我有一个SQL查询，它选择一个枚举列并执行连接。以下是涉及的数据库表和相关数据：
提供表
--
-- Table structure for table `offer`
--
CREATE TABLE `offer` (
`id` int(11) NOT NULL auto_increment,
`companyID` int(11) NOT NULL,
`categoryID` int(11) NOT NULL,
`dateAdded` timestamp NOT NULL default CURRENT_TIMESTAMP,
`details` text NOT NULL,
`amount` decimal(11,0) NOT NULL,
`maxAmount` decimal(11,0) NOT NULL default '0',
`dateExpires` int(11) NOT NULL default '0',
`active` enum('YES','NO') NOT NULL,
`featured` enum('YES','NO') NOT NULL default 'NO',
`adType` enum('OFFER','URL') NOT NULL default 'OFFER',
`URL` varchar(255) NOT NULL,
`address1` varchar(50) NOT NULL,
`address2` varchar(50) NOT NULL,
`city` varchar(50) NOT NULL,
`state` char(2) NOT NULL,
`zip` varchar(10) NOT NULL,
PRIMARY KEY (`id`),
KEY `companyID` (`companyID`),
KEY `categoryID` (`categoryID`),
KEY `zip` (`zip`)
) ENGINE=MyISAM AUTO_INCREMENT=66 DEFAULT CHARSET=utf8 AUTO_INCREMENT=66 ;
--
-- Dumping data for table `offer`
--
INSERT INTO `offer` (`id`, `companyID`, `categoryID`, `dateAdded`, `details`, `amount`, `maxAmount`, `dateExpires`, `active`, `featured`, `adType`, `URL`, `address1`, `address2`, `city`, `state`, `zip`) VALUES
(24, 10, 6, '2011-05-19 14:55:54', 'Computer Software & Books', 25, 0, 0, 'NO', 'NO', 'OFFER', '', '912 N. Avenue 57', '', 'Los Angeles', 'CA', '90043')
付款表
--
-- Table structure for table `payments`
--
CREATE TABLE `payments` (
`id` int(11) NOT NULL auto_increment,
`date` timestamp NOT NULL default CURRENT_TIMESTAMP,
`offerID` int(11) NOT NULL,
`amount` decimal(11,2) NOT NULL,
`memberID` int(11) NOT NULL,
`validationNumber` varchar(20) NOT NULL,
`status` enum('NEW','USED') NOT NULL default 'NEW',
`dateRedeemed` datetime NOT NULL,
PRIMARY KEY (`id`),
KEY `validationNumber` (`validationNumber`),
KEY `offerID` (`offerID`),
KEY `memberID` (`memberID`)
) ENGINE=MyISAM AUTO_INCREMENT=33 DEFAULT CHARSET=utf8 AUTO_INCREMENT=33 ;
--
-- Dumping data for table `payments`
--
INSERT INTO `payments` (`id`, `date`, `offerID`, `amount`, `memberID`, `validationNumber`, `status`, `dateRedeemed`) VALUES
(1, '2011-03-28 16:33:24', 24, 11.00, 8, '1A2B3-1', 'NEW', '2011-07-18 15:00:41'),
(2, '2011-04-12 12:47:58', 16, 81.10, 8, '2C3D4-2', 'NEW', '0000-00-00 00:00:00'),
(3, '2011-05-19 19:50:58', 24, 22.15, 14, 'ABCDE-3', 'USED', '2011-07-18 15:03:00'),
(4, '2011-05-19 19:50:58', 24, 44.30, 5, 'FGHIJK-4', 'USED', '2011-07-18 15:03:45');
交易表
--
-- Table structure for table `transactions`
--
CREATE TABLE `transactions` (
`id` int(11) NOT NULL auto_increment,
`companyID` int(11) NOT NULL,
`offerID` int(11) NOT NULL,
`date` timestamp NOT NULL default CURRENT_TIMESTAMP,
`amount` decimal(10,2) NOT NULL,
`type` enum('CREDIT','DEBIT') NOT NULL,
PRIMARY KEY (`id`),
KEY `companyID` (`companyID`,`offerID`)
) ENGINE=MyISAM AUTO_INCREMENT=38 DEFAULT CHARSET=utf8 AUTO_INCREMENT=38 ;
--
