希望大家都做得很好。總和列只有ID不同
我有以下的輸出...
+---------+--------------+--------------+-----------+---------+----------+ | ord_num | signoff_date | program_name | prod_desc | tx_comp | priority | +---------+--------------+--------------+-----------+---------+----------+ | 1234567 | 2012-08-12 | ilearn | run | 1 | 1 | | 1234567 | 2012-08-12 | ilearn | plan | 1 | 1 | | 1234568 | 2012-08-12 | other | run | 1 | 1 | | 1234569 | 2012-08-12 | other | run | 0 | 1 | +---------+--------------+--------------+-----------+---------+----------+
我想這樣做是SUM
的tx_comp
每列獨特的「ord_num」一次。現在
我不能使用GROUP BY ord_num
我還任務的類型做了sum
。
它喜歡我需要知道以前的ord_num
然後總結如果有什麼不同?
任何想法將不勝感激。
謝謝。
*編輯*
SELECT
signoff_date,
SUM(IF(prod_desc = 'run', 1, 0)) AS run,
SUM(IF(prod_desc = 'plan', 1, 0)) AS plan,
SUM(tx_comp) AS tx_comp
FROM
(
SELECT
ord_num,
signoff_date,
program_name,
prod_desc,
tx_comp,
priority
FROM
test.orders
LEFT JOIN test.tx_comp USING (ord_num)
) AS grp
顯然不是期望的輸出
+--------------+------+------+---------+ | signoff_date | run | plan | tx_comp | +--------------+------+------+---------+ | 2012-08-12 | 3 | 1 | 3 | +--------------+------+------+---------+
我之後...
+--------------+------+------+---------+ | signoff_date | run | plan | tx_comp | +--------------+------+------+---------+ | 2012-08-12 | 3 | 1 | 2 | +--------------+------+------+---------+
所以,你想要兩個不同的總和,從不同的標準,在一個單一的查詢? –
@MarcB是的,這是正確的...... – mybigman
「只有在ord_num不同」 - 從什麼?前一行? –