Question

有两个集合：

销售

{
  "_id" : ObjectId("5ba0bfb8d1acdc0de716e839"),
  "invoiceNumber" : 1,
  "saleDate" : ISODate("2018-09-01T00:00:00.000Z"),
  "totalTaxAmount" : 613,
  "subTotalAmount" : 2000,
  "totalAmount" : 2613,
  "balance" : 2613,
  "financialYear" : "2018-2019",
  "modeOfPayment" : "Digital Transfer",
  "customerName": "Acme Inc"
}

交易

{
  "_id" : ObjectId("5bbb4e131fb8af0dc645212d"),
  "transactionNumber" : 1    
  "transactionDate" : ISODate("2018-09-03T00:00:00.000Z"),
  "transactionType" : "Income",
  "partyName" : "Acme Inc",
  "transactionMode" : "Digital Transfer",
  "amount" : 2613,
  "paidItems" : [ 
      {
          "orderId" : "5b90a7d62bb5a21be4ff97e3",
          "invoiceNumber" : "1",
          "orderType" : "sale",
          "totalAmount" : 2613,
          "balance" : 613,
          "payingAmount" : 2000
      }
   ]
}

我需要按日期排序的两个日期（即saleDate，transactionDate）之间检索特定方（即customerName，partyName）的销售和交易作为“标题”；如下：

[
  {
    "date": ISODate("2018-09-01T00:00:00.000Z"),
    "heading": "Sale",
    "particulars": "Invoice # 1",
    "amount": 2613
  },
  {
    "date": ISODate("2018-09-03T00:00:00.000Z"),
    "heading": "Payment by Digital Transfer",
    "particulars": "Transaction # 1",
    "amount": 2000
  }
]

我研究并尝试了aggregation，$lookup，但没有返回期望的结果。

从SQL切换到MongoDB。在SQL中，以下查询工作正常：

select sale_date as dated, 'Sale' as heading, 'Invoice # ' + 
convert(varchar(12),invoice_number) as particulars, 
convert(varchar(12), total) as amount, 
from sales where sale_date between @from_date AND @to_date AND 
customer_name=@customer_name
UNION ALL
select transaction_date as dated, 'Payment by ' + transaction_mode as 
heading, 'Transaction # ' + convert(varchar(12), transaction_id) as 
particulars, convert(varchar(12), amount) as amount from transactions 
where transaction_date between @from_date AND @to_date AND 
party_name=@customer_name
order by dated DESC

在MongoDB社区中有一个feature request档案，它是“未解决的”。

我想知道mongoShell或MongoDB驱动程序（mongoose / JS）中对此有什么办法。使用当前稳定的MongoDB，nodejs，express和mongoose版本。谢谢！

Answer 1

我不认为您可以在mongodb中对2个不同的集合执行联合。

但是，您可以使用q.all从这两个集合中获取数据，然后使用自己的函数从它们中进行合并，或者可以是像lodash这样的第三方模块。

您可以在mongodb中执行联合的唯一方法是此处所述。 https://docs.mongodb.com/manual/reference/operator/aggregation/setUnion/

Answer 2

您可以尝试以下汇总

db.sales.aggregate([
  { "$limit": 1 },
  { "$facet": {
    "collection1": [
      { "$limit": 1 },
      { "$lookup": {
        "from": "sales",
        "pipeline": [
          { "$match": {
            "date": { "$gte": ISODate("2018-09-01"), "$lte": ISODate("2018-09-10") },
            "customer.name": customerName
          }},
          { "$project": {
            "_id":0, "dated": "$saleDate", "heading": "Sale", "particulars": "$invoiceNumber",
            "amount": "$totalAmount", "modeOfPayment": null
          }}
        ],
        "as": "collection1"
      }}
    ],
    "collection2": [
      { "$limit": 1 },
      { "$lookup": {
        "from": "transactions",
        "pipeline": [
          { "$match": {
            "transactionDate": { "$gte": ISODate("2018-09-01"), "$lte": ISODate("2018-09-10") },
            "userId": userId, "partyName": customerName
          }},
          { "$project": {
            "_id":0, "dated": "$transactionDate", "heading": "Payment","particulars": "$transactionNumber",
            "amount": "$amount", "paymentMode": "$transactionMode"
          }}
        ],
        "as": "collection2"
      }}
    ]
  }},
  { "$project": {
    "data": {
      "$concatArrays": [
        { "$arrayElemAt": ["$collection1.collection1", 0] },
        { "$arrayElemAt": ["$collection2.collection2", 0] },
      ]
    }
  }},
  { "$unwind": "$data" },
  { "$replaceRoot": { "newRoot": "$data" } },
  { "$sort": { "dated": -1 }}
])

Answer 3

免责声明：下列技术并非完全可取。 ;）尤其是在处理大型馆藏时。但是，它可用于实现与MongoDB v3.6及更高版本中的SQL UNION ALL相同的效果。

给出一个集合first和一个集合second：

db.first.aggregate([{
    $group: { // create an array to hold all documents of the first collection
        "_id": null,
        "first": {
            $push: "$$ROOT"
        }
    }
}, {
    $lookup: { // perform some kind of ridiculous lookup which will return all documents from the second collection in an array
        from: "second",
        let: { /* we do not need any variables */ },
        pipeline: [ { $match: { /* this filter will match every document */ } } ],
        as: "second"
    }
}, {
    $project: {
        "all": { $concatArrays: [ "$first", "$second" ] } // merge the two collections
    }
}, {
    $unwind: "$all" // flatten the resulting array
}, {
    $replaceRoot: { "newRoot": "$all" } // move content of "all" field all the way up 
}], {
    allowDiskUse: true // make sure we do not run into memory issues
})

SQL'UNION ALL'在MongoDB中的实现

3 个答案: