Question

我有一张桌子：

id   value   date
1      1     2014-08-10 10:00:00
1      2     2014-08-10 10:05:00
1      1     2014-08-10 12:00:00
1      1     2014-08-11 12:05:00
1      2     2014-08-11 12:06:00
2      2     2014-08-10 10:00:00
2      2     2014-08-11 10:05:00
2      1     2014-08-11 12:00:00
2      1     2014-08-20 12:05:00
2      2     2014-08-20 12:06:00

我希望获得第一天每个ID的总值。我可以使用以下命令在MySQL中执行此操作：

SELECT test.id, COUNT(CASE WHEN value=1 THEN 1 END) AS 'total_value_1',
                COUNT(CASE WHEN value=2 THEN 1 END) AS 'total_value_2'
FROM test,    
(SELECT id, date(min(date)) as firstDay
 FROM test
 GROUP BY id) AS temp
WHERE  test.id = temp.id AND date(test.date) = temp.firstDay
GROUP BY test.id;

结果是：

id total_value_1 total_value_2
1         2           1
2         0           1

我如何在MongoDB中执行此操作？

Answer 1

这可以通过聚合框架实现，您可以使用 a collection of operators 来定义和操作管道阶段中的文档，从而为您提供所需的结果。您的第一个管道阶段是$project运算符，您可以在其中转换值字段，以便在后续管道阶段中更轻松地进行操作。此操作的等效SQL是：

SELECT id, date, CASE WHEN value=1 THEN 1 END AS 'value_1',
                 CASE WHEN value=2 THEN 1 END AS 'value_2'
FROM test

下一个管道阶段使用$group和$project运算符来计算日期字段中月份的最小日期以及转换后的值字段上的总和汇总：

db.test.aggregate([
    {        
        "$project": {  
            "id": 1,            
            "date": 1,            
            "value_1": { "$cond": [{ "$eq": [ "$value", 1 ] }, 1, 0] },
            "value_2": { "$cond": [{ "$eq": [ "$value", 2 ] }, 1, 0] }
        }
    },
    {
        "$group": {
            "_id": {
                "id": "$id",
                "date": "$date"
            },
            "minimumDayofMonth": {
                "$min": { "$dayOfMonth": "$date"}
            },
            "total1" : {
                "$sum": "$value_1"
            },
            "total2" : {
                "$sum": "$value_2"
            }             
        }
    },
    {
        "$group": {
            "_id": {
                "id": "$_id.id",
                "day": "$minimumDayofMonth"
            }, 
            "firstDay": { "$min": "$minimumDayofMonth"  },            
            "total_value_1" : {
                "$sum": "$total1"
            },
            "total_value_2" : {
                "$sum": "$total2"
            }            
        }
    },
    {
        "$group": {
            "_id": "$_id.id",             
            "result": { "$first": "$$ROOT" }                
        }
    },
    {        
        "$project": {  
            "_id": 0,
            "id": "$_id",
            "total_value_1": "$result.total_value_1",
            "total_value_2": "$result.total_value_2"
        }
    }    
])

<强>输出：

/* 0 */
{
    "result" : [ 
        {
            "id" : 2,
            "total_value_1" : 0,
            "total_value_2" : 1
        }, 
        {
            "id" : 1,
            "total_value_1" : 2,
            "total_value_2" : 1
        }
    ],
    "ok" : 1
}

mongodb group by min

1 个答案: