合併行 ，行轉列
SELECT * FROM test.`dd`;output:
id verdor_id x y z date
------ --------- ------ ------ ------ ------------
1 A 11 789 24 2016-06-23
2 B 1 11 28 2016-06-23
3 B 1 11 28 2016-06-22
4 B 1 11 28 2016-06-21
5 A 11 789 24 2016-06-22# 合併行 ，行轉列
SELECT
dd.`date`,
MAX(CASE WHEN verdor_id = 'A' THEN dd.`x` ELSE 0 END) 'A_vendor_x',
MAX(CASE WHEN verdor_id = 'B' THEN dd.`x` ELSE 0 END) 'B_verdor_x',
MAX(CASE WHEN verdor_id = 'A' THEN dd.`y` ELSE 0 END) 'A_vendor_y',
MAX(CASE WHEN verdor_id = 'B' THEN dd.`y` ELSE 0 END) 'B_verdor_y',
MAX(CASE WHEN verdor_id = 'A' THEN dd.`z` ELSE 0 END) 'A_vendor_z',
MAX(CASE WHEN verdor_id = 'B' THEN dd.`z` ELSE 0 END) 'B_verdor_z'
FROM `dd`
GROUP BY dd.`date`output:
date A_vendor_x B_verdor_x A_vendor_y B_verdor_y A_vendor_z B_verdor_z
---------- ---------- ---------- ---------- ---------- ---------- ------------
2016-06-21 0 1 0 11 0 28
2016-06-22 11 1 789 11 24 28
2016-06-23 11 1 789 11 24 28如何理解上面這條 sql 是如何工作的？
首先：
SELECT
dd.`date`,
CASE WHEN verdor_id = 'A' THEN dd.`x` ELSE 0 END 'A_vendor_x',
CASE WHEN verdor_id = 'B' THEN dd.`x` ELSE 0 END 'B_verdor_x',
CASE WHEN verdor_id = 'A' THEN dd.`y` ELSE 0 END 'A_vendor_y',
CASE WHEN verdor_id = 'B' THEN dd.`y` ELSE 0 END 'B_verdor_y',
CASE WHEN verdor_id = 'A' THEN dd.`z` ELSE 0 END 'A_vendor_z',
CASE WHEN verdor_id = 'B' THEN dd.`z` ELSE 0 END 'B_verdor_z'
FROM `dd`output:
date A_vendor_x B_verdor_x A_vendor_y B_verdor_y A_vendor_z B_verdor_z
---------- ---------- ---------- ---------- ---------- ---------- ------------
2016-06-23 11 0 789 0 24 0
2016-06-23 0 1 0 11 0 28
2016-06-22 0 1 0 11 0 28
2016-06-21 0 1 0 11 0 28
2016-06-22 11 0 789 0 24 0然後，通過 max + group by 去掉 0 即當初未匹配到的情況。