Sequelize - 加入多列

Sequelize - Join with multiple column

我喜欢将以下查询转换为 sequelize 代码

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

我尝试了很多方法并遵循了很多提供的解决方案,但我无法通过 sequelize 代码实现所需的查询。

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

第二个条件我也要

and table_a.column_2 = table_b.column_2

有什么合适的方法可以实现吗?

您需要定义 JOIN 语句的自己的 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...
});

关于@TophatGordon 在已接受答案的评论中的疑问:是否需要在模型中设置任何关联。
还经历了github issue raised back in 2012,还处于open状态。
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,它没有用。 它抛出一个错误:

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

我的用例是在左外连接中将表 1 中的两个 non-primary-key 列匹配到表 2 中的两个列。我将展示我是如何实现的以及实现了什么:


不要被 table 名称和列名弄糊涂了,因为我不得不更改它们,而不是我使用的原始名称。

所以我必须在 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, 
    });
};

因此查找查询如下所示:

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(...)得到,用在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 时的问题,即当 Table1 显示为第 2 级 table 或被其他 table 包含时 - 例如 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, 
    });
};

所以从 Table0 中查找的查询看起来像这样:而且 foreignKey 和 sourceKey 也不会被考虑,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 table 已经创建,请跳过下面的部分...


将约束设置为 false,就好像 sequelize 试图创建第二个 table(SubTask) 它可能会抛出错误 (DatabaseError [SequelizeDatabaseError]: there is no unique constraint matching given keys for referenced table "task") 由于以下查询:

create table if not exists "sub_task" ("some_id" INTEGER, "some_id_2" INTEGER references "task" ("some_id") on delete cascade on update cascade, "data" INTEGER);

如果我们设置 constraint: false,它会创建下面的查询,当我们引用 non-primary 列时不会抛出唯一约束错误:

create table if not exists "sub_task" ("some_id" INTEGER, "some_id_2" INTEGER, "data" INTEGER);