Typescript: TS7006: Parameter 'xxx' implicitly has an 'any' type

Question

In testing my UserRouter, I am using a json file

data.json

[
  {
    "id": 1,
    "name": "Luke Cage",
    "aliases": ["Carl Lucas", "Power Man", "Mr. Bulletproof", "Hero for Hire"],
    "occupation": "bartender",
    "gender": "male",
    "height": {
      "ft": 6,
      "in": 3
    },
    "hair": "bald",
    "eyes": "brown",
    "powers": [
      "strength",
      "durability",
      "healing"
    ]
  },
  {
  ...
  }
]

Building my app, I get the following TS error

ERROR in ...../UserRouter.ts
(30,27): error TS7006: Parameter 'user' implicitly has an 'any' type.

UserRouter.ts

import {Router, Request, Response, NextFunction} from 'express';
const Users = require('../data');

export class UserRouter {
  router: Router;

  constructor() {
  ...
  }

  /**
   * GET one User by id
   */
  public getOne(req: Request, res: Response, _next: NextFunction) {
    let query = parseInt(req.params.id);
 /*[30]->*/let user = Users.find(user => user.id === query);
    if (user) {
      res.status(200)
        .send({
          message: 'Success',
          status: res.status,
          user
        });
    }
    else {
      res.status(404)
        .send({
          message: 'No User found with the given id.',
          status: res.status
        });
    }
  }


}

const userRouter = new UserRouter().router;
export default userRouter;

Answer 1

You are using the --noImplicitAny and TypeScript doesn't know about the type of the Users object. In this case, you need to explicitly define the user type.

Change this line:

let user = Users.find(user => user.id === query);

to this:

let user = Users.find((user: any) => user.id === query); 
// use "any" or some other interface to type this argument

Or define the type of your Users object:

//...
interface User {
    id: number;
    name: string;
    aliases: string[];
    occupation: string;
    gender: string;
    height: {ft: number; in: number;}
    hair: string;
    eyes: string;
    powers: string[]
}
//...
const Users = <User[]>require('../data');
//...

Answer 2

In your tsconfig.json file set the parameter "noImplicitAny": false under compilerOptions to get rid of this error.

Answer 3

if you get an error as Parameter 'element' implicitly has an 'any' type.Vetur(7006) in vueJs

with the error:

 exportColumns.forEach(element=> {
      if (element.command !== undefined) {
        let d = element.command.findIndex(x => x.name === "destroy");

you can fixed it by defining thoes variables as any as follow.

corrected code:

exportColumns.forEach((element: any) => {
      if (element.command !== undefined) {
        let d = element.command.findIndex((x: any) => x.name === "destroy");

Answer 4

I encounted this error and found that it was because the "strict" parameter was set to true in the tsconfig.json file. Just set it "false" (obviously). In my case I had generated the tsconfig file from the cmd prompt and simply missed the "strict" parameter, which was located further down in the file.

Answer 5

go to tsconfig.json and comment the line the //strict:true this worked for me

Answer 6

Minimal error reproduction

export const users = require('../data'); // presumes @types/node are installed
const foundUser = users.find(user => user.id === 42); 
// error: Parameter 'user' implicitly has an 'any' type.ts(7006)

Recommended solution: --resolveJsonModule

--resolveJsonModule

import users from "./data.json" // `import` instead of `require`
const foundUser = users.find(user => user.id === 42); // user is strongly typed, no `any`!

---

There are some alternatives for other cases than static JSON import.

Option 1: Explicit user type (simple, no checks)

type User = { id: number; name: string /* and others */ }
const foundUser = users.find((user: User) => user.id === 42)

Option 2: Type guards (middleground)

Type guards

function isUserArray(maybeUserArr: any): maybeUserArr is Array<User> {
  return Array.isArray(maybeUserArr) && maybeUserArr.every(isUser)
}

function isUser(user: any): user is User {
  return "id" in user && "name" in user
}

if (isUserArray(users)) {
  const foundUser = users.find((user) => user.id === 42)
}

assertion functionsif

function assertIsUserArray(maybeUserArr: any): asserts maybeUserArr is Array<User> {
  if(!isUserArray(maybeUserArr)) throw Error("wrong json type")
}

assertIsUserArray(users)
const foundUser = users.find((user) => user.id === 42) // works

Option 3: Runtime type system library (sophisticated)

A runtime type check library like io-ts or ts-runtime can be integrated for more complex cases.

---

Not recommended solutions

noImplicitAny: false

function add(s1, s2) { // s1,s2 implicitely get `any` type
  return s1 * s2 // `any` type allows string multiplication and all sorts of types :(
}
add("foo", 42)

Also better provide an explicit User type for user. This will avoid propagating any to inner layer types. Instead typing and validating is kept in the JSON processing code of the outer API layer.

Answer 7

    "noImplicitAny": false
    

make these changes in your compilerOptions section of tsconfig.json file this worked for me

no need to set

    "strict":false

And please wait 1 or two minutes compilation is very slow some pcs

Answer 8

Parameter 'post' implicitly has an 'any' type in Angularcli Perhaps on project creation you've enabled Angular's strict mode in your app? Max recommends to disable strict mode If you've enabled strict mode, please disable it for this course by setting the strict property in tsconfig.json to false

Answer 9

I found this issue in Angular to arguments of function.

Before my code giving error

Parameter 'event' implicitly has an 'any' type

Here Is code

changeInpValue(event)
{
    this.inp = event.target.value;
}

Here is the change, after the argument write : any and the error is solved

changeInpValue(event : any)
{
    this.inp = event.target.value;
}

Working fine for me.

Answer 10

Add the below rule in your tsconfig.json file:- "strict": false /* does not allow all strict type-checking options. */,

Answer 11

This did the trick for me. Check how I have used the x variable

 From

     const user = users.find(x => x.username === username && x.password === password);

   To
const user = users.find((x:any) => x.username === username && x.password === password);