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SELECT語句的結果與VIEW中的SELECT結果不同。如何解決問題並從視野中獲得相同的結果?不同的SELECT和VIEW中的SELECT結果
動作表:
+--+---------+--------------+-----------+------+
|id|person_id|action_type_id|currency_id|sum |
+--+---------+--------------+-----------+------+
|1 |1 |1 |1 | 1.00 |
|2 |1 |1 |1 | 5.00 |
|3 |1 |1 |2 |10.00 |
|4 |1 |2 |1 | 2.00 |
|5 |2 |1 |1 |20.00 |
|6 |2 |2 |2 | 5.00 |
+--+---------+--------------+-----------+------+
選擇:
SELECT person_id AS p, currency_id AS c,
(
CAST(
COALESCE(
(SELECT SUM(sum) FROM actions WHERE action_type_id=1 AND person_id=p AND currency_id=c)
, 0)
AS DECIMAL(11,2)) -
CAST(
COALESCE(
(SELECT SUM(sum) FROM actions WHERE action_type_id=2 AND person_id=p AND currency_id=c)
, 0)
AS DECIMAL(11,2))
) AS sum
FROM actions
GROUP BY currency_id, person_id
ORDER BY person_id, currency_id;
結果:
+--+--+------+
|p |c |sum |
+--+--+------+
|1 |1 | 4.00 |
|1 |2 |10.00 |
|2 |1 |20.00 |
|2 |2 |-5.00 |
+--+--+------+
選擇內部視圖:
CREATE VIEW p_sums AS
SELECT person_id AS p, currency_id AS c,
(
CAST(
COALESCE(
(SELECT SUM(sum) FROM actions WHERE action_type_id=1 AND person_id=p AND currency_id=c)
, 0)
AS DECIMAL(11,2)) -
CAST(
COALESCE(
(SELECT SUM(sum) FROM actions WHERE action_type_id=2 AND person_id=p AND currency_id=c)
, 0)
AS DECIMAL(11,2))
) AS sum
FROM actions
GROUP BY currency_id, person_id
ORDER BY person_id, currency_id;
SELECT * FROM p_sums;
結果:
+--+--+------+
|p |c |sum |
+--+--+------+
|1 |1 |29.00 |
|1 |2 |29.00 |
|2 |1 |29.00 |
|2 |2 |29.00 |
+--+--+------+
你使用的是哪個版本的mysql? – Jason 2011-01-06 13:17:06
只是FYI,對於新手(我也是一個),當有人幫助提供解決方案時,它是讓你超越你的殘局的一種方法,點擊他們答案下的複選框可以讓他們相信,以便其他人知道解決方案已經解決。 – DRapp 2011-01-06 13:29:43