2
我已經指派我的收藏裏的人,地位簡單地像下面MongoDB的總結嵌套組
[
{"ASSIGN_ID": "583f84bce58725f76b322398", "STATUS": 1},
{"ASSIGN_ID": "583f84bce58725f76b322398","STATUS": 4},
{"ASSIGN_ID": "583f84bce58725f76b322398","STATUS": 4},
{"ASSIGN_ID": "583f84bce58725f76b322398","STATUS": 3},
{"ASSIGN_ID": "583f84bce58725f76b322311","STATUS": 1},
{"ASSIGN_ID": "583f84bce58725f76b322311","STATUS": 3},
{"ASSIGN_ID": "583f84bce58725f76b322322","STATUS": 1},
{"ASSIGN_ID": "583f84bce58725f76b322322","STATUS": 4}
]
我希望將這些數據通過ASSIGN_ID和內部的每個狀態的,通過狀態計數相同如下。
[
{
"ASSIGN_ID":"583f84bce58725f76b322398",
"STATUS_GROUP":[
{
"STATUS":1,
"COUNT":1
},
{
"STATUS":3,
"COUNT":1
},
{
"STATUS":4,
"COUNT":2
}
]
},
{
"ASSIGN_ID":"583f84bce58725f76b322311",
"STATUS_GROUP":[
{
"STATUS":1,
"COUNT":1
},
{
"STATUS":3,
"COUNT":1
}
]
},
{
"ASSIGN_ID":"583f84bce58725f76b322322",
"STATUS_GROUP":[
{
"STATUS":1,
"COUNT":1
},
{
"STATUS":4,
"COUNT":1
}
]
}
]
不過,我已經寫代碼,它只能通過狀態只有分組。請在下面找到查詢。
Inspection.aggregate([
{$group: {
"_id": '$STATUS',
"count" : { $sum : 1 }}}], function (err, result) {
}});
請幫我解決這個問題。
謝謝,它的工作!還假設該集合有另一列。我如何添加特定的列到這個結果。希望你有我的問題 –