filter out unneeded object properties for interface in typescript

问题

As background, I'm using Prisma (graphql), mysql2 (nodejs) and typescript.

I'm using an interactive command line script to connect to mysql2.

This is the warning I am seeing:

Ignoring invalid configuration option passed to Connection: type. This is currently a warning, but in future versions of MySQL2, an error will be thrown if you pass an invalid configuration options to a Connection

This is how I instantiate mysql2, in my MysqlConnector class:

this.connectionPromise = await this.mysql.createConnection(this.connectionOptions)

"connectionOptions" is set in the class constructor:

constructor(connectionDetails: MysqlConnectionDetails) {

This is my type definition:

export interface MysqlConnectionDetails {
  host: string
  port: number
  user: string
  password: string
  database?: string
}

This is the type definition of the object that I pass into my MysqlConnector class:

export interface DatabaseCredentials {
  type: DatabaseType
  host: string
  port: number
  user: string
  password: string
  database?: string
  alreadyData?: boolean
  schema?: string
  ssl?: boolean
  filter?: any
}

So, I am passing in an object that has additional parameters that mysql2 does not want/need. I am new to Typescript. I tried this, before passing the object in to the MysqlConnector class:

let forDeletion = ['type', 'alreadyData']
connector = new MysqlConnector(credentials.filter(item => !forDeletion.includes(item)))

But I get an error, saying "filter" is not a property on the type "DatabaseCredentials", and I concluded this is probably not the right way to do this.

I assumed there is a way in Typescript (through tsconfig) automatically to filter out properties that are not in the receiving type. "MysqlConnectionDetails" doesn't have a property "type", in its definition, so it gets it dynamically, when I pass in another type.

回答1:

Not sure if there's a specific TypeScript way to deal with this, but you can use destructuring/spread for this:

const { type, ...validCredentials } = credentials;

This essentially picks type out of credentials, and validCredentials will contain the rest of the properties.

The reason .filter didn't work is because filter is a prototype method on arrays, not objects.

回答2:

If you insist on using Array methods, .map() is probably most helpful:

let obj2big = { a: 1, b: 2, c: 3 }

// let's assume what you want from this is { a: 1, c: 3 }

let arr = [obj2big]

let result = arr.map(i=>({a:i.a, c: i.c}))

console.log(result[0])

来源：https://stackoverflow.com/questions/52190091/filter-out-unneeded-object-properties-for-interface-in-typescript

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