Sequelize - 使用 where 和 limit 查询多对多关系
Sequelize - Query with Many-to-Many relationship using where and limit
我有这样的关系:
Clients -> ProgramsClients <- Programs
我想做的基本上是:
SELECT * FROM Programs p JOIN ProgramsClients pc on p.id = pc.programId WHERE pc.clientId = 1 LIMIT 0, 100;
我已经通过以下代码成功实现了这样的目标:
query = {
include: [{
model: models.Clients,
attributes: [],
require: true,
}],
where: { '$Clients.id$': 1 }
}
models.Programs.findAll(query) // This works
生成:
SELECT [...]
FROM `programs` AS `Programs` LEFT OUTER JOIN (
`ProgramsClients` AS `Clients->ProgramsClients`
INNER JOIN `clients` AS `Clients`
ON `Clients`.`id` = `Clients->ProgramsClients`.`ClientId`)
ON `Programs`.`id` = `Clients->ProgramsClients`.`ProgramId`
WHERE `Clients`.`id` = 1;
这有效,但是当我尝试限制它时,出现错误。
代码:
query = {
include: [{
model: models.Clients,
attributes: [],
require: true,
}],
limit: 0,
offset: 10,
where: { '$Clients.id$': 1 }
}
models.Programs.findAll(query) // This fails
生成:
SELECT [...]
FROM (SELECT `Programs`.`id`, `Programs`.`name`, `Programs`.`description`, `Programs`.`createdAt`, `Programs`.`updatedAt`
FROM `programs` AS `Programs` WHERE `Clients`.`id` = 1 LIMIT 0, 10) AS `Programs`
LEFT OUTER JOIN ( `ProgramsClients` AS `Clients->ProgramsClients`
INNER JOIN `clients` AS `Clients`
ON `Clients`.`id` = `Clients->ProgramsClients`.`ClientId`)
ON `Programs`.`id` = `Clients->ProgramsClients`.`ProgramId`;
错误:
DatabaseError [SequelizeDatabaseError]: Unknown column 'Clients.id' in 'where clause'
注意:我正在使用 MySQL 数据库。
有没有更简单的方法来解决这个问题并为 SQL 生成所需的(或类似的)结果?
提前致谢
我停了一下。当我回来时,我设法解决了它。
基本上,我误读了文档中的 super many-to-many section。
您可以简单地定义 One-to-many 关系(即使您正在使用 many-to-many 关系)与关联的 table(在本例中为 ProgramsClients),然后包括 ProgramsClients 和做你想做的。 (您必须为此声明 ProgramsClients 的 id 列)。
query = {
include: [{
model: models.ProgramsClients,
as: 'programsclient'
attributes: [],
require: true,
where: { clientId: 1 }
}],
limit: 0,
offset: 10,
}
我有这样的关系:
Clients -> ProgramsClients <- Programs
我想做的基本上是:
SELECT * FROM Programs p JOIN ProgramsClients pc on p.id = pc.programId WHERE pc.clientId = 1 LIMIT 0, 100;
我已经通过以下代码成功实现了这样的目标:
query = {
include: [{
model: models.Clients,
attributes: [],
require: true,
}],
where: { '$Clients.id$': 1 }
}
models.Programs.findAll(query) // This works
生成:
SELECT [...]
FROM `programs` AS `Programs` LEFT OUTER JOIN (
`ProgramsClients` AS `Clients->ProgramsClients`
INNER JOIN `clients` AS `Clients`
ON `Clients`.`id` = `Clients->ProgramsClients`.`ClientId`)
ON `Programs`.`id` = `Clients->ProgramsClients`.`ProgramId`
WHERE `Clients`.`id` = 1;
这有效,但是当我尝试限制它时,出现错误。 代码:
query = {
include: [{
model: models.Clients,
attributes: [],
require: true,
}],
limit: 0,
offset: 10,
where: { '$Clients.id$': 1 }
}
models.Programs.findAll(query) // This fails
生成:
SELECT [...]
FROM (SELECT `Programs`.`id`, `Programs`.`name`, `Programs`.`description`, `Programs`.`createdAt`, `Programs`.`updatedAt`
FROM `programs` AS `Programs` WHERE `Clients`.`id` = 1 LIMIT 0, 10) AS `Programs`
LEFT OUTER JOIN ( `ProgramsClients` AS `Clients->ProgramsClients`
INNER JOIN `clients` AS `Clients`
ON `Clients`.`id` = `Clients->ProgramsClients`.`ClientId`)
ON `Programs`.`id` = `Clients->ProgramsClients`.`ProgramId`;
错误:
DatabaseError [SequelizeDatabaseError]: Unknown column 'Clients.id' in 'where clause'
注意:我正在使用 MySQL 数据库。
有没有更简单的方法来解决这个问题并为 SQL 生成所需的(或类似的)结果?
提前致谢
我停了一下。当我回来时,我设法解决了它。
基本上,我误读了文档中的 super many-to-many section。
您可以简单地定义 One-to-many 关系(即使您正在使用 many-to-many 关系)与关联的 table(在本例中为 ProgramsClients),然后包括 ProgramsClients 和做你想做的。 (您必须为此声明 ProgramsClients 的 id 列)。
query = {
include: [{
model: models.ProgramsClients,
as: 'programsclient'
attributes: [],
require: true,
where: { clientId: 1 }
}],
limit: 0,
offset: 10,
}