mySql結合了兩個SELECT查詢和JOIN

Question

我有兩個mySql查詢來從數據庫中獲取結果。 我正在努力將他們團結在一起。

查詢1：

SELECT userEwallets.id as ewalletId, users.id as userId , money_repositories.money as money, a.nestedUserId
FROM userEwallets
JOIN users ON users.id = userEwallets.userId
JOIN money_repositories ON userEwallets.id = money_repositories.ewalletId
WHERE ewalletNumber = 'SHIRR937303656'

查詢2：

SELECT nested.id as nestedUserId
FROM userEwallets as nested
JOIN users ON users.id = nested.userId
JOIN money_repositories ON nested.id = money_repositories.ewalletId
WHERE ewalletNumber = 'SHIRR9122331743'

然后我的組合命令：

SELECT userEwallets.id as ewalletId, users.id as userId , money_repositories.money as money, a.nestedUserId
FROM (
    SELECT nested.id as nestedUserId
    FROM userEwallets as nested
    JOIN users ON users.id = nested.userId
    JOIN money_repositories ON nested.id = money_repositories.ewalletId
    WHERE ewalletNumber = 'SHIRR912233'
)    as a
JOIN users ON users.id = userEwallets.userId
JOIN money_repositories ON userEwallets.id = money_repositories.ewalletId
WHERE ewalletNumber = 'SHIRR93730'

我收到此錯誤：

#1054 - Unknown column 'userEwallets.id' in 'field list'

這兩個命令相同，但是它們在where子句中與ewalletNumber有簡單的區別

使用數據庫結構更新

money_repositories表：

CREATE TABLE IF NOT EXISTS `money_repositories` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `userId` int(11) NOT NULL,
  `ewalletId` int(11) NOT NULL,
  `money` int(11) NOT NULL,
  `createdAt` int(11) NOT NULL,
  `updatedAt` int(11) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_persian_ci AUTO_INCREMENT=4 ;

userEwallets表：

CREATE TABLE IF NOT EXISTS `userEwallets` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `userId` int(11) NOT NULL,
  `ewalletNumber` varchar(15) COLLATE utf8_persian_ci NOT NULL,
  `currencySymbol` varchar(5) COLLATE utf8_persian_ci NOT NULL,
  `createdAt` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `updatedAt` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_persian_ci AUTO_INCREMENT=7 ;

users表：

CREATE TABLE IF NOT EXISTS `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `username` varchar(20) COLLATE utf8_persian_ci NOT NULL,
  `password` varchar(65) COLLATE utf8_persian_ci NOT NULL,
  `name` varchar(20) COLLATE utf8_persian_ci NOT NULL,
  `family` varchar(20) COLLATE utf8_persian_ci NOT NULL,
  `birthDay` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00' ON UPDATE CURRENT_TIMESTAMP,
  `email` varchar(20) COLLATE utf8_persian_ci NOT NULL,
  `mobileNumber` varchar(15) COLLATE utf8_persian_ci NOT NULL,
  `verifyCode` varchar(5) COLLATE utf8_persian_ci NOT NULL,
  `photoUri` varchar(50) COLLATE utf8_persian_ci NOT NULL,
  `ebanNumber` varchar(20) COLLATE utf8_persian_ci NOT NULL,
  `status` tinyint(1) NOT NULL,
  `createdAt` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `updatedAt` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_persian_ci AUTO_INCREMENT=35 ;

Answer 1

userEwallets.id應a.id在選擇你的組合命令。 因為你得到的子查詢作為a 。

修改后的組合查詢；

SELECT a.id                     AS ewalletId, 
       users.id                 AS userId, 
       money_repositories.money AS money, 
       a.nesteduserid 
FROM   (SELECT nested.id AS nestedUserId, 
               nested.id, 
               nested.ewalletnumber 
        FROM   userewallets AS nested 
               JOIN users 
                 ON users.id = nested.userid 
               JOIN money_repositories 
                 ON nested.id = money_repositories.ewalletid 
        WHERE  nested.ewalletnumber = 'SHIRR912233') AS a 
       JOIN users 
         ON users.id = userewallets.userid 
       JOIN money_repositories 
         ON userewallets.id = money_repositories.ewalletid 
WHERE  a.ewalletnumber = 'SHIRR93730' 

camleCase查詢已修復：

SELECT a.id                     AS ewalletId, 
       users.id                 AS userId, 
       money_repositories.money AS money, 
       a.nestedUserId 
FROM   (SELECT nested.id AS nestedUserId, 
               nested.id, 
               nested.ewalletNumber 
        FROM   userEwallets AS nested 
               JOIN users                         ON users.id   = nested.userId 
               JOIN money_repositories  ON nested.id = money_repositories.ewalletId 
        WHERE  nested.ewalletNumber = 'SHIRR912233') AS a 

       JOIN users                         ON users.id            = nested.userId 
       JOIN money_repositories  ON userEwallets.id = money_repositories.ewalletId 

WHERE  a.ewalletNumber = 'SHIRR937303' 

Answer 2

因為您沒有在查詢中包括表userEwallets ，但在子查詢中將無法識別該表，但子查詢除外。 嘗試

更新

Select 
     users.id as userId, userEwallets.id As ewalletId,money_repositories.money
From
    users
Inner Join 
    userEwallets ON userEwallets.userId= users.id
Inner Join 
    money_repositories ON money_repositories.userId=users.id
And
    money_repositories.ewalletId = userEwallets.id
Where
    userEwallets.ewalletNumber = 'SHIRR93730'