: OperationSearchService) {
}
ngOnInit() {
this.filter = new FilterOperations(); //HERE GOES THE ERROR
}
onChangeMainFilter(newValue) {
this.mainSearchSelector = newValue;
}
findOperations() {
//some code
}
}
It keeps sending the same message when I try to compile it: "Supplied parameters do not match any signature of call target ". What am I doing wrong when initializing the value of filter:
当我尝试编译它时，它不断发送相同的消息：“提供的参数与调用目标的任何签名都不匹配”。初始化过滤器的值时我做错了什么：
this.filter = new FilterOperations();
回答by Nitzan Tomer
Your code for FilterOperations:
您的代码FilterOperations：
export class FilterOperations {
constructor(
public mainFilter : string,
public currencyType : string,
public status : string
) { }
}
is equivalent to:
相当于：
export class FilterOperations {
public mainFilter: string;
public currencyType: string;
public status: string;
constructor(mainFilter: string, currencyType: string, status: string) {
this.mainFilter = mainFilter;
this.currencyType = currencyType;
this.status = status;
}
}
Though you probably meant to do:
虽然你可能打算这样做：
export class FilterOperations {
public mainFilter: string;
public currencyType: string;
public status: string;
constructor() {}
}
In the first two the constructor expects 3 arguments, but when you instantiate it you're not passing any:
在前两个构造函数中需要 3 个参数，但是当您实例化它时，您没有传递任何参数：
this.filter = new FilterOperations();
If you want an empty ctor then use my 3rd code snippet.
You can also do:
如果您想要一个空的 ctor，请使用我的第三个代码片段。
你也可以这样做：
export class FilterOperations {
constructor(
public mainFilter?: string,
public currencyType?: string,
public status?: string
) { }
}
Now that all argument are optional you can call the ctor with no args.
现在所有参数都是可选的，您可以调用不带参数的 ctor。
