我怎麼能在sequelize中執行這個查詢

Question

SELECT *
FROM item_master
WHERE project_id IN (SELECT id FROM project_master
                     WHERE workspace_id in (SELECT id FROM workspace_master
                                            WHERE company_id = 4));

如何在sequelize - Node js 中執行此 mysql 查詢而不使用原始查詢？

Answer 1

如果您的模型結構類似於item_master 有許多project_master，並且project_master 有許多workspace_master，那麼以下sequelize查詢將與async/await一起應用。

當我編輯我的答案。正如您在評論中所說，您有模型結構，例如工作區有許多項目，項目有許多項目。然後 Sequelize 查詢將類似於：

const getResult = async (workpace_id, company_id = 4) => {
      const result = await workspace_master.findAll({
      subQuery: false,
      include: [
        {
          model: project_master,
          as: 'project_masters', // this is your alias defined in model
          attributes: ['id','foo','bar'],
          include: [
            {
              model: item_master,
              as: 'item_masters', // this is your alias defined in model
              attributes: ['id','foo','bar']
            }
          ]
        }
      ],
      where: {
         id: workspace_id,
         company_id: company_id
      }
    });
    
    if(!result || !result.length) return res.send('Something went wrong!');
    return res.send(result);
   }

現在嘗試一下，我希望這能解決您的問題。

Answer 2

const item = await Item.findAll({
    where: {
        status: { [Op.ne]: 99 },
        '$project.workspace.company_id$': { [Op.eq]: req.user.companyId }
    },
    include: {
        model: Project,
        as: 'project',
        attributes: ['id'],
        include: {
            model: Workspace,
            as: 'workspace',
            attributes: ['id', 'companyId'],
            where: {
                companyId: req.user.companyId
            },
        },
    }
})

這正在按我想要的方式工作。