前两天IT群里有人在问如何将员工的每个月的提成按月份汇总出来,时间是一个具体的时间点没有具体月份。问题和效果如下:
1.员工_提成表:
2.汇总效果:
3.查询语句如下:
SELECT
`nickname`,
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 01 THEN
`salary`
ELSE
0
END
) AS '一月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 02 THEN
`salary`
ELSE
0
END
) AS '二月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 03 THEN
`salary`
ELSE
0
END
) AS '三月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 04 THEN
`salary`
ELSE
0
END
) AS '四月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 05 THEN
`salary`
ELSE
0
END
) AS '五月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 06 THEN
`salary`
ELSE
0
END
) AS '六月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 07 THEN
`salary`
ELSE
0
END
) AS '七月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 08 THEN
`salary`
ELSE
0
END
) AS '八月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 09 THEN
`salary`
ELSE
0
END
) AS '九月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 10 THEN
`salary`
ELSE
0
END
) AS '十月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 11 THEN
`salary`
ELSE
0
END
) AS '十一月',
sum(
CASE DATE_FORMAT(belong_at, '%m')
WHEN 12 THEN
`salary`
ELSE
0
END
) AS '十二月'
FROM
ts_test_label
GROUP BY
`nickname`
4.创建员工_提成表:
SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for ts_test_label
-- ----------------------------
DROP TABLE IF EXISTS `ts_test_label`;
CREATE TABLE `ts_test_label` (
`nickname` varchar(20) DEFAULT NULL COMMENT '昵称',
`salary` float(7,2) DEFAULT NULL COMMENT '提成',
`belong_at` datetime DEFAULT NULL COMMENT '所属时间'
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
完毕!
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