Question

我正在编写一个带有sequelize和postgresl的Express API，postgresl托管在Heroku中。

当我尝试在下面的这些表之间创建关系时，它给了我这个错误

未处理的拒绝SequelizeDatabaseError：关系“问题”确实不存在

想法是，一个问题具有一个ID，一个标题，一个分数，具有多个答案和一个正确答案。

问题模型：

import { sequelize } from "../Util/db"
import { Users } from "./User"
import { Answer } from "./Answer"
const Sequelize = require("sequelize")
const Question = sequelize.define("question", {
  id: {
    type: Sequelize.INTEGER,
    primaryKey: true,
    autoIncrement: true
  },
  title: {
    type: Sequelize.STRING
  },
  score: {
    type: Sequelize.INTEGER,
    defaultValue: 0
  },
  content: {
    type: Sequelize.JSONB
  }
})

Question.belongsTo(Users)
Question.hasMany(Answer, { as: "Answers", foreignKey: "id"})
Question.hasOne(Answer, {as: "RightAnswer", foreignKey: "id"})
Question.sync({force: false}).then(() => {
  // Table created
  console.log("Question Table Synchronized")
})


interface QuestionType {
  id: number
  title: string
  score: number
  content: string
}


export { Question, QuestionType }

答案模型：

import { sequelize } from "../Util/db"
import { Users } from "./User"
import { Question } from "./Question"
const Sequelize = require("sequelize")
const Answer = sequelize.define("answer", {
  id: {
    type: Sequelize.INTEGER,
    primaryKey: true,
    autoIncrement: true
  },
  score: {
    type: Sequelize.INTEGER,
    defaultValue: 0
  },
  content: {
    type: Sequelize.JSONB
  }
})

Answer.belongsTo(Users)
Answer.sync({force: false}).then(() => {
  // Table created
  console.log("Answer Table Synchronized")
})


interface AnswerType {
  id: number
  score: number
  content: string
}


export { Answer, AnswerType }

创建脚本，由续集创建

Executing (default): CREATE TABLE IF NOT EXISTS "users" ("id"  SERIAL , "email" VARCHAR(255), "score" INTEGER DEFAULT 0, "name" VARCHAR(255), "createdAt" TIMESTAMP WITH TIME ZONE NOT NULL, "updatedAt" TIMESTAMP WITH TIME ZONE NOT NULL, PRIMARY KEY ("id"));
Executing (default): CREATE TABLE IF NOT EXISTS "answers" ("id"  SERIAL  REFERENCES "questions" ("id") ON DELETE CASCADE ON UPDATE CASCADE, "score" INTEGER DEFAULT 0, "content" JSONB, "createdAt" TIMESTAMP WITH TIME ZONE NOT NULL, "updatedAt" TIMESTAMP WITH TIME ZONE NOT NULL, "userId" INTEGER REFERENCES "users" ("id") ON DELETE SET NULL ON UPDATE CASCADE, PRIMARY KEY ("id"));
Executing (default): CREATE TABLE IF NOT EXISTS "questions" ("id"  SERIAL , "title" VARCHAR(255), "score" INTEGER DEFAULT 0, "content" JSONB, "createdAt" TIMESTAMP WITH TIME ZONE NOT NULL, "updatedAt" TIMESTAMP WITH TIME ZONE NOT NULL, "userId" INTEGER REFERENCES "users" ("id") ON DELETE SET NULL ON UPDATE CASCADE, PRIMARY KEY ("id"));
Executing (default): SELECT i.relname AS name, ix.indisprimary AS primary, ix.indisunique AS unique, ix.indkey AS indkey, array_agg(a.attnum) as column_indexes, array_agg(a.attname) AS column_names, pg_get_indexdef(ix.indexrelid) AS definition FROM pg_class t, pg_class i, pg_index ix, pg_attribute a WHERE t.oid = ix.indrelid AND i.oid = ix.indexrelid AND a.attrelid = t.oid AND t.relkind = 'r' and t.relname = 'users' GROUP BY i.relname, ix.indexrelid, ix.indisprimary, ix.indisunique, ix.indkey ORDER BY i.relname;
Unhandled rejection SequelizeDatabaseError: relation "questions" does not exist
    at Query.formatError (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/sequelize/lib/dialects/postgres/query.js:363:16)
    at query.catch.err (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/sequelize/lib/dialects/postgres/query.js:86:18)
    at tryCatcher (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/util.js:16:23)
    at Promise._settlePromiseFromHandler (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/promise.js:512:31)
    at Promise._settlePromise (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/promise.js:569:18)
    at Promise._settlePromise0 (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/promise.js:614:10)
    at Promise._settlePromises (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/promise.js:689:18)
    at Async._drainQueue (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/async.js:133:16)
    at Async._drainQueues (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/async.js:143:10)
    at Immediate.Async.drainQueues (/home/mazzardo/Codigo/pas/node-typescript-boilerplate/node_modules/bluebird/js/release/async.js:17:14)
    at runCallback (timers.js:651:20)
    at tryOnImmediate (timers.js:624:5)
    at processImmediate [as _immediateCallback] (timers.js:596:5)
Executing (default): SELECT i.relname AS name, ix.indisprimary AS primary, ix.indisunique AS unique, ix.indkey AS indkey, array_agg(a.attnum) as column_indexes, array_agg(a.attname) AS column_names, pg_get_indexdef(ix.indexrelid) AS definition FROM pg_class t, pg_class i, pg_index ix, pg_attribute a WHERE t.oid = ix.indrelid AND i.oid = ix.indexrelid AND a.attrelid = t.oid AND t.relkind = 'r' and t.relname = 'questions' GROUP BY i.relname, ix.indexrelid, ix.indisprimary, ix.indisunique, ix.indkey ORDER BY i.relname;
User Table Synchronized
Question Table Synchronized

旁注，我在项目上使用打字稿

Answer 1

这可能是由于每个sync的竞争情况。每次以异步方式调用sync时，由于我们具有外键约束，因此有可能未创建其中一个正在寻找关联的表。我认为您应该使用

sequelize.sync()应该可以解决您的问题。

从docs

因为Sequelize做很多魔术，所以你必须打电话设置关联后的Sequelize.sync！这样做可以让你以下：

关联关系错误

1 个答案: