Postgres 尝试返回我的模型中不存在的列答案

【问题标题】：Postgress tries to return a column that does not exist in my modelPostgres 尝试返回我的模型中不存在的列
【发布时间】：2021-02-03 01:56:29
【问题描述】：

我使用 graphql API 并尝试将数据插入 Postgress 表并收到错误：

"message": "column \"UserId\" does not exist",

还有我的原始查询：

Executing (default): INSERT INTO "Recipes" 
("id","title","ingredients","direction","createdAt","updatedAt","userId") VALUES 
(DEFAULT,$1,$2,$3,$4,$5,$6) 
 RETURNING 
"id","title","ingredients","direction","createdAt","updatedAt","userId","UserId";

问题是列 UserId 不在我的模型中，但 userId 是！而且我不知道为什么 Postgres 试图返回 UserId 列。我的模特。用户：

module.exports = (sequelize, DataTypes) => {
  class User extends Model {
    static associate(models) {
      User.hasMany(models.Recipe)
    }
  };
  User.init({
    name: {
        type: DataTypes.STRING,
        allowNull: false
    },
    email: {
        type: DataTypes.STRING,
        allowNull: false
    },
    password: {
        type: DataTypes.STRING,
        allowNull: false
    }
  }, {
    sequelize,
    modelName: 'User',
  });
  return User;
};

食谱：

module.exports = (sequelize, DataTypes) => {
  class Recipe extends Model {
    static associate(models) {
      Recipe.belongsTo(models.User, { foreignKey: 'userId' })
    }
  };
  Recipe.init({
    title: {
      type: DataTypes.STRING,
      allowNull: false
    },
    ingredients: {
        type: DataTypes.STRING,
        allowNull: false
    },
    direction: {
        type: DataTypes.STRING,
        allowNull: false
    }
  }, {
    sequelize,
    modelName: 'Recipe',
  });
  return Recipe;
};

我的 graphql 架构：

const typeDefs = gql`
    type User {
        id: Int!
        name: String!
        email: String!
        recipes: [Recipe!]!
      }

    type Recipe {
        id: Int!
        title: String!
        ingredients: String!
        direction: String!
        user: User!
    }

    type Query {
        user(id: Int!): User
        allRecipes: [Recipe!]!
        recipe(id: Int!): Recipe
    }

    type Mutation {
        createUser(name: String!, email: String!, password: String!): User!
        createRecipe(
          userId: Int!
          title: String!
          ingredients: String!
          direction: String!
        ): Recipe!
    }
`

graphql 解析器：

  Mutation: {
      async createRecipe (root, { userId, title, ingredients, direction }, { models }) {
          return models.Recipe.create({ userId, title, ingredients, direction })
      }
  }

还有我的 grapql 请求：

mutation {
  createRecipe(
    userId: 1
    title: "Sample 2"
    ingredients: "Salt, Pepper"
    direction: "Add salt, Add pepper"
  ) {
    id
    title
    ingredients
    direction
  }
}

【问题讨论】：

标签： node.js postgresql graphql sequelize.js database-migration

【解决方案1】：

我在做一个项目时遇到了类似的问题；指定父母和孩子的关系解决了我的问题：

用户：

module.exports = (sequelize, DataTypes) => {

class User extends Model {
    static associate(models) {
      User.hasMany(models.Recipe, {as: 'recipes', foreignKey:'userId'})
    }
  };
  User.init({
    name: {
        type: DataTypes.STRING,
        allowNull: false
    },
    email: {
        type: DataTypes.STRING,
        allowNull: false
    },
    password: {
        type: DataTypes.STRING,
        allowNull: false
    }
  }, {
    sequelize,
    modelName: 'User',
  });
  return User;
};

食谱：

module.exports = (sequelize, DataTypes) => {
  class Recipe extends Model {
    static associate(models) {
      Recipe.belongsTo(models.User, { foreignKey: 'userId' })
    }
  };
  Recipe.init({
    title: {
      type: DataTypes.STRING,
      allowNull: false
    },
    ingredients: {
        type: DataTypes.STRING,
        allowNull: false
    },
    direction: {
        type: DataTypes.STRING,
        allowNull: false
    }
  }, {
    sequelize,
    modelName: 'Recipe',
  });
  return Recipe;
};

【讨论】：