-- Dumping data for table `transactions`
--
INSERT INTO `transactions` (`id`, `companyID`, `offerID`, `date`, `amount`, `type`) VALUES
(1, 10, 24, '2011-05-20 11:25:35', 11.00, 'CREDIT'),
(2, 10, 24, '2011-05-20 11:25:35', 22.15, 'CREDIT'),
(3, 10, 24, '2011-05-20 11:26:11', 44.30, 'CREDIT'),
(4, 10, 0, '2011-05-20 14:55:13', -3.50, 'DEBIT'),
(5, 10, 0, '2011-05-20 14:59:50', -5.00, 'DEBIT'),
(6, 10, 0, '2011-05-20 15:45:29', -4.50, 'DEBIT'),
(36, 10, 0, '2011-07-08 15:03:06', -10.00, 'DEBIT'),
(37, 10, 0, '2011-07-08 15:03:19', -2.45, 'DEBIT');
以下是查询：
SELECT DISTINCT t.*, o.amount AS discount, p.status FROM transactions t
LEFT JOIN offer o ON o.id = t.offerID
LEFT JOIN payments p ON p.offerId = t.offerID
WHERE t.companyID = '10' ORDER BY date ASC
最后，这是结果集：
id companyID offerID date amount type discount status
1 10 24 2011-05-20 11:25:35 11.00 CREDIT 25 NEW
1 10 24 2011-05-20 11:25:35 11.00 CREDIT 25 USED
2 10 24 2011-05-20 11:25:35 22.15 CREDIT 25 NEW
2 10 24 2011-05-20 11:25:35 22.15 CREDIT 25 USED
3 10 24 2011-05-20 11:26:11 44.30 CREDIT 25 USED
3 10 24 2011-05-20 11:26:11 44.30 CREDIT 25 NEW
4 10 0 2011-05-20 14:55:13 -3.50 DEBIT NULL NULL
5 10 0 2011-05-20 14:59:50 -5.00 DEBIT NULL NULL
6 10 0 2011-05-20 15:45:29 -4.50 DEBIT NULL NULL
36 10 0 2011-07-08 15:03:06 -10.00 DEBIT NULL NULL
37 10 0 2011-07-08 15:03:19 -2.45 DEBIT NULL NULL
问题是CREDIT结果在NEW和USED枚举值中都出现两次。我只想要支付行设置的枚举值。如果有人能帮助我理解为什么这个查询不起作用，以及如何解决它，我将不胜感激。提前谢谢！
@michael，@ chopikadze：
我已将memberID添加到transactions表中。 我已经修改了我的查询：
SELECT t. * , o.amount AS discount, p.status
FROM transactions t
LEFT JOIN offer o ON o.id = t.offerID
LEFT JOIN payments p ON p.offerId = t.offerID
LEFT JOIN members m ON m.id = t.memberID
WHERE t.companyID = '10'
ORDER BY date ASC
这是我的新结果集：
id companyID offerID memberID date amount type discount status
1 10 24 8 2011-05-20 11:25:35 11.00 CREDIT 25 NEW
1 10 24 8 2011-05-20 11:25:35 11.00 CREDIT 25 USED
1 10 24 8 2011-05-20 11:25:35 11.00 CREDIT 25 USED
2 10 24 14 2011-05-20 11:25:35 22.15 CREDIT 25 NEW
2 10 24 14 2011-05-20 11:25:35 22.15 CREDIT 25 USED
2 10 24 14 2011-05-20 11:25:35 22.15 CREDIT 25 USED
3 10 24 5 2011-05-20 11:26:11 44.30 CREDIT 25 NEW
3 10 24 5 2011-05-20 11:26:11 44.30 CREDIT 25 USED
3 10 24 5 2011-05-20 11:26:11 44.30 CREDIT 25 USED
4 10 0 0 2011-05-20 14:55:13 -3.50 DEBIT NULL NULL
5 10 0 0 2011-05-20 14:59:50 -5.00 DEBIT NULL NULL
6 10 0 0 2011-05-20 15:45:29 -4.50 DEBIT NULL NULL
36 10 0 0 2011-07-08 15:03:06 -10.00 DEBIT NULL NULL
37 10 0 0 2011-07-08 15:03:19 -2.45 DEBIT NULL NULL
似乎添加memberID会导致更多重复结果。我做错了什么？请纠正我这个人。感谢。
答案 0 :(得分：0)
SELECT t. * , o.amount AS discount, p.status
FROM transactions t
LEFT JOIN offer o ON o.id = t.offerID
LEFT JOIN payments p ON p.offerId = t.offerID and p.memberID = t.memberID
WHERE t.companyID = '10'
ORDER BY date ASC
正如我所说，你不应该只是
将memberID字段添加到事务表
但也
不仅可以在OfferID上加入交易和付款，还可以在memberID上加入