提供的参数与调用目标的任何签名都不匹配 - Typescript答案

【问题标题】：Supplied parameters do not match any signature of call target - Typescript提供的参数与调用目标的任何签名都不匹配 - Typescript
【发布时间】：2017-04-28 18:04:12
【问题描述】：

我有一个名为 FilterOperation.ts 的模型，如下所示：

export class FilterOperations {
  constructor(
    public mainFilter : string,
    public currencyType : string,
    public status : string
    ) { }
}

然后像这样定义我的组件：

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
    }
}

当我尝试编译它时，它一直发送相同的消息：“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();

【问题讨论】：

标签： angular typescript

【解决方案1】：

FilterOperations 的代码：

export class FilterOperations {
    constructor(
        public mainFilter : string,
        public currencyType : string,
        public status : string
    ) { }
}

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;
    }
}

虽然你可能打算这样做：

export class FilterOperations {
    public mainFilter: string;
    public currencyType: string;
    public status: string;

    constructor() {}
}

在前两个中，构造函数需要 3 个参数，但是当你实例化它时你没有传递任何参数：

this.filter = new FilterOperations();

如果您想要一个空的 ctor，请使用我的第三个代码 sn-p。
你也可以这样做：

export class FilterOperations {
    constructor(
        public mainFilter?: string,
        public currencyType?: string,
        public status?: string
    ) { }
}

现在所有参数都是可选的，您可以在没有参数的情况下调用 ctor。

【讨论】：

public 关键词很重要