Postgres 尝试返回我的模型中不存在的列

Posted 2023-03-07

【中文标题】Postgres 尝试返回我的模型中不存在的列

【英文标题】：Postgress tries to return a column that does not exist in my model

【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
  

【参考方案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;
;