Element implicitly has an 'any' type because expression of type 'string' can't be used to index

Question

Trying out TypeScript for a React project and I'm stuck on this error:

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; }'

Which appears when I try to filter the array in my component

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

So far I looked at the article "Indexing objects in TypeScript" (https://dev.to/kingdaro/indexing-objects-in-typescript-1cgi) since it had a similar error, but I tried to add the index signature to type plotTypes and I still get the same error.

My component code:

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;

Answer 1

I want to add another edge case in which this error might occur. I have had this happen due to a broken dependency in node_modules that did not properly export its types. So, updating from one version to another, suddently this type of error would appear, because tsc could not fully resolve everything.

This led to a lot of

Element implicitly has an 'any' type

errors.

Ensure that the origin of this error is not with a dependent package whose package does not export correct types.

Answer 2

For use with a reduce function. Make sure you properly type the accumulator.

Example:

const getNestedLevels = (
  subjects: TutorSubject[] | TutorActivity[] | undefined,
) => {
  if (subjects === undefined) {
    return {};
  }
  return subjects.reduce(
    (acc, curr) => {
      let subject;
      let level;

      if (curr.specialization) {
        subject = curr.specialization.name;
      }

      if (curr.specializationLevel) {
        level = curr.specializationLevel.name;
      }

      if (subject && level) {
        acc[subject] = [...(subject in acc ? acc[subject] : []), level];
      }

      return acc;
    },
    {} as { [key: string]: string[] }, // 👈 type this acc
  );
};

Answer 3

I ran into this issue today as well, tried what mentioned above to use keyof, but it thorws another error saying "Type 'any' is not assignable to type 'never".

Did some research and turns out it's because ts thinks that the key I have could be any string, and such key string may not exist in my object

take this for example

interface MyObj {
    [abc as string]: string;
    [xyz as string]: boolean;
}
const flag: boolean = true;
const newObj: MyObj = cloneDeep(myObj);
const myKey: string = someKeyFromSomewhereXYZ; // for my situation my key was passed in as a child of an object prop from parent component
newObj[myKey as keyof MyObj] = !flag; // this is where my error pops, and doing keyof won't fix the issue.

so for me to fix the issue, what I did was to add the specific key to myKey, make it look like this

const myKey: 'xyz' = someKeyFromSomewhereXYZ

it seems like ts thinks that myKey could be any sting, hence doing newObj[myKey] could be undefined, which leads to 'never' in ts I guess.

Answer 4

You can also utilize switch with fall-through statements:

switch (prop) {
    case 'propOne':
    case 'propTwo':
    case 'propThree':
        return someObj[prop];
    default:
        return false;
}

As long as those props exist on someObj, you will not get a TypeScript error.

Answer 5

I fixed this issue by using keyof

messageMap = {
 "Hi": "Hello",
 "Who are you": "My name is Test Sat Bot",
 "What is your role": "Just guide for the user",
}

this ❌

let answer = this.messageMap[question];

replace with ✔️

let answer = this.messageMap[question as keyof typeof this.messageMap];

Here question is type of string

getBotMessage(question: string){    
  let answer = this.messageMap[question as keyof typeof this.messageMap];
}

Answer 6

This might help someone First the empty object need to be with the type Record<string, any> so it will accept key as string and value as any

const sortFields: Record<string, any> = {};
const keyName = `field`;

sortFields[keyName] = 'asc';

So here when we log sortFields

 console.log("sortFields==== ", sortFields); // sortFields==== {field: 'asc'}

Or if we want to add more than one element in the object we can do it like this

sortFields.email = "asc"; // Or sortFields["email"] = "asc"; but its better written in dot notation.
sortFields.name = "desc";
sortFields.phone = "desc";

So now when we log sortFields

 console.log("sortFields==== ", sortFields); // sortFields==== {field: 'asc', email: 'asc', name: 'desc', phone: 'desc'}

