我還是 mongodb 的新手，我有這個基本的注冊系統資料：
{ "_id" : ObjectId("62277d92a561e550d5ec73ca"), "sid" : 1, "sname" : "sad", "semail" : "dsa", "scourse" : "it", "enrolled" : [ { "subjid" : 3 } ] }
{ "_id" : ObjectId("6227875bdbcc41a56a863697"), "sid" : 2, "sname" : "daws", "semail" : "dws", "scourse" : "cs", "enrolled" : [ { "subjid" : 1, "grades" : [ { "prelim" : "A", "midterm" : "B", "prefinal" : "B", "final" : "A" } ] }, { "subjid" : 2, "grades" : [ { "prelim" : "D", "midterm" : "A", "prefinal" : "B", "final" : "F" } ] } ] }
我想顯示已注冊 subjid 1 的 sid 2 的成績。
我嘗試使用這條聚合線：
db.students2.aggregate( [{"$match":{"sid":{"$eq":2},"enrolled.subjid":{"$eq":2}}}, {$group: {_id:'$enrolled.subjid[1]', prelim:{$first:'$enrolled.grades.prelim'},midterm:{$first:'$enrolled.grades.midterm'},prefinal:{$first:'$enrolled.grades.prefinal'},"final":{$first:'$enrolled.grades.final'} } } ])
但這就是結果：
{ "_id" : [ ], "prelim" : [ [ "A" ], [ "D" ] ], "midterm" : [ [ "B" ], [ "A" ] ], "prefinal" : [ [ "B" ], [ "B" ] ], "final" : [ [ "A" ], [ "F" ] ] }
我只想得到 subjid 1 的成績，但它也得到了 subjid 2 的成績
uj5u.com熱心網友回復：
也許你需要這樣的東西：
db.collection.aggregate([
{
"$match": {
"sid": 2,
"enrolled.subjid": 1
}
},
{
"$addFields": {
"enrolled": {
"$filter": {
"input": "$enrolled",
"as": "en",
"cond": {
$eq: [
"$$en.subjid",
1
]
}
}
}
}
},
{
$unwind: "$enrolled"
},
{
$unwind: "$enrolled.grades"
},
{$limit:1}
,
{
$project: {
_id: "$enrolled.subjid",
prelim: "$enrolled.grades.prelim",
midterm: "$enrolled.grades.midterm",
prefinal: "$enrolled.grades.prefinal",
"final": "$enrolled.grades.final"
}
}
])
解釋：
- 匹配必要的檔案（sid=2,subjid=1）
- 根據 subjid ( subjid=1 ) 僅過濾已注冊的元素
- 展開兩個陣列
- 僅限于第一個結果檔案，僅在有更多檔案時可用。
- 投射必要的領域
操場
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