续集给出错误的表名

我有一个数据库。 我有一个“ user ”表

我正在尝试使用sequelize创build我的第一个REST api

然而，当它执行我的查询，我在控制台中得到以下内容：

 SELECT `id`, `username`, `password`, `name`, `organization_id`, `type_id`, `join_date` FROM `users` AS `user` WHERE `user`.`id` = '1'; 

正如你所看到的，它试图使用一个名为users的表，但是这个表不存在。

这是我的一些代码：

请告诉我，如果你需要更多我不确定哪里出错了？ ：S

  var User = sequelize.define('user', { id: DataTypes.INTEGER, username: DataTypes.STRING, password: DataTypes.STRING, name: DataTypes.STRING, organization_id: DataTypes.INTEGER, type_id: DataTypes.INTEGER, join_date: DataTypes.STRING }, { instanceMethods: { retrieveAll: function(onSuccess, onError) { User.findAll({}, {raw: true}) .ok(onSuccess).error(onError); }, retrieveById: function(user_id, onSuccess, onError) { User.find({where: {id: user_id}}, {raw: true}) .success(onSuccess).error(onError); }, add: function(onSuccess, onError) { var username = this.username; var password = this.password; var shasum = crypto.createHash('sha1'); shasum.update(password); password = shasum.digest('hex'); User.build({ username: username, password: password }) .save().ok(onSuccess).error(onError); }, updateById: function(user_id, onSuccess, onError) { var id = user_id; var username = this.username; var password = this.password; var shasum = crypto.createHash('sha1'); shasum.update(password); password = shasum.digest('hex'); User.update({ username: username,password: password},{where: {id: id} }) .success(onSuccess).error(onError); }, removeById: function(user_id, onSuccess, onError) { User.destroy({where: {id: user_id}}).success(onSuccess).error(onError); } } }); 

Answer 1

为了解决你的问题，你需要在你的options对象中设置freezeTableName = true。

例如

  var User = sequelize.define('user', { id: DataTypes.INTEGER, username: DataTypes.STRING, password: DataTypes.STRING, name: DataTypes.STRING, organization_id: DataTypes.INTEGER, type_id: DataTypes.INTEGER, join_date: DataTypes.STRING }, { freezeTableName = true, instanceMethods: { retrieveAll: function(onSuccess, onError) { User.findAll({}, {raw: true}) .ok(onSuccess).error(onError); }, retrieveById: function(user_id, onSuccess, onError) { User.find({where: {id: user_id}}, {raw: true}) .success(onSuccess).error(onError); }, add: function(onSuccess, onError) { var username = this.username; var password = this.password; var shasum = crypto.createHash('sha1'); shasum.update(password); password = shasum.digest('hex'); User.build({ username: username, password: password }) .save().ok(onSuccess).error(onError); }, updateById: function(user_id, onSuccess, onError) { var id = user_id; var username = this.username; var password = this.password; var shasum = crypto.createHash('sha1'); shasum.update(password); password = shasum.digest('hex'); User.update({ username: username,password: password},{where: {id: id} }) .success(onSuccess).error(onError); }, removeById: function(user_id, onSuccess, onError) { User.destroy({where: {id: user_id}}).success(onSuccess).error(onError); } } }); 

Answer 2

我之前没有使用过sequelize，但从他们的文档中读取 – 可能是你错过了 –

User.sync({force: true}).then(function () { // Table created return User.create({ firstName: 'John', lastName: 'Hancock' }); });

即表创build部分…

谢谢