Question

我有一个小问题，我不确定如何解决 - 我以前从未遇到过。我使用的是MS SQL，并且有一个查询两个表的SQL JOIN查询 - 保险和债权人。保险表包含所有保单细节，而债权人表保存公司详情（保险公司/经纪人姓名等）。

Insurance表包含两个名为CreditorID和BrokerID的外键。这两个外键都与相同的字段相关，即债权人表中的CreditorID字段（因为一些债权人既是保险公司又是保险经纪人）。我不确定如何编写查询。

E.g。保险表可能如下所示：

lInsuranceID   lCreditorID  lBrokerID  mLastPremium  dDatePaid  sPolicyNumber
1              1            null       1000.00       28/03/2014 12345
2              14           1          2000.00       17/03/2014 67891

债权人表可能如下所示：

lCreditorID   sCreditorName
1             Frank's Insurance COMPANY
14            Frank's Insurance BROKER

所以问题是，当我检索特定属性的信息时，我需要从Creditor表中提取两次信息 - 一次使用Insurance.CreditorID = Creditor.CreditorID，再次使用Insurance.CreditorID = Creditor.BrokerID。 / p>

但我不知道如何在一次加入中做到这一点。我试过这个，但它没有用（我得到的错误是＆＃34; Creditor.sCreditorName＆＃34;无法绑定：

 SELECT Insurance.sPolicyNumber,
        Insurance.dDatePaid,
        Insurance.dRenewal,
        Insurance.mLastPremium,
        Creditor.sCreditorName
   FROM Insurance
        INNER JOIN Creditor cred ON Insurance.lCreditorID = cred.lCreditorID
        INNER JOIN Creditor cred1 ON Insurance.lBrokerID = cred1.lCreditorID
  WHERE Insurance.lOwnersCorporationID = '1'

有什么建议吗？

修改好的，谢谢所有的提示，窥视。我将查询修改为看起来像这样，它似乎正在工作（我就是这样一个JOIN noob。即使它有效，我也不确定为什么）

SELECT Insurance.sPolicyNumber, Insurance.dDatePaid, Insurance.dRenewal, Insurance.mLastPremium, cred.sCreditorName as Company, cred1.sCreditorName as Broker
FROM Insurance
LEFT JOIN Creditor cred ON Insurance.lCreditorID = cred.lCreditorID
LEFT JOIN Creditor cred1 ON Insurance.lBrokerID = cred1.lCreditorID
WHERE Insurance.lOwnersCorporationID = '1'

Answer 1

我认为你想要的是这样的

 SELECT Insurance.sPolicyNumber,
        Insurance.dDatePaid,
        Insurance.dRenewal,
        Insurance.mLastPremium,
        cred.sCreditorName AS Creditor,
        cred1.sCreditorName AS Broker
   FROM Insurance
        INNER JOIN Creditor cred ON Insurance.lCreditorID = cred.lCreditorID
        INNER JOIN Creditor cred1 ON Insurance.lBrokerID = cred1.lCreditorID
  WHERE Insurance.lOwnersCorporationID = '1'

我添加了SQLFiddle example来演示（减去select语句中不存在于示例表中的列）。这会使用您的原始查询，但请参阅下面的另一个LEFT OUTER JOIN示例。

修改

如下所述，LEFT JOIN会很有用，因此您不会过滤BrokerID为NULL的行。更新了SQLFiddle上的示例以显示此内容。

Answer 2

也许左连接可能有帮助？由于您可能在所有实例中都没有Broker ID。

 SELECT Insurance.sPolicyNumber,
        Insurance.dDatePaid,
        Insurance.dRenewal,
        Insurance.mLastPremium,
        Creditor.sCreditorName
   FROM Insurance
        INNER JOIN Creditor cred ON Insurance.lCreditorID = cred.lCreditorID
         LEFT JOIN Creditor cred1 ON Insurance.lBrokerID = cred1.lCreditorID
  WHERE Insurance.lOwnersCorporationID = '1'

Answer 3

您只需要在SELECT表达式列表中使用表别名：

 SELECT i.sPolicyNumber,
        i.dDatePaid,
        i.dRenewal,
        i.mLastPremium,
        c.sCreditorName,
        b.sCreditorName
   FROM Insurance i
        LEFT JOIN Creditor  c ON c.lCreditorID = i.lCreditorID
        LEFT JOIN Creditor  b ON b.lCreditorID = i.lBrokerID
  WHERE Insurance.lOwnersCorporationID = '1'

请注意，我切换到LEFT OUTER JOIN。这是因为您已经证明引用值可能是NULLable。如果您不使用OUTER JOIN，您将从Insurance表中删除行，我认为这对您不利。

Answer 4

 SELECT Insurance.sPolicyNumber,
        Insurance.dDatePaid,
        Insurance.dRenewal, 
        Insurance.mLastPremium, 
        -- the changed columns
        cred.sCreditorName AS CREDITOR,
        cred1.sCreditorName AS BROKER,  
   FROM Insurance
        INNER JOIN Creditor cred ON Insurance.lCreditorID = cred.lCreditorID
        LEFT OUTER JOIN Creditor cred1 ON Insurance.lBrokerID = cred1.lCreditorID
  WHERE Insurance.lOwnersCorporationID = '1'

当您使用Alias（cred和cred1）引用该表时，您必须在列列表中使用别名

Answer 5

在原始查询中，您将引用“Creditor.sCreditorName”，但是您在联接中使用了别名。

另外，如果FK为空，则使用内部联接，这可能会导致一些条目。例如，您的lInsuranceID = 1的条目将被遗漏。

正确的查询可能是

 SELECT ins.sPolicyNumber,
        ins.dDatePaid,
        ins.dRenewal,
        ins.mLastPremium,
        cred.sCreditorName as CreditorName,
        brok.sCreditorName as BrokerName
   FROM Insurance ins
        LEFT JOIN Creditor cred ON ins.lCreditorID = cred.lCreditorID
        LEFT JOIN Creditor brok ON ins.lBrokerID = brok.lCreditorID
  WHERE ins.lOwnersCorporationID = '1'

MSSQL中的一些测试：

declare @Insurance table
(
    lInsuranceID int,
    lCreditorID  int,
    lBrokerID  int,
    mLastPremium  money,
    dDatePaid  date,
    sPolicyNumber int
)

 INSERT INTO @Insurance
        (lInsuranceID, lCreditorID, lBrokerID, mLastPremium, dDatePaid,
         sPolicyNumber)
 SELECT 1, 1, null, 1000.00, '2014-03-28', 12345
  UNION ALL
 SELECT 2, 14, 1, 2000.00, '2014-03-17', 67891

declare @Creditor table
(
    lCreditorID int,
    sCreditorName varchar(64)
)

 INSERT INTO @Creditor
        (lCreditorID, sCreditorName)
 SELECT 1, 'Frank''s Insurance COMPANY'
  UNION ALL
 SELECT 14, 'Frank''s Insurance BROKER'


 SELECT ins.sPolicyNumber,
        ins.dDatePaid,
        ins.mLastPremium,
        brok.sCreditorName as BrokerName,
        cred.sCreditorName as CreditorName
   FROM @Insurance ins
        LEFT JOIN @Creditor cred on cred.lCreditorID = ins.lCreditorID
        LEFT JOIN @Creditor brok on brok.lCreditorID = ins.lBrokerID

SQL Join Query - 两个FK，一个字段

5 个答案: