问题描述
我是 Mongodb 的新手 &在我使用 MEAN 堆栈构建的 Web 应用程序中使用它.我的目标是通过连接两个表并对它们应用过滤条件来查询它们.例如:我有两个表 - Bike-BikeID、注册号、品牌、型号和;Appointment - Appointment Date,Status,Bike(ref Bike object) 我只想显示那些没有预约 status='Booked' 的自行车.我想在Mongoose中完成以下SQL.
I'm new to Mongodb & using it in web application that I'm building using MEAN stack. My goal is to query two tables by joining them and applying a filter condition on them. For eg: I have two tables - Bike-BikeID,Registration No.,Make,Model & Appointment - Appointment Date,Status,Bike(ref Bike object) and I want to show only those bikes who do not have an appointment with status='Booked'. I want to accomplish the following SQL in Mongoose.
Select bike.* from Bike inner join Appointment on Bike.BikeID = Appointment.BikeID and Appointment.Status != 'Booked'
我正在使用以下代码,但没有得到想要的结果.有人可以帮助我进行此查询.
I'm using the following code, but I'm not getting the desired results. Can someone help me with this query.
app.get('/api/getbikeappo*',function(req,res){
var msg="";
//.find({cust:req.query._id})
//.populate('cust','email')
ubBike.aggregate([
{
$match:
{
cust : req.query._id
}
},
{
$lookup:
{
from: "appos",
localField: "_id",
foreignField: "bike",
as : "appointments"
}
},
{
$match:
{
"appointments" : {$eq : []}
}
}
])
.exec(function(err,bikes){
res.send(bikes);
if(err) throw err;
});
});
bikes - collection
{
"_id": {
"$oid": "57fb600fdd9070681de19c18"
},
"brand": "Splendor",
"model": "Splendor",
"year": "2002",
"kms": 0,
"regno": "TN02M8937",
"cust": {
"$oid": "57f8c44466cab97c1355a09a"
},
"__v": 0
}
{
"_id": {
"$oid": "57fb6025dd9070681de19c19"
},
"brand": "Activa",
"model": "Activa",
"year": "2016",
"kms": 0,
"regno": "TN14M3844",
"cust": {
"$oid": "57f8c44466cab97c1355a09a"
},
"__v": 0
}
appointment collection
----------------------
{
"_id": {
"$oid": "57fb6040dd9070681de19c1a"
},
"appoidt": {
"$date": "2016-10-15T18:30:00.000Z"
},
"reqdt": {
"$date": "2016-10-10T09:32:48.694Z"
},
"status": "Booked",
"bike": {
"$oid": "57fb600fdd9070681de19c18"
},
"cust": {
"$oid": "57f8c44466cab97c1355a09a"
},
"__v": 0
}
-----------------
Expected output is
{
"_id": {
"$oid": "57fb6025dd9070681de19c19"
},
"brand": "Activa",
"model": "Activa",
"year": "2016",
"kms": 0,
"regno": "TN14M3844",
"cust": {
"$oid": "57f8c44466cab97c1355a09a"
},
"__v": 0
}
推荐答案
您就快到了,您只需要合适的 $match
查询如下:
You were almost there, you just needed the right $match
query which follows:
ubBike.aggregate([
{ "$match": { "cust": req.query._id } },
{
"$lookup": {
"from": "appos",
"localField": "_id",
"foreignField": "bike",
"as": "appointments"
}
},
{ "$match": { "appointments.status": { "$ne": "Booked" } } }
]).exec(function(err, bikes){
if(err) throw err;
res.send(bikes);
});
这篇关于使用 joins & 查询猫鼬中的过滤器的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!