[英]MongoDB find all docs where field doesn't exists, plus if exists apply field operator ($max) condition
我正在為我的聚合中的$match
階段尋找一個查詢,它的作用幾乎與此問題相同,但是..
結果應該是這樣的:
[
{
"method": 3,
"item": 1,
"rank": 3 //because it has field named rank, and suits condition {rank: $max}
},
{
"method": 4,
"item": 1 //we need this, because document doesn't have rank field at all
},
{
"method": 5,
"item": 1 //we need this, because document doesn't have rank field at all
}
]
我已經嘗試過的事情:
{
$match: {
$or: [
{item: id, rank: {$exists: true, $max: "$rank"}}, //id === 1
{item: id, rank: {$exists: false}} //id === 1
]
}
}
UPD:就目前而言,我可能不僅僅限制
$match
階段,默認匹配后$project
也是相關的,所以我可以在$match
階段通過id
請求每個文檔,無論是否具有 docrank
字段,然后,在$project
階段按排名$exists
進行“分離”
試試這個:
db.collection.aggregate([
{
$match: {
item: id
}
},
{
$group: {
_id: "$item", //<- Change here your searching field
max: {
$max: "$rank" //<- Change here your field to apply $max
},
data: {
$push: "$$ROOT"
}
}
},
{
$unwind: "$data"
},
{
$match: {
$expr: {
$or: [
{
$eq: [
{
$type: "$data.rank"
},
"missing"
]
},
{
$eq: [
"$data.rank",
"$max"
]
}
]
}
}
},
{
$replaceWith: "$data"
}
])
我找到了一個答案,與@Valijon 的方法分開,但它也是基於上面的邏輯。 我的查詢是:
db.collection.aggregate([
{
$match: {
item: id
}
},
{
$project: {
method: 1,
item: 1,
rank: {
$ifNull: [
"$rank",
0
]
}
}
},
{
$group: {
_id: "$item",
data: {
$addToSet: "$$ROOT"
},
min_value: {
$min: "$rank"
},
max_value: {
$max: "$rank"
}
}
},
{
$unwind: "$data"
},
{
$match: {
$or: [
{
$expr: {
$eq: [
"$data.rank",
"$max_value"
]
}
},
{
$expr: {
$eq: [
"$data.rank",
"$min_value"
]
}
},
]
}
}
])
我的查詢基於$project
階段,它給出空字段值 0。它也可以是 -1,或任何未在集合中使用的值。 然后我將結果分開。
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