SELECT語句子查詢

編輯：我需要在那裏減去直接和間接分鐘的總數。SELECT語句子查詢

我想要SUM M.分鐘作爲別名「dminutes」。然後，再次取M.minutes的SUM並減去具有「間接」列值的M.minutes（並給它「inminutes」別名）。但是，它顯示爲空，所以語法錯誤。建議？

table = tasks 
column = task_type 


Example: 
M.minutes total = 60 minutes 
M. minutes (with "direct" task_type column value) = 50 minutes (AS dminutes) 
M. minutes (with "indirect" task_type column value) = 10 minutes (AS inminutes)

SQL語句：

SELECT 
U.user_name, 
SUM(M.minutes) as dminutes, 
ROUND(SUM(M.minutes))-(SELECT (SUM(M.minutes)) from summary s WHERE ta.task_type='indirect') as inminutes 
FROM summary S 
JOIN users U ON U.user_id = S.user_id 
JOIN tasks TA ON TA.task_id = S.task_id 
JOIN minutes M ON M.minutes_id = S.minutes_id 
WHERE DATE(submit_date) = curdate() 
AND TIME(submit_date) BETWEEN '00:00:01' and '23:59:59' 
GROUP BY U.user_name 
LIMIT 0 , 30

來源

2012-10-28 Mike

我覺得這樣的事情應該工作。

您可能需要調整一下。

SELECT direct.duser_id, indirect.iminutes, direct.dminutes, 
    direct.dminutes - indirect.iminutes FROM 
    (SELECT U.user_id AS iuser_id, SUM(M.minutes) AS iminutes 
    FROM summary S 
    JOIN users U 
    ON U.user_id = S.user_id 
    JOIN minutes M 
    ON M.minutes_id = S.minutes_id 
    JOIN tasks TA 
    ON TA.task_id = S.task_id 
    WHERE TA.task_type='indirect' 
    AND DATE(submit_date) = curdate() 
    AND TIME(submit_date) BETWEEN '00:00:01' and '23:59:59' 
    GROUP BY U.user_id) AS indirect 
JOIN 
    (SELECT U.user_id AS duser_id, SUM(M.minutes) AS dminutes 
    FROM summary S 
    JOIN users U 
    ON U.user_id = S.user_id 
    JOIN minutes M 
    ON M.minutes_id = S.minutes_id 
    JOIN tasks TA 
    ON TA.task_id = S.task_id 
    WHERE TA.task_type='direct' 
    AND DATE(submit_date) = curdate() 
    AND TIME(submit_date) BETWEEN '00:00:01' and '23:59:59' 
    GROUP BY U.user_id) AS direct 
WHERE indirect.iuser_id = direct.duser_id

來源

2012-10-28 20:40:30

我認爲你是正確的，在正確的軌道上！雖然，我得到一個語法錯誤..我試過調整它，但它不喜歡臨時表加入「ON iuser_id = duser_id」。錯誤：＃1064 - 你的SQL語法錯誤;檢查與你的MySQL服務器版本相對應的手冊，在'ON iuser_id = duser_id'附近使用正確的語法'' – Mike

你是我的朋友，是某種巫師！哈哈它完美的工作，謝謝你！ – Mike

這對我來說很瘋狂，需要多少編碼才能在Mysql中進行簡單的減法工作。不要誤解我的意思，我喜歡它，因爲我只需要做一次，然後聲明就會永遠使用，但是人......大聲笑 – Mike

SUM是一個討厭的小功能：

http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_sum

Returns the sum of expr. If the return set has no rows, SUM() returns NULL. The DISTINCT keyword can be used to sum only the distinct values of expr.

SUM() returns NULL if there were no matching rows.

嘗試包裝SUM的COALESCE或IFNULL：

... COALESCE(SUM(whatever), 0) ...

來源

2012-10-28 20:43:05 biziclop

那張照片是什麼？ – Kermit

這是一個複製粘貼故障:) – biziclop

+1的包裝想法 – inhan

SELECT語句子查詢

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