投入单词是一件相当尴尬的事情，但是当项目首次切换到关税/客户时我想要一行。 如果物品切换客户，则应返回开关日期，而不管关税的差异。如果项目切换项目，则返回的日期不得更改，除非新项目的关税与旧项目的关税不同。
我不太清楚如何让它更清晰，但我愿意接受建议。
我的查询如下：
SET @id = 1;
SELECT DISTINCT
ip.ItemID,
ip.ProjectID,
p.TariffID,
p.CustomerID,
cs.Date
FROM item_project ip
LEFT JOIN item_project ip1
ON ip.ItemID = ip1.ItemID
AND ip.Date < ip1.Date
LEFT JOIN project p
ON ip.ProjectID = p.ProjectID
LEFT JOIN (
SELECT
ip.ItemID,
ip.Date
FROM item_project ip
LEFT JOIN item_project ip1
ON ip.ProjectID = ip1.ProjectID
AND ip.Date > ip1.Date
LEFT JOIN project p
ON ip.ProjectID = p.ProjectID
WHERE ip.ItemID = @id
AND ip1.ItemID IS NULL
AND p.CustomerID = (
SELECT p.CustomerID
FROM project p
LEFT JOIN item_project ip
ON p.ProjectID = ip.ProjectID
LEFT JOIN item_project ip1
ON ip.ItemID = ip1.ItemID
AND ip.Date < ip1.Date
WHERE ip.ItemID = @id
AND ip1.ItemID IS NULL
)
AND p.TariffID = (
SELECT p.TariffID
FROM project p
LEFT JOIN item_project ip
ON p.ProjectID = ip.ProjectID
LEFT JOIN item_project ip1
ON ip.ItemID = ip1.ItemID
AND ip.Date < ip1.Date
WHERE ip.ItemID = @id
AND ip1.ItemID IS NULL
)
) AS cs
ON ip.ItemID = cs.ItemID
WHERE ip.ItemID = @id
AND ip1.ItemID IS NULL
给了我
"ItemID","ProjectID","TariffID","CustomerID","Date"
"1","2","1","1","2010-11-10 00:00:00"
这是错误的日期
SET @id=2给了我：
"2","2","1","1",NULL
除了日期
之外，这是正确的 SET @id=3给了我：
"3","2","1","1",NULL
除日期外，这也是正确的。
这是数据库
CREATE TABLE IF NOT EXISTS `item_project` (
`ID` int(10) unsigned NOT NULL auto_increment,
`ItemID` varchar(10) NOT NULL,
`ProjectID` int(10) unsigned NOT NULL,
`Date` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1;
INSERT INTO `item_project` (`ID`, `ItemID`, `ProjectID`, `Date`) VALUES
(1, '1', 1, '2010-11-05 00:00:00'),
(2, '1', 2, '2010-11-10 00:00:00'),
(3, '1', 3, '2010-11-20 00:00:00'),
(4, '2', 2, '2010-11-21 00:00:00'),
(5, '3', 4, '2010-11-21 00:00:00'),
(6, '3', 2, '2010-11-22 00:00:00'),
(7, '1', 2, '2010-11-23 00:00:00'),
CREATE TABLE IF NOT EXISTS `project` (
`ProjectID` int(10) unsigned NOT NULL auto_increment,
`Name` varchar(45) NOT NULL,
`TariffID` varchar(45) NOT NULL,
`CustomerID` varchar(45) NOT NULL,
PRIMARY KEY (`ProjectID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;
INSERT INTO `project` (`ProjectID`, `Name`, `TariffID`, `CustomerID`) VALUES
(1, 'Test', '2', '1'),
(2, 'Another test', '1', '1'),
(3, 'Project1', '1', '1'),
(4, 'Main project', '2', '2');
CREATE TABLE IF NOT EXISTS `tariff` (
`TariffID` int(10) unsigned NOT NULL auto_increment,
`Tariff` varchar(45) NOT NULL,
PRIMARY KEY (`TariffID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;
INSERT INTO `tariff` (`TariffID`, `Tariff`) VALUES
(1, 'Tariff 1'),
(2, 'Tariff 2');
修改：商品可以返回旧的关税或客户。在这种情况下，显示的日期应该是发生这种情况的日期。
答案 0 :(得分：2)
有趣的情况。这是我想出来的
