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能一次性用数据库查询出来吗?
能一次性用数据库查询出来吗?
我很在意!!!!
1. first step
<code>SELECT t.* FROM ( SELECT t2.`etime` AS `日期`, SUM(CASE WHEN t2.`epid` = t1_1.`pid` THEN t2.`ecost` ELSE 0 END) AS `A`, SUM(CASE WHEN t2.`epid` = t1_2.`pid` THEN t2.`ecost` ELSE 0 END) AS `B`, SUM(CASE WHEN t2.`epid` = t1_3.`pid` THEN t2.`ecost` ELSE 0 END) AS `C`, SUM(t2.`ecost`) AS `总计` FROM table2 t2 LEFT JOIN table1 t1_1 ON t1_1.`pid` = t2.`epid` AND t1_1.`pid` = 1 LEFT JOIN table1 t1_2 ON t1_2.`pid` = t2.`epid` AND t1_2.`pid` = 2 LEFT JOIN table1 t1_3 ON t1_3.`pid` = t2.`epid` AND t1_3.`pid` = 3 GROUP BY t2.`etime` ) t ORDER BY t.`日期` DESC </code>
output:
<code>+------------+----+----+----+------+ | 日期 | A | B | C | 总计 | +------------+----+----+----+------+ | 2015-08-04 | 88 | 44 | 0 | 132 | | 2015-08-03 | 88 | 77 | 66 | 231 | +------------+----+----+----+------+ </code>
<code>SELECT IFNULL(tt.`日期`, '总计') AS `日期`, SUM(tt.`A`) AS `A`, SUM(tt.`B`) AS `B`, SUM(tt.`C`) AS `C`,SUM(tt.`总计`) AS `总计` FROM ( SELECT t.* FROM ( SELECT t2.`etime` AS `日期`, SUM(CASE WHEN t2.`epid` = t1_1.`pid` THEN t2.`ecost` ELSE 0 END) AS `A`, SUM(CASE WHEN t2.`epid` = t1_2.`pid` THEN t2.`ecost` ELSE 0 END) AS `B`, SUM(CASE WHEN t2.`epid` = t1_3.`pid` THEN t2.`ecost` ELSE 0 END) AS `C`, SUM(t2.`ecost`) AS `总计` FROM table2 t2 LEFT JOIN table1 t1_1 ON t1_1.`pid` = t2.`epid` AND t1_1.`pid` = 1 LEFT JOIN table1 t1_2 ON t1_2.`pid` = t2.`epid` AND t1_2.`pid` = 2 LEFT JOIN table1 t1_3 ON t1_3.`pid` = t2.`epid` AND t1_3.`pid` = 3 GROUP BY t2.`etime` ) t ORDER BY t.`日期` DESC ) tt GROUP BY tt.`日期` WITH ROLLUP </code>
output:
<code>+------------+-----+-----+----+------+ | 日期 | A | B | C | 总计 | +------------+-----+-----+----+------+ | 2015-08-03 | 88 | 77 | 66 | 231 | | 2015-08-04 | 88 | 44 | 0 | 132 | | 总计 | 176 | 121 | 66 | 363 | +------------+-----+-----+----+------+ </code>
附原始数据表:
<code>mysql> select * from table1; +-----+-------+ | pid | pname | +-----+-------+ | 1 | A | | 2 | B | | 3 | C | +-----+-------+ mysql> select * from table2; +-----+------------+------+-------+ | eid | etime | epid | ecost | +-----+------------+------+-------+ | 1 | 2015-08-03 | 1 | 88 | | 2 | 2015-08-03 | 2 | 77 | | 3 | 2015-08-03 | 3 | 66 | | 4 | 2015-08-04 | 1 | 55 | | 5 | 2015-08-04 | 2 | 44 | | 6 | 2015-08-04 | 1 | 33 | +-----+------------+------+-------+ 6 rows in set </code>
总感觉数据哪里有问题,是题主算错了,还是我算错了。
要做转置,处理起来相当的麻烦,基本思路是先用一个SQL查出列(A、B、C等),再循环,生成另一个SQL,再执行生成的SQL得到结果……太麻烦不写了,自己百度搜转置
简单的方法就是用一般的SQL统计查询,然后再在程序里来拼出你想要的表。
表1为主表 左联接加入表2
SELECT b.etime,a.pname,SUM(b.ecost) as sum_cost FROM table1 as a,table2 as b WHERE a.pid=b.epid GROUP BY b.etime,b.epid
执行后结果如图:
拿回来自己再算总计。
楼主的图画得不错,差点把我绕进去了...
select * from ( select * from table2 left join table1 on table2.epid =table1.pid order by eid asc,etime asc) as sumTable group by etime,eid;
得
panme etime ecost
A 08-04 x
A 08-05 x
B 08-04 x
B 08-05 x
好了,到了这步,楼主不会用php转成那种格式吗?要用sql也可以,不过也是需要一样的逻辑,PHP难道不比sql好使吗?PHP是世界上最好的语言喔
SELECT etime,SUM(IF(epid=1,ecost,0)) AS A,
SUM(IF(epid=2,ecost,0)) AS B,
SUM(IF(epid=3,ecost,0)) AS C, SUM(ecost) AS Total
FROM mytable GROUP BY etime;
没测试。