您好,
我有一个非常大的MySQL数据库,其表格的结构如下:
在exmaple中,DATE采用unix时间戳格式。所以它需要
转换为正常的美国日期格式,这只是我的数据库中的一些记录
ID日期
REG_TYPE
--------------------------------------
1 1251917888
0
2 1251917888
1
3 1251917888
1
4 1251917888
0
5 1251917888
2
6 1251917888
3
7 1251917888
2
8 1251917888
4
9 1251917888
0
10 1251917888
0
问题是我想得到按日期排序的每个REG_TYPE的计数。
下面显示了我想要得到的内容:
日期REGTYPE(0)
REGTYPE(1)REGTYPE(2)REGTYPE(3)
REGTYPE(4)
-------------------------------------------------- ------------------------------------------------ <无线电通信/>
XXXX
4
2
2
1
1
XXXX
X
X
X
X
X
我想为数据库中的每个日期得到这个,就像摘要一样
每个日期。
任何人都可以建议一个可能的解决方案吗?我必须转换此输出
从MYSQL获取数据后,在PHP中的数组。不使用循环的原因
在PHP中,这是因为数据库太大而且会导致超时
最诚挚的问候
答案 0 :(得分:5)
您要做的是一个数据透视操作,SQL语法不直接支持它。但是,它并不太复杂,概念上涉及两个步骤:
我正在使用此数据集作为示例:
mysql> select * from foo;
+----+------------+----------+
| id | thedate | reg_type |
+----+------------+----------+
| 1 | 1251917888 | 0 |
| 2 | 1251917888 | 1 |
| 3 | 1251917888 | 1 |
| 4 | 1251917888 | 0 |
| 5 | 1251917888 | 2 |
| 6 | 1251917888 | 3 |
| 7 | 1251917888 | 2 |
| 8 | 1251917888 | 4 |
| 9 | 1251917888 | 0 |
| 10 | 1251917888 | 0 |
| 11 | 1251831488 | 1 |
| 12 | 1251831488 | 2 |
| 13 | 1251831488 | 2 |
| 14 | 1251831488 | 1 |
| 15 | 1251831488 | 3 |
| 16 | 1251831488 | 4 |
| 17 | 1251831488 | 3 |
| 18 | 1251831488 | 5 |
| 19 | 1251831488 | 1 |
| 20 | 1251831488 | 1 |
+----+------------+----------+
第1步是“炸毁”数据集:
select id
, thedate
, case when reg_type = 0 then 1 else 0 end as reg_type_0
, case when reg_type = 1 then 1 else 0 end as reg_type_1
, case when reg_type = 2 then 1 else 0 end as reg_type_2
, case when reg_type = 3 then 1 else 0 end as reg_type_3
, case when reg_type = 4 then 1 else 0 end as reg_type_4
, case when reg_type = 5 then 1 else 0 end as reg_type_5
from foo;
给出:
+----+------------+------------+------------+------------+------------+------------+------------+
| id | thedate | reg_type_0 | reg_type_1 | reg_type_2 | reg_type_3 | reg_type_4 | reg_type_5 |
+----+------------+------------+------------+------------+------------+------------+------------+
| 1 | 1251917888 | 1 | 0 | 0 | 0 | 0 | 0 |
| 2 | 1251917888 | 0 | 1 | 0 | 0 | 0 | 0 |
| 3 | 1251917888 | 0 | 1 | 0 | 0 | 0 | 0 |
| 4 | 1251917888 | 1 | 0 | 0 | 0 | 0 | 0 |
| 5 | 1251917888 | 0 | 0 | 1 | 0 | 0 | 0 |
| 6 | 1251917888 | 0 | 0 | 0 | 1 | 0 | 0 |
| 7 | 1251917888 | 0 | 0 | 1 | 0 | 0 | 0 |
| 8 | 1251917888 | 0 | 0 | 0 | 0 | 1 | 0 |
| 9 | 1251917888 | 1 | 0 | 0 | 0 | 0 | 0 |
| 10 | 1251917888 | 1 | 0 | 0 | 0 | 0 | 0 |
| 11 | 1251831488 | 0 | 1 | 0 | 0 | 0 | 0 |
| 12 | 1251831488 | 0 | 0 | 1 | 0 | 0 | 0 |
