在我的后端，我將 mongoDB 與 nodejs 和 mongoose 一起使用
我在 mongodb 中有很多具有這種結構的記錄：
{
..fields
type: 'out',
user: 'id1', <--mongodb objectID,
orderPayment: [
{
_id: 'id1',
paid: true,
paymentSum: 40
},
{
_id: 'id2',
paid: true,
paymentSum: 60,
},
{
_id: 'id3',
paid: false,
paymentSum: 50,
}
]
},
{
..fields
type: 'in',
user: 'id1', <--mongodb objectID
orderPayment: [
{
_id: 'id1',
paid: true,
paymentSum: 10
},
{
_id: 'id2',
paid: true,
paymentSum: 10,
},
{
_id: 'id3',
paid: false,
paymentSum: 77,
}
]
}
我需要按“型別”對這些記錄進行分組，并獲得帶條件的總和。需要獲取“付費”記錄的總和和 noPaid 記錄的總和。
為了更好地理解，這是我需要得到的結果
輸出是：
{
out { <-- type field
paid: 100, <-- sum of paid
noPaid: 50 <-- sum of noPaid
},
in: { <-- type field
paid: 20, <-- sum of paid
noPaid: 77 <-- sum of noPaid
}
}
uj5u.com熱心網友回復：
不同的解決方案是這個。它可能比@YuTing 的解決方案提供更好的性能：
db.collection.aggregate([
{
$project: {
type: 1,
paid: {
$filter: {
input: "$orderPayment",
cond: "$$this.paid"
}
},
noPaid: {
$filter: {
input: "$orderPayment",
cond: { $not: "$$this.paid" }
}
}
}
},
{
$set: {
paid: { $sum: "$paid.paymentSum" },
noPaid: { $sum: "$noPaid.paymentSum" }
}
},
{
$group: {
_id: "$type",
paid: { $sum: "$paid" },
noPaid: { $sum: "$noPaid" }
}
}
])
蒙戈游樂場
uj5u.com熱心網友回復：
使用$cond中$group
db.collection.aggregate([
{
"$unwind": "$orderPayment"
},
{
"$group": {
"_id": "$type",
"paid": {
"$sum": {
$cond: {
if: { $eq: [ "$orderPayment.paid", true ] },
then: "$orderPayment.paymentSum",
else: 0
}
}
},
"noPaid": {
"$sum": {
$cond: {
if: { $eq: [ "$orderPayment.paid", false ] },
then: "$orderPayment.paymentSum",
else: 0
}
}
}
}
}
])
mongoplayground
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