MySQL多表查询

问题描述 投票:0回答:1

具有标准的问题:

“具有条件的问题”“ >>

提供有关创建满足此条件的SELECT语句的任何帮助。我认为正在使用的表如下。

CREATE TABLE `client` (
  `ClientID` varchar(7) NOT NULL,
  `ClientName` varchar(45) DEFAULT NULL,
  `Street` varchar(100) DEFAULT NULL,
  `City` varchar(45) DEFAULT NULL,
  `State` char(2) DEFAULT NULL,
  `Zip` char(5) DEFAULT NULL,
  PRIMARY KEY (`ClientID`)
)

CREATE TABLE `contact` (
  `ClientID` varchar(7) NOT NULL,
  `ContactName` varchar(45) NOT NULL,
  `ContactPhone` char(17) DEFAULT NULL,
  `ContactEmail` varchar(45) DEFAULT NULL,
  PRIMARY KEY (`ClientID`,`ContactName`),
  CONSTRAINT `FK_ClientContact` FOREIGN KEY (`ClientID`) REFERENCES `client` (`ClientID`)
)

这是我到目前为止所拥有的。

SELECT ClientName, contact.ContactName, ContactPhone, ContactEmail, count()
FROM client, contact
GROUP BY ClientName, ContactName, ContactPhone, ContactEmail
ORDER BY ClientName;

附加信息:

CREATE TABLE `event_contact` (
  `ClientID` varchar(7) NOT NULL,
  `ContactName` varchar(45) NOT NULL,
  `EventCode` varchar(12) NOT NULL,
  PRIMARY KEY (`ClientID`,`ContactName`,`EventCode`),
  KEY `FK_EC_Event_idx` (`EventCode`),
  CONSTRAINT `FK_EC_Contact` FOREIGN KEY (`ClientID`, `ContactName`) REFERENCES `contact` (`ClientID`, `ContactName`),
  CONSTRAINT `FK_EC_Event` FOREIGN KEY (`EventCode`) REFERENCES `events` (`EventCode`)
)

CREATE TABLE `events` (
  `EventCode` varchar(12) NOT NULL,
  `EventName` varchar(45) NOT NULL,
  `Description` varchar(150) DEFAULT NULL,
  `EventDate` date DEFAULT NULL,
  `StartTime` time DEFAULT NULL,
  `EndTime` time DEFAULT NULL,
  `Ticket` tinyint(4) DEFAULT NULL,
  `VenueID` char(7) NOT NULL,
  `ClientID` varchar(7) NOT NULL,
  PRIMARY KEY (`EventCode`),
  KEY `FK_Events_Venue_idx` (`ClientID`),
  KEY `FK_Events_Venue` (`VenueID`),
  CONSTRAINT `FK_Events_Client` FOREIGN KEY (`ClientID`) REFERENCES `client` (`ClientID`),
  CONSTRAINT `FK_Events_Venue` FOREIGN KEY (`VenueID`) REFERENCES `venue` (`VenueID`)
) 

具有标准的问题:创建满足此标准的SELECT语句的任何帮助将不胜感激。我认为正在使用的表格如下。创建表`client`(...

mysql mysql-workbench multiple-tables
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