<p><pre class="brush:php;toolbar:false;">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));</pre>
<p>如何在sequelize - Node js 中执行此 mysql 查询而不使用原始查询？</p>
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);
}
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如果您的模型结构类似于
item_master有许多project_master，并且project_master有许多workspace_master，那么以下sequelize查询将与async/await一起应用。当我编辑我的答案时。正如您在评论中所说，您有模型结构，例如工作区有许多项目，项目有许多项目。然后 Sequelize 查询将类似于：
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); }现在尝试一下，我希望这能解决您的问题。
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 }, }, } })这正在按我想要的方式工作。