SELECT i_general.ItemID, ProjectID, TariffID, CustomerID, the_date
FROM
(
SELECT
ip.ItemID,
p.ProjectID,
p.TariffID,
p.CustomerID
FROM item_project ip
INNER JOIN project p ON ip.ProjectID = p.ProjectID
INNER JOIN tariff t ON p.TariffID = t.TariffID
INNER JOIN (
SELECT
ip.ItemID, MAX(Date) AS max_date
FROM item_project ip
GROUP BY ip.ItemID
) ip_max ON ip_max.ItemID=ip.ItemID AND ip_max.max_date=ip.Date
) i_general
INNER JOIN (
SELECT ip1.ItemID, IF(MIN(ip2.Date) IS NULL,MIN(ip1.Date),MIN(ip2.Date)) AS the_date
FROM item_project ip1
INNER JOIN project p1 ON ip1.ProjectID = p1.ProjectID
LEFT JOIN item_project ip2 ON ip1.ItemID=ip2.ItemID AND ip1.Date < ip2.Date
LEFT JOIN project p2 ON ip2.ProjectID = p2.ProjectID AND (p2.TariffID!=p1.TariffID OR p2.CustomerID!=p1.CustomerID)
GROUP BY ip1.ItemID
) i_date_info ON i_date_info.ItemID = i_general.ItemID
当然，您可以根据需要插入一些WHERE ItemID = @id。内部查询越多越好。
无论如何，它会导致
+--------+-----------+----------+------------+---------------------+
| ItemID | ProjectID | TariffID | CustomerID | the_date |
+--------+-----------+----------+------------+---------------------+
| 1 | 3 | 1 | 1 | 2010-11-10 00:00:00 |
| 2 | 2 | 1 | 1 | 2010-11-21 00:00:00 |
| 3 | 2 | 1 | 1 | 2010-11-22 00:00:00 |
+--------+-----------+----------+------------+---------------------+
因此，它似乎与现有数据集一起使用。 Lemme知道你是否可以提供不起作用的测试数据。
答案 1 :(得分：1)
我之前的回答似乎也适用于新数据集，但我想我理解你所看到的问题。新提出的解决方案
SELECT i_general.ItemID, ProjectID, TariffID, CustomerID, the_date
FROM
(
SELECT ip.ItemID, p.ProjectID, p.TariffID, p.CustomerID
FROM item_project ip
INNER JOIN project p ON ip.ProjectID = p.ProjectID
INNER JOIN tariff t ON p.TariffID = t.TariffID
INNER JOIN (
SELECT ip.ItemID, MAX(Date) AS max_date
FROM item_project ip
GROUP BY ip.ItemID
) ip_max ON ip_max.ItemID=ip.ItemID AND ip_max.max_date=ip.Date
) i_general
INNER JOIN
(
SELECT ItemID_1 AS ItemID, IF(MAX(Next_Change_Date) IS NULL, MIN(Date_1), MAX(Next_Change_Date)) AS the_date
FROM
(
SELECT ItemID_1, Date_1, MIN(Date_2) AS Next_Change_Date
FROM
(
SELECT ip1.ItemID AS ItemID_1, ip1.Date AS Date_1, p1.TariffID AS TariffID_1, p1.CustomerID AS CustomerID_1
FROM item_project ip1
INNER JOIN project p1 ON ip1.ProjectID = p1.ProjectID
) ipp1
LEFT JOIN
(
SELECT ip2.ItemID AS ItemID_2, ip2.Date AS Date_2, p2.TariffID AS TariffID_2, p2.CustomerID AS CustomerID_2
FROM item_project ip2
INNER JOIN project p2 ON ip2.ProjectID = p2.ProjectID
) ipp2 ON ItemID_1=ItemID_2 AND Date_1 < Date_2 AND ((TariffID_1!=TariffID_2 OR CustomerID_1!=CustomerID_2) AND Date_2 IS NOT NULL)
GROUP BY ItemID_1, Date_1
) i_date_pair_info
GROUP BY ItemID
) i_date_info ON i_date_info.ItemID = i_general.ItemID
我认为它提供了您正在寻找的结果。
内部子查询中的一个i_date_pair_info明确地将每个日期与所有后续更改（如果有的话）配对，这样更健壮。然后，分组会消除除最快的变化之外的所有变化。