Answer 7

In situation of nested object, if want for in in for in , here is a working example without error warning:

export const X = {
  aa: {
    a: '1',
    b: '2',
  },
  bb: {
    a: '3',
    b: '4',
  },
} as const

export const f1 = () => {
  let k1: keyof typeof X
  for (k1 in X) {
    console.log(k1)

    let k2: keyof (typeof X)[keyof typeof X]
    for (k2 in X[k1]) {
      console.log(X[k1][k2])
    }
  }
}

Answer 8

as a last resort, you can mute this error by setting "suppressImplicitAnyIndexErrors": true in tsconfig.json

{
  "compilerOptions": {
    "suppressImplicitAnyIndexErrors": true,
  }
}

Answer 9

TypeScript needs to be sure those values exist in trainInfo, else it reads all as string

instead of doing this

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

do this

interface trainInfo {
  name: "train_1" | "train_2" | "train_3"| "train_4";
  x: Array<number>;
  y: Array<number>;
  type: string;
  mode: string;
}

Answer 10

For anyone who stumbles upon this in the future:

If you're getting the TypeScript error

'...expression of type string cannot be used to index...'

then simply specify that the 'expression of type string' is a key of the type of that object. For example,

const someObj:ObjectType = data;
const field = 'username';

// This gives an error
const temp = someObj[field];

// Solution 1: When the type of the object is known
const temp = someObj[field as keyof ObjectType]

// Solution 2: When the type of the object is not known
const temp = someObj[field as keyof typeof someObj]

Answer 11

It worked for me with keyof and as operators:

const keys: [keyof ITrainInfo] = Object.keys(this.trainInfo) as [
    keyof ITrainInfo,
]
keys.forEach((property) => {
    // console.log(tmpUser[property])
    if (this.trainInfo === undefined) return
    if (this.trainInfo[property] !== undefined) {
        // your code here
        /*const trainsToSet = trains.find((field) => field.name === property)
        if (trainsToSet != undefined)
            trainsToSet.value = this.trainInfo[property]?.toString()
        */
    }
})

Answer 12

public getUserName(): string {

const accessToken = this.getAccessToken();
const claims:any = this.getUserClaims();
console.log('access token ',accessToken);
this.getUserInfo();
return claims['sub'].split('@')[0];

}

//give any type to the variable

Answer 13

With out typescript error

    const formData = new FormData();
    Object.keys(newCategory).forEach((k,i)=>{  
        var d =Object.values(newCategory)[i];
        formData.append(k,d) 
    })

Answer 14

This is not a answer to the original question, but a generic work around to this problem.

Original problem: person[cr.field] causes this error

---

I'm doing a generic advanced search form where user can select a field, comparator and the desired value. When trying to read the value from the object based on the key, I get this error (altought the field value is type of string and I think it should be just fine)

So what I do is I extract the [key, value] like this

const x: [string, any] = Object.entries(person).find(([key, _]) => key === cr.field);

For example if my criterion (cr) is { field: 'name', value: 'John' } and field name actually exists in a person obj., it should return the field name and the value as tuple (x is [string, any] or undef). If not found, undefined.

Answer 15

I know it's a little too late, but all it's needed is to add a little type conversion, I wrote a static function that safely-returns the array of keys with the correct typing. All you need is to define the type and pass the object as a parameter:

export class ObjectUtil {
  public static getObjectKeys<T>(obj: Object) {
    if (!obj) {
      return [];
    }

    return Object.keys(obj).map((key: string) => key as keyof T);
  }
}

Below is a simple example:

ObjectUtil.getObjectKeys<Address>(address).forEach((key) => {
  console.log(address[key]);
});

Answer 16

I use this:

interface IObjectKeys {
  [key: string]: string | number;
}

interface IDevice extends IObjectKeys {
  id: number;
  room_id: number;
  name: string;
  type: string;
  description: string;
}

NOTE: "[key: string]" what is it? An object in JavaScript is primarily just a collection of properties made up of key-value pairs. Moreover, the key can only be a string (even for array elements), but the value can be any data type.

