TypeScript 无法编译，因为“any”的类型声明失去了类型安全性

Question

我应该如何解决以下错误消息：

无法编译 /.../SoftwareLicenseCodes/index.tsx (14,20)：“any”的类型声明失去类型安全性。考虑用更精确的类型替换它。此错误发生在构建期间，无法消除。

请看下面的代码：

import * as React from 'react';
import './SoftwareLicenseCodes.css';

interface SoftwareLicenseCodesProps {
}

interface SoftwareLicenseCodesState {
    count: string;
    oneTimeUsage: boolean;
    duration: string;
    validFrom: string;
    validTo: string;
    distributor: string;
    [key: string]: any;
}

class SoftwareLicenseCodes extends React.Component<SoftwareLicenseCodesProps, SoftwareLicenseCodesState> {
    constructor(props: SoftwareLicenseCodesProps) {
        super(props);

        this.state = {
            distributor: '',
            count:'',
            oneTimeUsage: false,
            duration: '',
            validFrom: '',
            validTo: ''
        };

        this.onInputChange = this.onInputChange.bind(this);
    }

    handleSubmit(event: React.FormEvent<HTMLFormElement>) {
        alert('submit');
        event.preventDefault();
    }

    onInputChange = (event: React.FormEvent<HTMLInputElement>) => {
        const value = event.currentTarget.type === 'checkbox' ? event.currentTarget.checked : event.currentTarget.value;

        this.setState({
            [name]: value
        });
    }

    render() {
        return (
            <div className="user-container software-codes">
                <div className="user-single-container">
                    <h1>Software License Codes</h1>

                    <form className="software-codes__form" onSubmit={this.handleSubmit}>
                        <label>
                            <span className="software-codes__input-element">Count</span>
                            <input
                                name="count"
                                type="number"
                                value={this.state.count}
                            />
                        </label>

                        <label>
                            <span className="software-codes__input-element">Distributor</span>
                            <input
                                name="distributor"
                                type="text"
                                value={this.state.distributor}
                            />
                        </label>

                        <label>
                            <span className="software-codes__input-element">One time usage</span>
                            <input
                                name="oneTimeUsage"
                                type="checkbox"
                                checked={this.state.oneTimeUsage}
                            />
                        </label>

                        <label>
                            <span className="software-codes__input-element">Duration</span>
                            <input
                                name="duration"
                                type="number"
                                value={this.state.duration}
                            />
                        </label>
                        <input className="software-codes__input-element" type="submit" value="Submit" />
                    </form>
                </div>
            </div>
        );
    }
}

export default SoftwareLicenseCodes;

Answer 1

您的代码仅设置字符串或布尔值，因此您可以将其锁定更多：

interface SoftwareLicenseCodesState {
    count: string;
    oneTimeUsage: boolean;
    duration: string;
    validFrom: string;
    validTo: string;
    distributor: string;
    [key: string]: string|boolean;
    // ------------^^^^^^^^^^^^^^
}

或者，如果您想要完全类型安全，您可以删除字符串索引签名并编写额外的代码来切换输入名称，然后使用显式属性名称。这最大化了类型检查的使用，同时（显然）增加了代码大小/复杂性：

function setNamed(target: SoftwareLicenseCodesState, name: string, value: string|boolean): SoftwareLicenseCodesState {
    if (name === "oneTimeUsage") {
        // Probably add assertion here that value is a boolean
        target.oneTimeUsage = value as boolean;
    } else {
        // Probably add assertion here that value is a string
        const strValue = value as string;
        switch (name) {
            case "count":
                target.count = strValue;
                break;
            case "duration":
                target.duration = strValue;
                break;
            case "validFrom":
                target.validFrom = strValue;
                break;
            case "validTo":
                target.validTo = strValue;
                break;
            case "distributor":
                target.distributor = strValue;
                break;
            default:
                // Failed assertion here
        }
    }
    return target;
}

然后

this.setState(setNamed({}, name, value));

一切都很笨拙，但最大限度地进行了类型检查。

我真的很想找到一种方法让您使用索引类型，但是由于名称来自元素name的属性input，如果没有上述内容，我看不到如何做到这一点switch。keyof这让我很困扰，因为我似乎想起了一些为名称构建联合类型的超级聪明的方法......