Question

如果文档集合包含字段field1，field2，field3等，我需要找到

field3
对于field3的每个不同值，需要获取field3中每个不同值的第一个文档

对于＃1，我可以db.myCollection.distinct("field3")

我如何进行＃2？

样品采集：

[
    { "field1": 11, "field2": "toyota", "field3": "camry" },
    { "field1": 22, "field2": "toyota", "field3": "corolla" },
    { "field1": 33, "field2": "toyota", "field3": "camry" },
    { "field1": 44, "field2": "honda", "field3": "accord" },
    { "field1": 55, "field2": "honda", "field3": "accord" },
    { "field1": 66, "field2": "honda", "field3": "city" }
]

期望的结果：

[
    { "field1": 11, "field2": "toyota", "field3": "camry" },
    { "field1": 22, "field2": "toyota", "field3": "corolla" },
    { "field1": 44, "field2": "honda", "field3": "accord" },
    { "field1": 66, "field2": "honda", "field3": "city" }
]

Answer 1

您需要运行一个聚合操作，按field3对所有文档进行分组，并使用 $first 累加器与 $$ROOT 系统变量带来的第一个文件，如下所示：

db.myCollection.aggregate([
    {
        "$group": {
            "_id": "$field3",
            "doc": { "$first": "$$ROOT" }
        }
    }
])

或确切输出：

db.myCollection.aggregate([
    {
        "$group": {
            "_id": "$field3",
            "field1": { "$first": "$field1" },
            "field2": { "$first": "$field2" }
        }
    },
    {
        "$project": {
            "_id": 0,
            "field3": "$_id",
            "field2": 1,
            "field1": 1
        }
    }
])

Answer 2

聚合使用$ group aggregation operator

根据字段的不同值对记录进行分组

根据上述说明，请尝试在MongoDB shell中执行以下查询

db.myCollection.aggregate(

  // Pipeline
  [
    // Stage 1
    {
      $group: {
        _id:{field3:'$field3'},
        data:{$first:'$$ROOT'}
      }
    }

  ]


);

在上面提到的查询中，每个组的第一个文档的数据是通过使用$first聚合运算符获取的，而 $$ ROOT 是指当前正在通过聚合操作处理的文档

找到一个字段

2 个答案: