php - SQL SELECT id WHERE 电子邮件相同

Question

我不知道这是否可能，但我有两个表，userBasic和carPlateConfidence，在carPlateConfidence中我想在电子邮件所在的位置插入userBasic的id匹配。

$query .= "INSERT IGNORE INTO userBasic (id_uM, userNameG, userEmailG) values ((SELECT id_uM FROM userMore WHERE userEmailG='$userEmailG'),'$userNameG', '$userEmailG');";

$query .= "INSERT IGNORE INTO carPlateConfidence (emailConfid, id_uB,plateNumber, confidencePlate, plateNumberUn) values ('$userEmailG', (SELECT id_uB FROM userBasic WHERE userEmailG='(SELECT max(emailConfid) FROM carPlateConfidence)'), '$plateNumber','$confidencePlate', '$plateNumberUn');";

所以如果我有:

用户基本:

id_uM = 555;
userNameG = BlaBla;
userEmailG = blabla@blabla.com

在这张表中我想要

汽车板置信度:

emailConfid = blabla@blabla.com;
id_uB = 555
plateNumber = 1111
confidencePlate = 70 
plateNumberUn = 2222

如果电子邮件不匹配:

emailConfid = blabla2@blabla.com;
id_uB = NULL
plateNumber = 1111
confidencePlate = 70
plateNumberUn = 222

P>S> 目前我已经尝试过这个，从 userBasic 中选择 id:

(SELECT id_uB FROM userBasic WHERE userEmailG='(SELECT max(emailConfid) FROM carPlateConfidence)')

carPlateConfidence中的

id_uB设置为外键；

表格:

--
-- Table structure for table `carPlateConfidence`
--

DROP TABLE IF EXISTS `carPlateConfidence`;
CREATE TABLE IF NOT EXISTS `carPlateConfidence` (
  `id_cof` int(11) NOT NULL AUTO_INCREMENT,
  `id_uB` int(11) NOT NULL,
  `emailConfid` varchar(50) NOT NULL,
  `plateNumber` varchar(10) NOT NULL,
  `confidencePlate` varchar(10) DEFAULT NULL,
  `plateNumberUn` varchar(10) DEFAULT NULL,
  PRIMARY KEY (`id_cof`),
  KEY `id_uB` (`id_uB`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;

-- --------------------------------------------------------

--
-- Table structure for table `userBasic`
--

DROP TABLE IF EXISTS `userBasic`;
CREATE TABLE IF NOT EXISTS `userBasic` (
  `id_uB` int(11) NOT NULL AUTO_INCREMENT,
  `id_uM` int(11) NOT NULL,
  `userNameG` varchar(50) NOT NULL,
  `userEmailG` varchar(50) NOT NULL,
  PRIMARY KEY (`id_uB`),
  UNIQUE KEY `userEmailG` (`userEmailG`),
  KEY `id_uM` (`id_uM`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=119 ;

--
-- Constraints for dumped tables
--

--
-- Constraints for table `carPlateConfidence`
--
ALTER TABLE `carPlateConfidence`
  ADD CONSTRAINT `carPlateConfidence_ibfk_1` FOREIGN KEY (`id_uB`) REFERENCES `userBasic` (`id_uB`);

--
-- Constraints for table `userBasic`
--
ALTER TABLE `userBasic`
  ADD CONSTRAINT `userBasic_ibfk_1` FOREIGN KEY (`id_uM`) REFERENCES `userMore` (`id_uM`);

Answer 1

所以你想要更新，而不是插入:

UPDATE carPlateConfidence t
SET t.id_uB = (SELECT distinct s.id_uM FROM userBasic s
               WHERE s.userEmailG = t.emailConfid)

只有在只能有 1 个匹配项时，这才有效，如果可以有多个匹配项，您应该指定您想要的匹配项，如果不重要，可以使用 MAX()或限制:

UPDATE carPlateConfidence t
SET t.id_uB = (SELECT max(s.id_uM) FROM userBasic s
               WHERE s.userEmailG = t.emailConfid)