Drizzle Team•2y ago

InferModel including relations?

Hi, How do you use InferModel to obtain a type that includes relations? The results for an executed query has the inferred types but I'm not sure how to infer the type more generally? Thanks

13 Replies

gerty1•17mo ago

Bumping this, I'd love to know as well

ChronicStone•16mo ago

Hi @danielsharvey @gerty1 Did one of you happen to have found a workaround for this ?

Ꚃimon•16mo ago

@chronicstone You can get the returntype of the function where you fetch the data and do the following

export type InitialPostsType = NonNullable<
  Awaited<ReturnType<typeof getDataFunction>>
>

export type InitialPostsType = NonNullable<
  Awaited<ReturnType<typeof getDataFunction>>
>

ChronicStone•16mo ago

@intelligently Good point, thanks

danielOP•16mo ago

@gerty1 @chronicstone Sorry for the slow response! Similar to @intelligently 's answer:

export type TypeIncludingWith = Awaited<ReturnType<MyController['getItems']>>;
class MyController {
  public async getItems() {
    const rows = await this.db.query.inventoryItems.findMany({
      with: {
        designInfo: true,
      },
      where: ...
      offset: offset,
      limit: limit,
    });
    return rows;
  }
}

export type TypeIncludingWith = Awaited<ReturnType<MyController['getItems']>>;
class MyController {
  public async getItems() {
    const rows = await this.db.query.inventoryItems.findMany({
      with: {
        designInfo: true,
      },
      where: ...
      offset: offset,
      limit: limit,
    });
    return rows;
  }
}

fermentfan•16mo ago

That's not really a good fitting solution everywhere right? I don't want to write code just so I can define my types correctly... Is there another way? In my case I want to define a function parameter that is the type of a single relation in my table.

Angelelz•16mo ago

The Relations API can output too many types. You can have drizzle give you whatever you want. Depending on the shape you give it, the type will be different. How could you get a type if you don't write it? Now, what we've done where we need the relational API and we need to export the types, we just create them like so:

export const vehicles = mysqlTable(
  ...
);
export const inspections = mysqlTable(
  ...
);
export type Vehicle = InferModel<typeof vehicles>;
export type Inspection = InferModel<typeof inspections>;
export type VehicleWithInspections = Vehicle & { inspections: Inspection[] };

export const vehicles = mysqlTable(
  ...
);
export const inspections = mysqlTable(
  ...
);
export type Vehicle = InferModel<typeof vehicles>;
export type Inspection = InferModel<typeof inspections>;
export type VehicleWithInspections = Vehicle & { inspections: Inspection[] };

InferModel is a type helper imported from drizzle-orm.

danielOP•16mo ago

👍🏻 I've also used this mechanism in places.

Ꚃimon•16mo ago

Issue is that this will give you the wrong typehints if you're only selecting certain columns instead of all of them

Andrii Sherman•16mo ago

I guess @Dan Kochetov can help you with a proper type for it

Dan•16mo ago

https://github.com/drizzle-team/drizzle-orm/issues/695

GitHub

Implement infering table model with relations · Issue #695 · drizzl...

danielOP•15mo ago

@Andrew Sherman @Dan Kochetov Related question: Is there an official way to get the relation metadata? I'm currently doing the following:

import { createTableRelationsHelpers, extractTablesRelationalConfig, Table } from 'drizzle-orm';

/// Drizzle relations metadata
export const { tables: t } = extractTablesRelationalConfig(
  s,
  createTableRelationsHelpers,
);

import { createTableRelationsHelpers, extractTablesRelationalConfig, Table } from 'drizzle-orm';

/// Drizzle relations metadata
export const { tables: t } = extractTablesRelationalConfig(
  s,
  createTableRelationsHelpers,
);

Dan•15mo ago

no official way for now, but this should work fine

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InferModel including relations?