Spring Data Mongodb 查询分组方式答案

【问题标题】：Spring Data Mongodb Query Group BySpring Data Mongodb 查询分组方式
【发布时间】：2020-10-07 23:25:46
【问题描述】：

我将 Spring 与 mongodb 一起使用，我想构建一个查询以使用 GROUP BY （如 SQL）按标题对我的文档进行分组，我有 3 个文档：

{
  _id: 5eeabfda75204e17242d3ff4,
  title: "Review",
  message: "hi this is a review 1",
  created: 2020-06-18T00: 18: 35.608+00: 00,
  _class: "com.guestnhousebackpfe.models.Notifications"
},
{
  _id: 5eeabfda75204e17242d3ff5,
  title: "Review",
  message: "hi this is a review 2",
  created: 2020-06-15T00: 18: 35.608+00: 00,
  _class: "com.guestnhousebackpfe.models.Notifications"
},
{
  _id: 5eeabfda75204e17242d3ff6,
  title: "Reservation",
  message: "hi this is a a reservation",
  created: 2020-06-15T00: 18: 35.608+00: 00,
  _class: "com.guestnhousebackpfe.models.Notifications"
}

预期结果：我想让我的文档按标题分组：

{
  "Review": [
    {
      _id: 5eeabfda75204e17242d3ff4,
      title: "Review",
      message: "hi this is a review 1",
      created: 2020-06-18T00: 18: 35.608+00: 00,
      _class: "com.guestnhousebackpfe.models.Notifications"
    },
    {
      _id: 5eeabfda75204e17242d3ff5,
      title: "Review",
      message: "hi this is a review 2",
      created: 2020-06-15T00: 18: 35.608+00: 00,
      _class: "com.guestnhousebackpfe.models.Notifications"
    },

  ],
  "Reservation": [
    {
      _id: 5eeabfda75204e17242d3ff6,
      title: "Reservation",
      message: "hi this is a a reservation",
      created: 2020-06-15T00: 18: 35.608+00: 00,
      _class: "com.guestnhousebackpfe.models.Notifications"
    }
  ]
}

我怎样才能做到这一点？

【问题讨论】：

您可以发布您尝试过的查询吗？
您使用带有$group 阶段的聚合查询来获取结果。您需要按title 字段分组并使用$push 聚合运算符将文档存储到数组中。
它有效，但将 id 作为对象返回：“_id”：{“timestamp”：1592442842，“counter”：2965492，“date”：“2020-06-18T01:14:02.000+0000” , "时间": 1592442842000, "timeSecond": 1592442842, "processIdentifier": 5924, "machineIdentifier": 7675982 },

标签： mongodb spring-boot group-by mongodb-query aggregation-framework

【解决方案1】：

更新

我设法让它工作：

TypedAggregation<Notifications> aggregation = newAggregation(Notifications.class,
                 match(Criteria.where("idUser").is(idUser)),
                 group("title").push("$$ROOT").as("notifications")
            );
        AggregationResults<Notifications> results = mongoOperations.aggregate(aggregation, Notifications.class);

        return results.getRawResults();

【讨论】：