| 13 | 1251831488 | 0 | 0 | 1 | 0 | 0 | 0 |
| 14 | 1251831488 | 0 | 1 | 0 | 0 | 0 | 0 |
| 15 | 1251831488 | 0 | 0 | 0 | 1 | 0 | 0 |
| 16 | 1251831488 | 0 | 0 | 0 | 0 | 1 | 0 |
| 17 | 1251831488 | 0 | 0 | 0 | 1 | 0 | 0 |
| 18 | 1251831488 | 0 | 0 | 0 | 0 | 0 | 1 |
| 19 | 1251831488 | 0 | 1 | 0 | 0 | 0 | 0 |
| 20 | 1251831488 | 0 | 1 | 0 | 0 | 0 | 0 |
+----+------------+------------+------------+------------+------------+------------+------------+
接下来,我们在每个日期的输出中折叠为一行,并将每个reg_type_ *列相加,使用或初始查询作为内联视图(也称为“子查询”):
select thedate
, sum(i.reg_type_0) as reg_type_0
, sum(i.reg_type_1) as reg_type_1
, sum(i.reg_type_2) as reg_type_2
, sum(i.reg_type_3) as reg_type_3
, sum(i.reg_type_4) as reg_type_4
, sum(i.reg_type_5) as reg_type_5
from (
select id
, thedate
, case when reg_type = 0 then 1 else 0 end as reg_type_0
, case when reg_type = 1 then 1 else 0 end as reg_type_1
, case when reg_type = 2 then 1 else 0 end as reg_type_2
, case when reg_type = 3 then 1 else 0 end as reg_type_3
, case when reg_type = 4 then 1 else 0 end as reg_type_4
, case when reg_type = 5 then 1 else 0 end as reg_type_5
from foo
) i
group by thedate
order by thedate asc;
(另请注意,您可以将这两个查询合并为一个,但为了清楚起见,我在这里单独显示它们;至少在MySQL中,这似乎导致更简单的执行计划,这通常意味着更快的执行 - 如总是,在真实的数据集上测试你的SQL性能,不要相信我的话!)
这给了我们:
+------------+------------+------------+------------+------------+------------+------------+
| thedate | reg_type_0 | reg_type_1 | reg_type_2 | reg_type_3 | reg_type_4 | reg_type_5 |
+------------+------------+------------+------------+------------+------------+------------+
| 1251831488 | 0 | 4 | 2 | 2 | 1 | 1 |
| 1251917888 | 4 | 2 | 2 | 1 | 1 | 0 |
+------------+------------+------------+------------+------------+------------+------------+
这是期望的结果。您可以使用MySQL函数FROM_UNIXTIME将日期转换为DATE,并且在第2部分查询中执行此操作可能是最有效的(评估函数的次数较少,并且对整数组进行比较,而不是DATE - 不确定这在MySQL中是否真的有任何不同。)
答案 1 :(得分:1)
除了你输出的输出(你发布的示例输出是一个数据透视表)之外,你必须使用COUNT BY语句
例如:
SELECT `DATE`, COUNT(*)
FROM `TABLE_NAME`
GROUP BY `REG_TYPE`
答案 2 :(得分:1)
您可以使用以下内容在mysql中模拟数据透视表:
SELECT
SUM(REG_TYPE = 0) AS reg_type_0,
SUM(REG_TYPE = 1) AS reg_type_1,
SUM(REG_TYPE = 2) AS reg_type_2,
SUM(REG_TYPE = 3) AS reg_type_3,
SUM(REG_TYPE = 4) AS reg_type_4,
TO_DAYS(FROM_UNIXTIME(date)) AS day_number
FROM my_table
GROUP BY TO_DAYS(FROM_UNIXTIME(date))
日期操作位可能需要一些调整。
答案 3 :(得分:0)
你可以查询REG_TYPE的所有值(如果你事先不知道它们)然后组装这样的东西(抱歉,我的MySQL生锈了):
select date,
SUM(IF(REGTYPE=0, 0, 1) AS REGTYPE0,
SUM(IF(REGTYPE=1, 0, 1) AS REGTYPE1,
SUM(IF(REGTYPE=2, 0, 1) AS REGTYPE2,
SUM(IF(REGTYPE=3, 0, 1) AS REGTYPE3,
SUM(IF(REGTYPE=4, 0, 1) AS REGTYPE4
FROM table
GROUP BY date