If you use the optional property in your object:

interface IDevice extends IObjectKeys {
  id: number;
  room_id?: number;
  name?: string;
  type?: string;
  description?: string;
}

... you should add 'undefined' value into the IObjectKeys interface:

interface IObjectKeys {
  [key: string]: string | number | undefined;
}

Answer 17

I have made a simulation of the problem. looks like the issue is how we should Access Object Properties Dynamically Using Bracket Notation in Typescript

interface IUserProps {
  name: string;
  age: number;
}

export default class User {
  constructor(private data: IUserProps) {}

  get(propName: string): string | number {
    return this.data[propName as keyof IUserProps];
  }
}

I found a blog that might be helpful to understand this better.

here is a link https://www.nadershamma.dev/blog/2019/how-to-access-object-properties-dynamically-using-bracket-notation-in-typescript/

Answer 18

I made some small changes to Alex McKay's function/usage that I think make it a little easier to follow why it works and also adheres to the no-use-before-define rule.

First, define this function to use:

const getKeyValue = function<T extends object, U extends keyof T> (obj: T, key: U) { return obj[key] }

In the way I've written it, the generic for the function lists the object first, then the property on the object second (these can occur in any order, but if you specify U extends key of T before T extends object you break the no-use-before-define rule, and also it just makes sense to have the object first and its' property second. Finally, I've used the more common function syntax instead of the arrow operators (=>).

Anyways, with those modifications you can just use it like this:

interface User {
  name: string;
  age: number;
}

const user: User = {
  name: "John Smith",
  age: 20
};

getKeyValue(user, "name")

Which, again, I find to be a bit more readable.

Answer 19

When using Object.keys, the following works:

Object.keys(this)
    .forEach(key => {
      console.log(this[key as keyof MyClass]);
    });

Answer 20

Thanks to Alex Mckay I had a resolve for dynamic setting a props:

  for(let prop in filter)
      (state.filter as Record<string, any>)[prop] = filter[prop];

Answer 21

// bad
const _getKeyValue = (key: string) => (obj: object) => obj[key];

// better
const _getKeyValue_ = (key: string) => (obj: Record<string, any>) => obj[key];

// best
const getKeyValue = <T extends object, U extends keyof T>(key: U) => (obj: T) =>
  obj[key];

Bad - the reason for the error is the object type is just an empty object by default. Therefore it isn't possible to use a string type to index {}.

Better - the reason the error disappears is because now we are telling the compiler the obj argument will be a collection of string/value (string/any) pairs. However, we are using the any type, so we can do better.

Best - T extends empty object. U extends the keys of T. Therefore U will always exist on T, therefore it can be used as a look up value.

Here is a full example:

I have switched the order of the generics (U extends keyof T now comes before T extends object) to highlight that order of generics is not important and you should select an order that makes the most sense for your function.

const getKeyValue = <U extends keyof T, T extends object>(key: U) => (obj: T) =>
  obj[key];

interface User {
  name: string;
  age: number;
}

const user: User = {
  name: "John Smith",
  age: 20
};

const getUserName = getKeyValue<keyof User, User>("name")(user);

// => 'John Smith'

Alternative Syntax

const getKeyValue = <T, K extends keyof T>(obj: T, key: K): T[K] => obj[key];

Answer 22

When we do something like this obj[key] Typescript can't know for sure if that key exists in that object. What I did:

Object.entries(data).forEach(item => {
    formData.append(item[0], item[1]);
});

Answer 23

This is what it worked for me. The tsconfig.json has an option noImplicitAny that it was set to true, I just simply set it to false and now I can access properties in objects using strings.

Answer 24

This happens because you try to access plotOptions property using string name. TypeScript understands that name may have any value, not only property name from plotOptions. So TypeScript requires to add index signature to plotOptions, so it knows that you can use any property name in plotOptions. But I suggest to change type of name, so it can only be one of plotOptions properties.

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

Now you'll be able to use only property names that exist in plotOptions.

You also have to slightly change your code.

First assign array to some temp variable, so TS knows array type:

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"
    },
    // ...
}

Then filter:

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

If you're getting data from API and have no way to type check props at compile time the only way is to add index signature to your plotOptions:

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

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