mongodb - 为一个字段的每个不同值找到第一个文档

Question

对于包含字段 field1、field2、field3 等的文档集合，我需要查找

1. field3 的不同值
2. 对于 field3 的每个不同值，需要获取第一个包含 field3 中每个不同值的文档

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

我该如何处理#2？

样本收集:

[
    { "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，将所有文档按 field3 分组，并使用 $first 累加器与 $$ROOT 带第一个文档的系统变量，如下所示:

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

或精确输出:

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