mongodb - $cond 聚合中的嵌套条件

Question

我正在尝试创建一个计算 状态 我的 Mongo 查询中的字段(状态:已创建、已收到付款、已发货、已收到、已完成)。

db.orders.aggregate( [
   { $project: { status: {
      $cond: { if: { $ne: ["$feedback", null] }, 
      then: 'finished', else: {
         $cond: { if: { $ne: ["$received", null] }, 
         then: 'received', else: {
            $cond: { if: { $ne: ["$shipped", null] }, 
            then: 'shipped', else: {
               $cond: { if: { $ne: ["$payment", null] }, 
                  then: 'payment received', else: 'created' }
            } }
         } } 
      } }
   } } },
   { $match: {  } }
] )

示例数据:

{
    "_id" : "xxxxxx0",
    "payment" : ISODate("2016-02-03T10:45:00.011Z"),
    "shipped" : ISODate("2016-02-03T11:55:00.011Z"),
    "received" : ISODate("2016-02-03T12:45:00.011Z"),
    "feedback" : ISODate("2016-02-03T14:34:00.011Z")
},
{
    "_id" : "xxxxxx1",
    "payment" : ISODate("2016-02-03T10:45:00.011Z"),
    "shipped" : ISODate("2016-02-03T11:55:00.011Z"),
    "received" : ISODate("2016-02-03T12:45:00.011Z")
},
{
    "_id" : "xxxxxx2",
    "payment" : ISODate("2016-02-03T10:45:00.011Z"),
    "shipped" : ISODate("2016-02-03T11:55:00.011Z")
},
{
    "_id" : "xxxxxx3",
    "payment" : ISODate("2016-02-03T10:45:00.011Z")
},
{
    "_id" : "xxxxxx4"
}

出于某种原因，我所有的结果都显示为“完成”，我使用 $cond 错了吗？它是否支持嵌套的 $cond ？

Answer 1

如果要检查该字段是否存在，则不能使用 $eq null，它将始终返回 true
有一个技巧可以用 $gt 做到这一点。您可以在此处查看完整说明 ( https://docs.mongodb.com/manual/reference/bson-types/#bson-types-comparison-order )

db.orders.aggregate( [
   { $project: { status: {
      $cond: { if: { $gt: ["$feedback", null] }, 
      then: 'finished', else: {
         $cond: { if: { $gt: ["$received", null] }, 
         then: 'received', else: {
            $cond: { if: { $gt: ["$shipped", null] }, 
            then: 'shipped', else: {
               $cond: { if: { $gt: ["$payment", null] }, 
                  then: 'payment received', else: 'created' }
            } }
         } } 
      } }
   } } },
   { $match: {  } }
] )