需要计算MongoDB中以字符串格式保存的datetime字段的datetime差异答案

【问题标题】：Need to calculate the datetime difference for the datetime field saved in string format in MongoDB需要计算MongoDB中以字符串格式保存的datetime字段的datetime差异
【发布时间】：2020-05-16 04:12:20
【问题描述】：

示例文档：

{
    "_id" : ObjectId("50ed90a55502684f440001ac"),
    "time" : "2020-05-14T14:12:46.058-0400",
    "ServerTimeStamp" : "2020-05-14T14:12:46.058-0400"

}

{
    "_id" : ObjectId("50ed90a55502684f440001ad"),
    "time" : "2020-05-14T15:12:46.058-0400",
    "ServerTimeStamp" : "2020-05-14T14:12:46.058-0400"
}

{
    "_id" : ObjectId("50ed90a55502684f440001ae"),
    "time" : "2020-05-14T18:12:46.058-0400",
    "ServerTimeStamp" :"2020-05-14T14:12:46.058-0400"
}

我想计算以上行的时间差（以秒为单位）如下，这里时间字段是字符串格式而不是日期。

所需输出：

{

    "_id" : ObjectId("50ed90a55502684f440001ac"),
    "time" : "2020-05-14T14:12:46.058-0400",
    "time_difference" :null
}

{
    "_id" : ObjectId("50ed90a55502684f440001ad"),
    "time" : "2020-05-14T15:12:46.058-0400",
    "time_difference" : 3600  
}

{
    "_id" : ObjectId("50ed90a55502684f440001ae"),
    "time" : "2020-05-14T18:12:46.058-0400",
    "time_difference" : 10800
}

我尝试了以下查询，但出现错误，并且我已经将时间从字符串转换为日期：

db.hello.aggregate(
    {$sort: {$dateFromString:{time: 1}}},
    {$group: {_id: 0, document: {$push: '$$ROOT'}}},
    {$project: {documentAndPrevTime: {$zip: {inputs: ['$document', {$concatArrays: [[null], '$document.time']}]}}}},
    {$unwind: {path: '$documentAndPrevTime'}},
    {$replaceWith: {$mergeObjects: [{$arrayElemAt: ['$documentAndPrevTime', 0]}, {prevTime: {$arrayElemAt: ['$documentAndPrevTime', 1]}}]}},
    {$set: {time_difference: {$trunc: [{$divide: [{$subtract: ['$time', '$prevTime']}, 1000]}]}}},
    {$unset: 'prevTime'}
);

【问题讨论】：

标签： mongodb mongodb-query aggregation-framework

【解决方案1】：

您已经编写了所有阶段，但这里的问题是彼此之间没有关系，这意味着在先前设置中所做的更改不会反映在后续步骤中。几个问题：

{$sort: {$dateFromString:{time: 1}}} // You can't use `$dateFromString` in sort stage. Also syntax of `$dateFromString` is incorrect.

让我们假设它是否有效（它不会但假设）您实际上并没有转换time 并将转换后的时间存储到一个变量以供以后在$group 或更进一步的阶段使用。因此，您需要使用$addFields 或$project 将其存储到相应文档中的变量中。我没有走得更远，但您可以尝试以下查询：

查询：

db.collection.aggregate([
    /** sort on `time` field */
    { $sort: { time: 1 } },
    /** Convert string format of `time` field to milliseconds & store to `convertedTime` field for each doc */
    { $addFields: { convertedTime: { $toLong: { $dateFromString: { dateString: "$time" } } } } },
    /** Group without condition to push all documents into `docs` array */
    {
      $group: { _id: "", docs: { $push: "$$ROOT" } }
    },
    /** re-creating `docs` array */
    {
      $project: {
        _id: 0,
        docs: {
          $reduce: {
            input: { $slice: [ "$docs", 1, { $size: "$docs" } ] }, /** Pick `docs` array without first element for iteration */
            initialValue: {  docObj: [ { $arrayElemAt: [ "$docs", 0 ] } ], previousTime: { $arrayElemAt: [ "$docs.convertedTime", 0 ] } },
            in: {
              docObj: { $concatArrays: [ "$$value.docObj", [
                    { $mergeObjects: [ "$$this", { time_difference: { $divide: [ { $subtract: [ "$$this.convertedTime", "$$value.previousTime" ] }, 1000 ] } } ] }
                  ]
                ]
              },
              previousTime: "$$this.convertedTime" // Store current doc's time to `previousTime` to utilize for next record
            }
          }
        }
      }
    },
    {
      $unwind: { path: "$docs.docObj" }
    },
    /** Remove additionally added field */
    {
      $project: { "docs.docObj.convertedTime": 0 }
    },
    /** Replace root of the doc with `docs.docObj` */
    {
      $replaceRoot: { newRoot: "$docs.docObj" }
    }
  ])

测试： mongoplayground

参考：aggregation-pipeline

注意：此查询不会为第一个文档添加"time_difference" :null，但如果需要，请在initialValue 中尝试此操作：docObj: [ {$mergeObjects :[ { $arrayElemAt: [ "$docs", 0 ] }, { "time_difference" :null } ] ]。另外，我建议将此操作限制为使用$match 作为第一阶段的集合中的某些文档，而不是对所有文档进行此查询原因

{
   $group: { _id: "", docs: { $push: "$$ROOT" } }
}

当它在具有庞大数据集的整个集合上完成时，它本身将是一件大事。

【讨论】：