Question

想象我有这样的集合：

{
  "_id": "10280",
  "city": "NEW YORK",
  "state": "NY",
  "departments": [
             {"departmentType":"01",
              "departmentHead":"Peter"},
             {"departmentType":"02",
              "departmentHead":"John"}
  ]
},
{
  "_id": "10281",
  "city": "LOS ANGELES",
  "state": "CA",
  "departments": [
             {"departmentType":"02",
              "departmentHead":"Joan"},
             {"departmentType":"03",
              "departmentHead":"Mary"}
  ]
},
{
  "_id": "10284",
  "city": "MIAMI",
  "state": "FL",
  "department": [
  "departments": [
             {"departmentType":"01",
              "departmentHead":"George"},
             {"departmentType":"02",
              "departmentHead":"Harry"}
  ]
}

我希望获得每个departmentType的计数，例如：

[{"departmentType":"01", "dCount":2},
 {"departmentType":"02", "dCount":3},
 {"departmentType":"03", "dCount":1}
]

为此，我已经尝试了几乎所有内容，但我在网上找到的所有示例都更容易，其中group by是在文档的根级别的字段上完成的。相反，在这里我尝试按departmentType进行分组，这似乎打破了我到目前为止发现的所有内容。

有关如何使用Mongoose的聚合实现或mapreduce执行此操作的任何想法？

理想情况下，我希望排除所有显示＆lt; = 1的departmentTypes，并按departmentType对结果进行排序。

提前谢谢大家！

Answer 1

您需要展开departments数组，该数组将为数组中的每个条目创建一个文档，以便您可以在管道中聚合它们。

不幸的是，您无法预先过滤departmentTypes＆lt; = 1，因为$ size只会获取一个确切的值，但您可以将其从结果中过滤掉。它不是很好，但它的确有效。此示例仅预先过滤那些具有完全2个部门的记录，但它仅用于演示，您可能希望删除第一个$匹配，因为我们在稍后的结果上使用第二个$匹配过滤掉＆lt; = 1;

db.runCommand({
    aggregate: "so",
    pipeline: [
        {   // filter out only records with 2 departments
            $match: {
                departments: { $size: 2 }
            }
        },
        // unwind - create a doc for each department in the array
        { $unwind: "$departments" },
        {   // aggregate sum of departments by type
            $group: {
                _id: "$departments.departmentType",
                count: { $sum: 1 },
            }
        },
        {   // filter out departments with <=1
            $match: {
                count: { $gt: 1 },
            }
        },
        {   // rename fields as per example
            $project: {
                _id: 0,
                departmentType: "$_id",
                dCount: "$count",
            }
        }
    ]
});

请注意，我还假设您之前的json样本有拼写错误，而“部门”实际上并不存在。假设所有文档与前两个文档具有相同的模式，此代码将起作用。

如果您对所获得的实际字段名称不感兴趣，请随意删除第一个$ match和最后一个$ project。

mapReduce / Aggregation：按嵌套文档中的值分组

1 个答案: