没有子节的GraphQL变异

您将字段的类型定义为UserType.即使它是一个变异,它仍然遵循与查询相同的规则和行为.因为UserType是对象类型,所以它需要嵌套字段.

mutation _ {
  updateCurrentUser(fullName: "Syava", email: "fake@gmail.com") {
    fullName
    email
  }
}
// would respond with { fullName: 'Syava', email: 'fake@gmail.com' }

如果您不希望变异返回User,则可以将其类型声明为GraphQLBoolean - 例如,这是一个标量,并且没有任何嵌套字段.

{
  type: GraphQLBoolean,
  args: {
    fullName: { type: GraphQLString },
    email: { type: new GraphQLNonNull(emailType) },
    password: { type: GraphQLString },
  },
  resolve: async (root, { fullName, email, password }, { rootValue }) => {
    const user = await User.findById(rootValue.req.user.id);
    user.fullName = fullName;
    user.password = password; // or hashed to not store plain text passwords
    return user.save(); // assuming save returns boolean; depends on the library you use
  }
}