【发布时间】:2017-05-18 23:29:47
【问题描述】:
我有这个问题:
Time_Sheet_Details.findAll({
include: [
{
model: timesheetNotesSubcon,
required: false,
attributes:["note","file_name", "id", "working_hrs", "timestamp", "has_screenshot", "notes_category", "userid"],
},
{
model: Timesheet,
attributes:["id","leads_id","userid","subcontractors_id"],
where: clientWhere, // Client
include:[
{
model: Lead_Info, attributes:["id","fname","lname","email","hiring_coordinator_id","status"],
where: scWhere, // SC
include:[{
model: adminInfoSchema,
required: false,
attributes:["admin_id","admin_fname", "admin_lname", "admin_email", "signature_contact_nos", "signature_company"],
}]
},
{
model:Personal_Info,attributes:["userid","fname","lname","email"],
}
]
}],
where: {
reference_date: filters.reference_date
},
order:[
["id","DESC"]
],
offset:((page-1)*limit),
limit : limit,
subQuery:false
}).then(function(foundObject){
willFulfillDeferred.resolve(foundObject);
});
我的设置是我有一个 Timesheet 有一个 TimesheetDetails 有很多 timesheetNotesSubcon 。我正在获取TimesheetDetails 一起将所有timesheetNotesSubcon 与它一起。
它工作正常,但是如果我只想获取 timesheetNotesSubcon 的那些 timesheetNotesSubcon 等于或多于 1 条记录,那么所有 TimesheetDetails 在 timesheetNotesSubcon 中的结果为零的所有 timesheetNotesSubcon 都不会被获取。
【问题讨论】:
标签: mysql node.js sequelize.js