也许是这样的：

db.collection.aggregate([
{
 $group: {
   _id: null,
   ar: {
    $push: "$$ROOT"
   },
   mind: {
    "$min": "$midNightTimeStamp"
   },
   maxd: {
    "$max": "$midNightTimeStamp"
   }
  }
 },
 {
  $project: {
  ar: {
    $map: {
      input: {
        $range: [
          "$mind",
          {
            "$sum": [
              "$maxd",
              86400
            ]
          },
          86400
        ]
      },
      as: "dateInRange",
      in: {
        $let: {
          vars: {
            dateIndex: {
              "$indexOfArray": [
                "$ar.midNightTimeStamp",
                "$$dateInRange"
              ]
            }
          },
          in: {
            $cond: {
              if: {
                $ne: [
                  "$$dateIndex",
                  -1
                ]
              },
              then: {
                $arrayElemAt: [
                  "$ar",
                  "$$dateIndex"
                ]
              },
              else: {
                midNightTimeStamp: "$$dateInRange",
                "waterDispensed": NumberInt(0)
              }
            }
          }
        }
       }
      }
     }
    }
  },
  {
   $unwind: "$ar"
  },
  {
   $project: {
    _id: 0,
    "waterDispensed": "$ar.waterDispensed",
    midNightTimeStamp: "$ar.midNightTimeStamp",
    "Date": {
      $toDate: {
         "$multiply": [
         "$ar.midNightTimeStamp",
         1000
       ]
      }
    }
   }
  }
])

解释道：
1.$分组文档以查找时间戳的最大值和最小值，$将所有元素推入名为“ar”的临时数组
1.$投影数组$Map，生成日期范围为最大值和最小值，步长为1x天（86400），用waterDispanced填充空元素：0
1.$展开数组$ar
1.$在最终输出中只投影我们需要的字段。
playground

这与另一个答案有一点不同，它只需要抓住所需的"uID"。MQL中的注解解释了这个过程。

db.collection.aggregate([
  { // The uID we want
    "$match": { "uID": "12345678" }
  },
  { // grab all the uID docs as "water"
    // keep uID
    "$group": {
      "_id": null,
      "water": { "$push": "$$CURRENT" },
      "uID": { "$first": "$uID" }
    }
  },
  { // create outArray
    "$set": {
      "outArray": {
        // by mapping time vals
        "$map": {
          "input": {
            "$range": [ NumberInt(1645381800), NumberInt(1645900200), 86400 ]
          },
          "in": {
            "$cond": [
              { // already have doc?
                "$in": [ "$$this", "$water.midNightTimeStamp" ]
              },
              { // yes!  Get it!
                "$arrayElemAt": [
                  "$water",
                  { "$indexOfArray": [ "$water.midNightTimeStamp", "$$this" ] }
                ]
              },
              { // no, create it
                "uID": "$uID",
                "midNightTimeStamp": "$$this",
                "waterDispensed": 0
              }
            ]
          }
        }
      }
    }
  },
  { // only need outArray now
    "$project": {
      "_id": 0,
      "outArray": 1
    }
  },
  { // create docs
    "$unwind": "$outArray"
  },
  { // hoist them
    "$replaceWith": "$outArray"
  },
  { // don't need _id
    "$unset": "_id"
  }
])

在mongoplayground.net上试试。

从MongoDB5.1开始，您可以使用$densify聚合操作符用平均值或默认值填充缺失的时间序列数据。
https://www.mongodb.com/docs/rapid/reference/operator/aggregation/densify/
在您的情况下，可能需要在聚合时将时间戳字段转换为日期，以便可以使用$densify。
您还可以在MongoDB World 2022上观看this presentation中$densify的快速解释。