根据不同的条件投影不同的字段

Question

我在MongoDB中使用聚合，现在我遇到了一个问题，我想根据条件匹配与否来投影我的字段。

例如，我有一个字段coupon_type，我将检查其值是否等于 1，然后我将投影字段["curr_ctr", "total_coupons"]否则，如果其值不等于 1，那么我将投影字段["curr_ctr", "total_coupons", "curr_ctr", "coupons"].

我

可以使用两个查询并并行运行它们，但我正在尝试使用一个查询来实现我的结果。

任何人都可以告诉我如何在一个查询中做到这一点？

更新

我的文件如下

[{ "_id" : ObjectId("5878e1edf1df5a2b69bcf60e"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1eff1df5a2b69bcf60f"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 0 } ,
{ "_id" : ObjectId("5878e1f1f1df5a2b69bcf610"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1f3f1df5a2b69bcf611"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1f5f1df5a2b69bcf612"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 11 } ,
{ "_id" : ObjectId("5878e1f7f1df5a2b69bcf613"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 110 },
{ "_id" : ObjectId("5878e1f9f1df5a2b69bcf614"), "curr_ctr": 12, "total_coupons":35, "coupons": ["hello", "hello2"], "coupon_type" : 0 } ]

现在，对于所有等于 0 coupon_type，我想投影字段["curr_ctr", "total_coupons", "curr_ctr", "coupons"]，对于所有不等于 0 coupon_type，我想投影字段["curr_ctr", "total_coupons"]

更新

实际上我想coupons数组从索引 n 投影到 n+1，其中 n 是用户的输入。

Answer 1

让我们考虑您的集合包含以下文档

[{ "_id" : ObjectId("5878e1edf1df5a2b69bcf60e"), "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1eff1df5a2b69bcf60f"), "coupon_type" : 0 } ,
{ "_id" : ObjectId("5878e1f1f1df5a2b69bcf610"), "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1f3f1df5a2b69bcf611"), "coupon_type" : 1 } ,
{ "_id" : ObjectId("5878e1f5f1df5a2b69bcf612"), "coupon_type" : 11 } ,
{ "_id" : ObjectId("5878e1f7f1df5a2b69bcf613"), "coupon_type" : 110 },
{ "_id" : ObjectId("5878e1f9f1df5a2b69bcf614"), "coupon_type" : 0 } ]

现在您需要在project中使用$eq作为：

db.collectoin.aggregate({
  "$project": {
    "result": {
      "$cond": {
    "if": {
      "$eq": ["$coupon_type", 1]
    },
    "then": ["curr_ctr", "total_coupons"],
    "else": ["curr_ctr", "total_coupons", "curr_ctr", "coupons"]
      }
    }
  }
})

根据新的更新，您应该像这样修改查询：

db.collection.aggregate({
  "$project": {
    "result": {
      "$cond": {
    "if": {
      "$eq": ["$coupon_type", 1]
    },
    "then": ["$curr_ctr", "$total_coupons", "$coupons"],
    "else": ["$curr_ctr", "$total_coupons"]
      }
    }
  }
})

更新

根据新的更新，让我们考虑用户给出的n，例如：var n = 1;

现在查询将如下所示：

db.coupon.aggregate({
  "$project": {
    "result": {
      "$cond": {
    "if": {
      "$eq": ["$coupon_type", 1]
    },
    "then": ["$curr_ctr", "$total_coupons", {
      "coupons": {
        "$slice": ["$coupons", n, n + 1]
      }
    }],
    "else": ["$curr_ctr", "$total_coupons"]
      }
    }
  }
})