MongoDB 根据所有其他元素删除元素（迭代）

Question

如果同一个对象具有相似的元素，我想从我的 MongoDB 对象中删除 ($reduce) 元素。我的对象：

  {
    "_id": "5eabf8b144345b36b00bfbaa",
    "ranktime": [
      {
        "pos": "15",
        "datum": "Mon May 01 2020 12:25:14 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "10",
        "datum": "Fri May 05 2020 12:25:14 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "15",
        "datum": "Mon May 01 2020 18:45:27 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "20",
        "datum": "Fri May 05 2020 18:45:27 GMT+0200 (GMT+02:00)",
        "source": "SOURCE1"
      },
      {
        "pos": "10",
        "datum": "Fri May 05 2020 12:25:14 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "15",
        "datum": "Mon May 01 2020 18:45:27 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      }
    ]
  }

因此，如果 ranktime.source == "SOURCE2" 并且日期与之前的对象相同，我想删除 ranktime 中的条目。实际上，我必须遍历 ranktime 的单个元素。这在 MongoDB 中可行吗？

预期结果是：

  {
    "_id": "5eabf8b144345b36b00bfbaa",
    "ranktime": [
      {
        "pos": "15",
        "datum": "Mon May 01 2020 12:25:14 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "10",
        "datum": "Fri May 05 2020 12:25:14 GMT+0200 (GMT+02:00)",
        "source": "SOURCE2"
      },
      {
        "pos": "20",
        "datum": "Fri May 05 2020 18:45:27 GMT+0200 (GMT+02:00)",
        "source": "SOURCE1"
      }
    ]
  }

Answer 1

因此，根据您的示例，您要输出 ranktime 除非它是 SOURCE2 并且已将相同的日期添加到输出中（但仅为SOURCE2)。

您可以像以前一样使用$reduce，但您需要扫描可以使用$anyElementTrue 运算符实现的先前添加的元素，并且由于您的输出包含第三个元素，我假设重复日期仅在以下情况下才是停止条件为SORUCE2 添加了相同的日期，因此还需要$filter 来准备之前添加的SOURCE2s 集：

db.col.updateMany({}, [
    {
        $set: {
            ranktime: {
                $reduce: {
                    input: "$ranktime",
                    initialValue: [],
                    in: {
                        $cond: [ 
                            { 
                                $and: [ 
                                    { "$eq": [ "$$this.source", "SOURCE2" ] },
                                    {
                                        $anyElementTrue: {
                                            $map: {
                                                input: { $filter: { input: "$$value", as: "prev", cond: { $eq: { "$$prev.source", "SOURCE2" } } } }, // already added SOURCE2 elements
                                                as: "addedElement",
                                                in: { "$eq": [ { $substr: [ "$$addedElement.datum", 0, 15 ] }, { $substr: [ "$$this.datum", 0, 15 ] } ] }
                                            }                        
                                        }
                                    }
                                ]
                            },
                            "$$value", // skip current element ($$this) 
                            { $concatArrays: [ "$$value", [ "$$this" ] ] } // add current element to the output
                        ]
                    }
                }
            }
        }
    }
])

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