Sequelize - 加入多列

Question

我喜欢把下面的查询转换成sequelize code

select * from table_a 
inner join table_b 
on table_a.column_1 = table_b.column_1
and table_a.column_2 = table_b.column_2

我尝试了许多方法并遵循了许多提供的解决方案，但我无法从 sequelize 代码中实现所需的查询。

我达到的最大值如下：

select * from table_a 
inner join table_b 
on table_a.column_1 = table_b.column_1

我也想要第二个条件。

and table_a.column_2 = table_b.column_2

有什么合适的方法来实现吗？

Answer 1

关于@TophatGordon 在接受的答案评论中的疑问：如果我们需要在模型中设置任何关联。
还通过了仍处于 open 状态的github issue raised back in 2012。
所以我也遇到了同样的情况，并试图为左外连接设置我自己的ON 条件。
当我直接尝试在Table1.findAll(...include Table2 with ON condition...) 中使用on: {...} 时，它不起作用。 它抛出了一个错误：

EagerLoadingError [SequelizeEagerLoadingError]: Table2 is not associated to Table1!

我的用例是将 Table1 中的两个非主键列与左外连接中 Table2 中的两个列进行匹配。我将展示我是如何以及取得了什么成就的：

---

不要被表名和列名弄糊涂了，因为我必须从我使用的原始名称中更改它们。

所以我不得不在 Table1(Task) 中创建一个关联，例如：

Task.associate = (models) => {    

Task.hasOne(models.SubTask, {
        foreignKey: 'someId', // <--- one of the column of table2 - SubTask: not a primary key here in my case; can be primary key also
        sourceKey: 'someId', // <---  one of the column of table1 - Task: not a primary key here in my case; can be a primary key also
        scope: {
            [Op.and]: sequelize.where(sequelize.col("Task.some_id_2"),
                // '=',
                Op.eq, // or you can use '=',
                sequelize.col("subTask.some_id_2")),
        },
        as: 'subTask',
        // no constraints should be applied if sequelize will be creating tables and unique keys are not defined, 
        //as it throws error of unique constraint            
        constraints: false, 
    });
};

所以查找查询看起来像这样：

Task.findAll({
    where: whereCondition,
    // attributes: ['id','name','someId','someId2'],
    include: [{
        model: SubTask, as: 'subTask', // <-- model name and alias name as defined in association 
        attributes: [], // if no attributes needed from SubTask - empty array
    },
    ],
});

结果查询：

• 一个匹配条件取自 [foreignKey] = [sourceKey]
• 第二个匹配条件由sequelize.where(...)获得，用于scope:{...}

select
  "Task"."id",
  "Task"."name",
  "Task"."some_id" as "someId",
  "Task"."some_id_2" as "someId2"
from
  "task" as "Task"
left outer join "sub_task" as "subTask" on
  "Task"."some_id" = "subTask"."some_id"
  and "Task"."some_id_2" = "subTask"."some_id_2";

---

另一种实现与上述相同的方法来解决在包含中使用 Table1 时出现的问题，即当 Table1 显示为 2 级表或包含在其他表中时 - 比如说 Table0

Task.associate = (models) => {    

Task.hasOne(models.SubTask, {
        foreignKey: 'someId', // <--- one of the column of table2 - SubTask: not a primary key here in my case; can be primary key also
        sourceKey: 'someId', // <---  one of the column of table1 - Task: not a primary key here in my case; can be a primary key also
        as: 'subTask',
        // <-- removed scope -->
        // no constraints should be applied if sequelize will be creating tables and unique keys are not defined, 
        //as it throws error of unique constraint            
        constraints: false, 
    });
};

所以来自 Table0 的查找查询看起来像这样：也不考虑foreignKey 和sourceKey，因为我们现在将使用自定义on: {...}

Table0.findAll({
    where: whereCondition,
    // attributes: ['id','name','someId','someId2'],
    include: {
        model: Task, as: 'Table1AliasName', // if association has been defined as alias name 
        include: [{
            model: SubTask, as: 'subTask', // <-- model name and alias name as defined in association 
            attributes: [], // if no attributes needed from SubTask - empty array
            on: {
                [Op.and]: [
                    sequelize.where(
                        sequelize.col('Table1AliasName_OR_ModelName.some_id'),
                        Op.eq, // '=',
                        sequelize.col('Table1AliasName_OR_ModelName->subTask.some_id')
                    ),
                    sequelize.where(
                        sequelize.col('Table1AliasName_OR_ModelName.some_id_2'),
                        Op.eq, // '=',
                        sequelize.col('Table1AliasName_OR_ModelName->subTask.some_id_2')
                    ),
                ],
            },
        }],
    }
});

如果您的表已经创建，请跳过以下部分...

---

将约束设置为 false，就像 sequelize 尝试创建第二个表（子任务）一样，由于以下查询，它可能会抛出错误 (DatabaseError [SequelizeDatabaseError]: there is no unique constraint matching given keys for referenced table "task")：

如果不存在则创建表 "sub_task" ("some_id" INTEGER, "some_id_2" INTEGER 在更新时删除级联时引用“任务”（“some_id”） 级联，“数据”整数）；

如果我们设置 constraint: false，它会在下面创建这个查询，而不是在我们引用非主列时抛出唯一约束错误：

如果不存在则创建表 "sub_task" ("some_id" INTEGER, "some_id_2" INTEGER, "data" INTEGER);

Answer 2

您需要在JOIN 语句中定义自己的on 子句

ModelA.findAll({
    include: [
        {
            model: ModelB,
            on: {
                col1: sequelize.where(sequelize.col("ModelA.col1"), "=", sequelize.col("ModelB.col1")),
                col2: sequelize.where(sequelize.col("ModelA.col2"), "=", sequelize.col("ModelB.col2"))
            },
            attributes: [] // empty array means that no column from ModelB will be returned
        }
    ]
}).then((modelAInstances) => {
    // result...
});