Question

我可以使用sequelize.js对开发数据库中的表执行INSERT INTO命令，但不能对测试数据库中的表执行操作。

尽管进行了深入研究，但仍无法解决问题。

一个类似的问题已经发布在这里，尽管我无法用答案来回答我的问题： sequelize with postgres database not working after migration from mysql

这是我相关的迁移文件：

'use strict';
module.exports = {
  up: (queryInterface, Sequelize) => {
    return queryInterface.createTable('Trees', {
      id: {
        allowNull: false,
        autoIncrement: true,
        primaryKey: true,
        type: Sequelize.INTEGER
      },
      title: {
        type: Sequelize.STRING,
        allowNull: false
      },
      content: {
        type: Sequelize.STRING(10000),
        allowNull: false
      },
      createdAt: {
        allowNull: false,
        type: Sequelize.DATE
      },
      updatedAt: {
        allowNull: false,
        type: Sequelize.DATE
      }
    });
  },
  down: (queryInterface, Sequelize) => {
    return queryInterface.dropTable('Trees');
  }
};

这是模型文件：

'use strict';
module.exports = (sequelize, DataTypes) => {
  var Tree = sequelize.define('Tree', {
    title: {
      type: DataTypes.STRING,
      allowNull: false
    },
    content: {
      type: DataTypes.STRING(10000),
      allowNull: false
    }
  }, {});
  Tree.associate = function(models) {
    // associations can be defined here
  };
  return Tree;
};

以下是访问数据库的代码：

const setTree = (treeVal, callback) => {
  console.log(`this would be the part`);
  Tree.create({
    title: 'Tree',
    content: JSON.stringify(treeVal)
  })
  .then((treeStr) => {
    let primaryTopics = JSON.parse(treeStr.content);
    callback(null, primaryTopics);
  })
  .catch((err) => {
    callback(err);
  });
}

这是通过module.exports方法导出的：

  callTree(callback) {
    return Tree.findOne({
      where: {
        title: 'Tree'
      }
    })
    .then((treeStr) => {
      if (treeStr === null) {
        return callback(`not defined yet`);
      }
      let primaryTopics = treeStr.content;
      primaryTopics = JSON.parse(primaryTopics);
      callback(null, primaryTopics);
    })
    .catch((err) => {
      callback(err);
    });
  }

我在这里将这种方法用于集成测试（PrimaryTopic表位于同一数据库中，尝试运行它时不会收到任何错误）：

  beforeEach((done) => {
    this.primaryTopic;
    sequelize.sync({force: true}).then((res) => {

      PrimaryTopic.create({
        title: 'Title: Hello World',
        content: '<p>Content: Hello World</p>'
      })
      .then((primaryTopic) => {
        this.primaryTopic = primaryTopic;
        treeQueries.buildTree((err, res) => {
          if (err) {
            console.error(err);
          }
        });
        done();
      })
      .catch((err) => {
        console.log(err);
        done();
      });

    });
  });

我已经在所有代码中搜索了可能的错误，但是还没有发现任何东西。

我可以使用psql访问测试数据库中的Trees表，尽管它是空的。

我可以使用相同的代码毫无问题地将值插入开发数据库中的Trees表中。

这是我尝试运行测试（使用jasmine.js进行测试）时收到的错误：

{ SequelizeDatabaseError: relation "Trees" does not exist
    at Query.formatError (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/sequelize/lib/dialects/postgres/query.js:363:16)
    at query.catch.err (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/sequelize/lib/dialects/postgres/query.js:86:18)
    at tryCatcher (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/util.js:16:23)
    at Promise._settlePromiseFromHandler (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/promise.js:512:31)
    at Promise._settlePromise (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/promise.js:569:18)
    at Promise._settlePromise0 (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/promise.js:614:10)
    at Promise._settlePromises (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/promise.js:689:18)
    at Async._drainQueue (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/async.js:133:16)
    at Async._drainQueues (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/async.js:143:10)
    at Immediate.Async.drainQueues [as _onImmediate] (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/bluebird/js/release/async.js:17:14)
    at runCallback (timers.js:756:18)
    at tryOnImmediate (timers.js:717:5)
    at processImmediate [as _immediateCallback] (timers.js:697:5)
  name: 'SequelizeDatabaseError',
  parent: 
   { error: relation "Trees" does not exist
    at Connection.parseE (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:553:11)
    at Connection.parseMessage (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:378:19)
    at Socket.<anonymous> (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:119:22)
    at Socket.emit (events.js:160:13)
    at addChunk (_stream_readable.js:269:12)
    at readableAddChunk (_stream_readable.js:256:11)
    at Socket.Readable.push (_stream_readable.js:213:10)
    at TCP.onread (net.js:599:20)
     name: 'error',
     length: 104,
     severity: 'ERROR',
     code: '42P01',
     detail: undefined,
     hint: undefined,
     position: '13',
     internalPosition: undefined,
     internalQuery: undefined,
     where: undefined,
     schema: undefined,
     table: undefined,
     column: undefined,
     dataType: undefined,
     constraint: undefined,
     file: 'parse_relation.c',
     line: '1160',
     routine: 'parserOpenTable',
     sql: 'INSERT INTO "Trees" ("id","title","content","createdAt","updatedAt") VALUES (DEFAULT,\'Tree\',\'[{"title":"Title: Hello World","id":1,"secondaryTopics":[]}]\',\'2018-08-14 06:38:37.243 +00:00\',\'2018-08-14 06:38:37.243 +00:00\') RETURNING *;' },
  original: 
   { error: relation "Trees" does not exist
    at Connection.parseE (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:553:11)
    at Connection.parseMessage (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:378:19)
    at Socket.<anonymous> (/home/siddhartha/Documents/01-Studio/01-Commercial-Public/01-Komodo/2018-resources/node_modules/pg/lib/connection.js:119:22)
    at Socket.emit (events.js:160:13)
    at addChunk (_stream_readable.js:269:12)
    at readableAddChunk (_stream_readable.js:256:11)
    at Socket.Readable.push (_stream_readable.js:213:10)
    at TCP.onread (net.js:599:20)
     name: 'error',
     length: 104,
     severity: 'ERROR',
     code: '42P01',
     detail: undefined,
     hint: undefined,
     position: '13',
     internalPosition: undefined,
     internalQuery: undefined,
     where: undefined,
     schema: undefined,
     table: undefined,
     column: undefined,
     dataType: undefined,
     constraint: undefined,
     file: 'parse_relation.c',
     line: '1160',
     routine: 'parserOpenTable',
     sql: 'INSERT INTO "Trees" ("id","title","content","createdAt","updatedAt") VALUES (DEFAULT,\'Tree\',\'[{"title":"Title: Hello World","id":1,"secondaryTopics":[]}]\',\'2018-08-14 06:38:37.243 +00:00\',\'2018-08-14 06:38:37.243 +00:00\') RETURNING *;' },
  sql: 'INSERT INTO "Trees" ("id","title","content","createdAt","updatedAt") VALUES (DEFAULT,\'Tree\',\'[{"title":"Title: Hello World","id":1,"secondaryTopics":[]}]\',\'2018-08-14 06:38:37.243 +00:00\',\'2018-08-14 06:38:37.243 +00:00\') RETURNING *;' }

Here's a link to the full repository.

Answer 1

我知道这是旧的，但它可以帮助其他人面临同样的问题。我从 mysql 迁移到 postgres 并遇到了同样的问题，因为我的表名以大写（“Home”）开头，我试图添加一个以小写开头的表名“home”的约束。

Answer 2

您需要在数据模型的选项中将FrozenTableName设置为true。否则，sequelize将尝试搜索Trees表

module.exports = (sequelize, DataTypes) => {
  var Tree = sequelize.define('Tree', {
    title: {
      type: DataTypes.STRING,
      allowNull: false
    },
    content: {
      type: DataTypes.STRING(10000),
      allowNull: false
    }
  }, {
freezeTableName: true
});
  Tree.associate = function(models) {
    // associations can be defined here
  };
  return Tree;
};

有关完整文档：https://sequelize.org/v3/docs/models-definition/#configuration

Answer 3

如果您使用扩展模型接口，您可以简单地将 tableName: "Trees" 附加到您的模型配置中。

我发现解决的最简单方法是在模型上显式设置 tableName。正如其他人所提到的，sequelize 默认使用模型的复数形式作为表名。例如 Tree，变成 Trees。

当您查询时，sequelize 会处理与您的模型 Tree 同名的表。通过在模型中定义 tableName，sequelize 应该搜索正确的表。将 tableName: "Trees" 附加到您的模型配置中，即：

module.exports = (sequelize, DataTypes) => {
    class Tree extends Model {
        ...
    }
    Tree.init(
        {
            title: DataTypes.STRING,
            ...
        },
        {
            sequelize,
            modelName: 'Tree',
            tableName: 'Trees',
        }
    );
    return Tree;
};

Sequelize错误：关系不存在

3 个答案: