我有這樣的關(guān)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;我已經(jīng)設(shè)法通過以下代碼達(dá)到了這樣的目的：query = {   include: [{       model: models.Clients,       attributes: [],       require: true,    }],   where: { '$Clients.id$': 1 }}models.Programs.findAll(query) // This works其中產(chǎn)生：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;這有效，但是當(dāng)我嘗試限制它時(shí)，出現(xiàn)錯(cuò)誤。代碼：query = {   include: [{       model: models.Clients,       attributes: [],       require: true,    }],   limit: 0,   offset: 10,   where: { '$Clients.id$': 1 }}models.Programs.findAll(query) // This fails其中產(chǎn)生：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`;錯(cuò)誤： DatabaseError [SequelizeDatabaseError]: Unknown column 'Clients.id' in 'where clause'注意：我使用的是 MySQL 數(shù)據(jù)庫(kù)。有沒有更簡(jiǎn)單的方法來解決這個(gè)問題并為 SQL 生成所需的（或類似的）結(jié)果？