case when 根据具体时间汇总员工每个月提成

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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