MongoDB根据所有其他元素删除元素(Iterating)

Question

我想从我的MongoDB对象中删除($reduce(元素，条件是同一个对象有相似的元素。我的目标：

{
"_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=="；来源2"；以及日期是否与之前的对象相同。实际上，我必须遍历ranktime的单个元素。这在MongoDB中可能吗？

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，您希望输出ranktime，除非是SOURCE2并且已经将相同的日期添加到输出中(但仅适用于SOURCE2(。

您可以像以前一样使用$reduce，但您需要扫描以前添加的元素，这可以使用$anyElementTrue运算符来实现。由于您的输出包含第三个元素，我假设只有在为SORUCE2添加了相同日期的情况下，重复日期才是停止条件，因此还需要$filter来准备以前添加的SOURCE2的集合：

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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