元素隐式具有“任意”类型，因为类型“字符串”的表达式不能用于索引

小编典典

元素隐式具有“任意”类型，因为类型“字符串”的表达式不能用于索引

reactjs

为React项目尝试TypeScript时，我陷入了这个错误：

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{ train_1: boolean; train_2: boolean; train_3: boolean; train_4: boolean; }'.
  No index signature with a parameter of type 'string' was found on type '{ train_1: boolean; train_2: boolean; train_3: boolean; train_4: boolean; }'

当我尝试过滤组件中的数组时出现

.filter(({ name }) => plotOptions[name]);

到目前为止，我看了文章“ TypeScript中的索引对象”（https://dev.to/kingdaro/indexing-objects-in-
typescript-1cgi），因为它存在类似的错误，但是我尝试将索引签名添加到类型plotTypes，我仍然收到相同的错误。

我的组件代码：

import React, { Component } from "react";
import createPlotlyComponent from "react-plotly.js/factory";
import Plotly from "plotly.js-basic-dist";
const Plot = createPlotlyComponent(Plotly);

interface IProps {
  data: any;
}

interface IState {
  [key: string]: plotTypes;
  plotOptions: plotTypes;
}

type plotTypes = {
  [key: string]: boolean;
  train_1: boolean;
  train_2: boolean;
  train_3: boolean;
  train_4: boolean;
};

interface trainInfo {
  name: string;
  x: Array<number>;
  y: Array<number>;
  type: string;
  mode: string;
}

class FiltrationPlots extends Component<IProps, IState> {
  readonly state = {
    plotOptions: {
      train_1: true,
      train_2: true,
      train_3: true,
      train_4: true
    }
  };
  render() {
    const { data } = this.props;
    const { plotOptions } = this.state;

    if (data.filtrationData) {
      const plotData: Array<trainInfo> = [
        {
          name: "train_1",
          x: data.filtrationData.map((i: any) => i["1-CumVol"]),
          y: data.filtrationData.map((i: any) => i["1-PressureA"]),
          type: "scatter",
          mode: "lines"
        },
        {
          name: "train_2",
          x: data.filtrationData.map((i: any) => i["2-CumVol"]),
          y: data.filtrationData.map((i: any) => i["2-PressureA"]),
          type: "scatter",
          mode: "lines"
        },
        {
          name: "train_3",
          x: data.filtrationData.map((i: any) => i["3-CumVol"]),
          y: data.filtrationData.map((i: any) => i["3-PressureA"]),
          type: "scatter",
          mode: "lines"
        },
        {
          name: "train_4",
          x: data.filtrationData.map((i: any) => i["4-CumVol"]),
          y: data.filtrationData.map((i: any) => i["4-PressureA"]),
          type: "scatter",
          mode: "lines"
        }
      ].filter(({ name }) => plotOptions[name]);
      return (
        <Plot
          data={plotData}
          layout={{ width: 1000, height: 1000, title: "A Fancy Plot" }}
        />
      );
    } else {
      return <h1>No Data Loaded</h1>;
    }
  }
}

export default FiltrationPlots;

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2020-07-22

共1个答案

小编典典

发生这种情况是因为您尝试plotOptions使用string
访问属性name。TypeScript理解它name可能具有任何值，不仅是来自的属性名称plotOptions。因此，TypeScript需要向其中添加索引签名plotOptions，因此它知道您可以在中使用任何属性名称plotOptions。但我建议更改的类型name，因此它只能是plotOptions属性之一。

interface trainInfo {
    name: keyof typeof plotOptions;
    x: Array<number>;
    y: Array<number>;
    type: string;
    mode: string;
}

现在，您将只能使用中存在的属性名称plotOptions。

您还必须稍微更改代码。

首先将数组分配给一些临时变量，因此TS知道数组类型：

const plotDataTemp: Array<trainInfo> = [
    {
      name: "train_1",
      x: data.filtrationData.map((i: any) => i["1-CumVol"]),
      y: data.filtrationData.map((i: any) => i["1-PressureA"]),
      type: "scatter",
      mode: "lines"
    },
    // ...
}

然后过滤：

const plotData = plotDataTemp.filter(({ name }) => plotOptions[name]);

如果您要从API获取数据并且无法在编译时键入检查道具，则唯一的方法是在您的索引中添加索引签名plotOptions：

type tplotOptions = {
    [key: string]: boolean
}

const plotOptions: tplotOptions = {
    train_1: true,
    train_2: true,
    train_3: true,
    train_4: true
}

2020-07-22