如何获得每天的count（）

时间：2018-01-26 04:21:52

标签： mysql

我正在使用分析，我想得到每天的计数，并且它将在当天返回0而没有计数。我正在使用MYSQL。

  

这是我的疑问：

SELECT DAY(a.Date)
FROM (
    SELECT LAST_DAY(CURDATE()) - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY AS DATE
    FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
    CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
) a
WHERE a.Date BETWEEN (CURDATE() - INTERVAL 31 DAY) AND CURDATE() GROUP BY a.Date;

我得到了获取31天的日期的查询，但我无法获得每天的计数（）。

  

示例：

     

我有一个包含以下列的货件表：

     

id = some id

     

status =＆＃39;已发送＆＃39;，＆＃39;已发送＆＃39;，＆＃39;已完成＆＃39;

     

date_dispatched =某个日期

     

我希望将所有带有调度状态的数据计算31天。

     

预期产出的例子：

+-----+--------+
| Day | Count  |
+-----+--------+
| 25  |   0    |
| 26  |   0    |
| 27  |   0    |
| 28  |   0    |
| 29  |   12   |
| 30  |   24   |
| 31  |   34   |
| 1   |   24   |
| 2   |   67   |
| 3   |   0    |
| 4   |   0    |
| 5   |   0    |
| 6   |   0    |
| 7   |   0    |
| 8   |   0    |
| 9   |   0    |
| 10  |   0    |
| 11  |   12   |
| 12  |   44   |
| 13  |   67   |
| 14  |   0    |
| 15  |   0    |
| 16  |   0    |
| 17  |   0    |
| 18  |   0    |
| 19  |   0    |
| 20  |   0    |
| 21  |   0    |
| 22  |   0    |
| 23  |   0    |
| 24  |   0    |
+--------------+

2 个答案:

答案 0 :(得分：0)

此示例从2015-01-01到2015-01-31 

select lstOfDays, date_dispatched,case when Dispatched_Count is null then 0 else Dispatched_Count end as Dispatched_Count from
(SELECT ADDDATE('2015-01-01', INTERVAL @i:=@i+1 DAY) AS lstOfDays
        FROM (
        SELECT a.a
        FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
        ) a
        JOIN (SELECT @i := -1) r1
        WHERE 
        @i < DATEDIFF('2015-01-31', '2015-01-01')) as a

LEFT JOIN

(select date_dispatched, Status, count(1) as Dispatched_Count from
 shipment_table where Status =  'Dispatched'  GROUP BY date_dispatched) as b

on a.lstOfDays = b.date_dispatched ORDER BY lstofDays

您必须输入StartDate和EndDate

这将返回如下输出：

lstofDays   date_dispatched   Dispatched_Count

2015-01-01                           0
2015-01-02                           0
2015-01-03                           0
2015-01-04                           0
2015-01-05                           0
2015-01-06                           0
2015-01-07    2015-01-07             1
2015-01-08                           0
2015-01-09    2015-01-09             2
2015-01-10                           0
2015-01-11                           0
2015-01-12                           0
2015-01-13                           0
2015-01-14                           0
2015-01-15                           0
2015-01-16                           0
2015-01-17                           0
2015-01-18                           0
2015-01-19                           0
2015-01-20                           0
2015-01-21                           0
2015-01-22                           0
2015-01-23                           0
2015-01-24                           0
2015-01-25                           0
2015-01-26                           0
2015-01-27                           0
2015-01-28                           0
2015-01-29                           0
2015-01-30                           0
2015-01-31                           0

您需要添加year和month，否则它将包含您所有的date_dispatched字段。

或者这将自动从Date Today - 31 days

生成 
set @edate = DATE_FORMAT(CURDATE(), "%Y-%m-%d");
set @sdate = DATE_ADD(DATE_FORMAT(CURDATE(), "%Y-%m-%d"),INTERVAL -31 DAY);

select case when date_dispatched is null then Date_format(date(lstOfDays),'%d') else Date_format(date(date_dispatched),'%d') end date_dispatched,
case when Dispatched_Count is null then 0 else Dispatched_Count end as Dispatched_Count from
(SELECT ADDDATE(@sdate, INTERVAL @i:=@i+1 DAY) AS lstOfDays
        FROM (
        SELECT a.a
        FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
        CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
        ) a
        JOIN (SELECT @i := -1) r1
        WHERE 
        @i < DATEDIFF(@edate,@sdate)) as a

LEFT JOIN

(select date_dispatched, Status, count(1) as Dispatched_Count from
shipment_table where Status =  'Dispatched'  GROUP BY date_dispatched) as b

on a.lstOfDays = b.date_dispatched ORDER BY lstofDays

结果：

26  0
27  0
28  0
29  0
30  0
31  0
01  0
02  0
03  0
04  0
05  0
06  0
07  0
08  0
09  0
10  0
11  0
12  0
13  0
14  0
15  0
16  0
17  1
18  0
19  2
20  0
21  0
22  0
23  0
24  0
25  0
26  0

答案 1 :(得分：0)

尝试这个：您应该GROUP BY天并根据条件计算值，如下所示：

SELECT DAY(date_dispatched) Days,
    COUNT(1) AS DispatchedCount 
--Place the column here if you have any particular like `dispatched_number`
--If we have separate rows for every transaction then `COUNT(1) will work fine`
FROM shipment
WHERE status = 'Dispatched'
    AND date_dispatched BETWEEN DATE_SUB(CURDATE(), INTERVAL 30 DAY) AND CURDATE()
GROUP BY DAY(date_dispatched)
ORDER by DAY(date_dispatched);

您可以查看每个状态的计数，因此您需要从选择

中的status = 'Dispatched'下面的CASE删除SUM(CASE WHEN status = 'Dispatched' THEN 1 ELSE 0 END) AS DispatchedCount, SUM(CASE WHEN status = 'Delivered' THEN 1 ELSE 0 END) AS DeliveredCount, SUM(CASE WHEN status = 'Completed' THEN 1 ELSE 0 END) AS CompletedCount, 
json = 
'[
{
    "id": "0001",
    "type": "donut",
    "name": "Cake",
    "ppu": 0.55,
    "batters":
        {
            "batter":
                [
                    { "id": "1001", "type": "Regular" },
                    { "id": "1002", "type": "Chocolate" }
                ]
        },
    "topping":
        [
            { "id": "5007", "type": "Powdered Sugar" },
            { "id": "5006", "type": "Chocolate with Sprinkles" },
            { "id": "5003", "type": "Chocolate" },
            { "id": "5004", "type": "Maple" }
        ]
},
{
    "id": "0002",
    "type": "donut",
    "name": "Raised",
    "ppu": 0.55,
    "batters":
        {
            "batter":
                [
                    { "id": "1005", "type": "DeleteMe" }
                ]
        },
    "topping":
        [
            { "id": "5001", "type": "None" },
            { "id": "5002", "type": "Glazed" }
        ]
},
{
    "id": "0003",
    "type": "donut",
    "name": "Old Fashioned",
    "ppu": 0.55,
    "batters":
        {
            "batter":
                [
                    { "id": "1001", "type": "Regular" },
                    { "id": "1002", "type": "Chocolate" }
                ]
        },
    "topping":
        [
            { "id": "5001", "type": "None" }
        ]
}
]';

objArray = JSON.parse(json);

objArray.reject! {|obj|
  obj['batters']['batter'].detect {|batter|
    batter['type'] == 'DeleteMe'
  }
};

result = objArray.to_json;
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