Question

我有一个数据库。我有一个“user”表

我正在尝试使用REST

创建我的第一个sequelize api

然而，当它执行我的查询时，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';

你可以看到它试图使用一个名为users的表但是这个表不存在。

以下是我的一些代码：

请告诉我你是否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

要解决您的问题，您需要在选项对象中设置freezeTableName = true。

e.g。

    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，但是从他们的文档中读取 - 可能是你遗失了 -

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

即。表创建部分......

谢谢

sequelize给出了错误的表名

2 个答案: