Why databases back GraphQL - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a GraphQL query that fetches all users.

GraphQL

type Query { users: [1] }

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A[User]

BUser

CString

DInt

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'User' instead of '[User]' which returns a single user, not a list.

Using scalar types like String or Int for a list of users.

The users query returns a list of User objects, so the type is [User].

2fill in blank

medium

Complete the resolver function to fetch users from the database.

GraphQL

const resolvers = { Query: { users: async () => await [1].find() } }

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AGraphQL

BUserModel

Cdatabase

DfetchUsers

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'database' or 'GraphQL' which are not the model names.

Using a function name like 'fetchUsers' instead of the model.

The resolver calls the database model UserModel to find all users.

3fill in blank

hard

Fix the error in the GraphQL schema to correctly define a user type with id and name.

GraphQL

type User { id: ID! name: [1]! }

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AInt

BFloat

CBoolean

DString

Attempts:

3 left

💡 Hint

Common Mistakes

Using Int or Boolean for name which are incorrect types.

Using Float which is for decimal numbers.

The name field should be a String type in GraphQL.

4fill in blank

hard

Fill both blanks to write a resolver that fetches a user by id from the database.

GraphQL

const resolvers = { Query: { user: async (_, [1]) => await UserModel.[2]({ id: id }) } }

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Aargs

Bparams

CfindOne

DfindAll

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'params' instead of 'args' for resolver arguments.

Using 'findAll' which returns multiple records instead of one.

The resolver receives arguments as 'args' and uses 'findOne' to get a single user by id.

5fill in blank

hard

Fill all three blanks to define a mutation that adds a new user with name and returns the created user.

GraphQL

type Mutation { addUser(name: [1]!): [2] } const resolvers = { Mutation: { addUser: async (_, [3]) => { const user = new UserModel({ name: name }); await user.save(); return user; } } }

Drag options to blanks, or click blank then click option'

AString

BUser

Cargs

DInt

Attempts:

3 left

💡 Hint

Common Mistakes

Using Int for name which should be String.

Returning a scalar instead of User type.

Using wrong parameter name instead of 'args'.

The mutation takes a name of type String, returns a User type, and receives arguments as 'args'.