Question

我使用SQL Server查询设计器尝试构建一个外部查询，该查询将返回每个受保人的全名和地址以及没有策略的主策略。我的create语句如下：

    CREATE TABLE Address (
    AddressID integer NOT NULL,
    HouseNumber Integer NOT NULL,
    Street varchar(20) NOT NULL,
    CityCounty varchar(20) NOT NULL,
    StateAbb char(2),
    CountryAbb char(2) NOT NULL,
    Zip char(5) NOT NULL,
    LastUpdatedBy varchar(20) NOT NULL,
    LastUpdated date NOT NULL,
    CONSTRAINT PK_Address PRIMARY KEY (AddressID));

    CREATE TABLE Insured(
    InsuredID integer NOT NULL,
    FirstName varchar(15) NOT NULL,
    LastName varchar(15) NOT NULL,
    MI char(1),
    DateOfBirth date NOT NULL,
    CreditScore integer NOT NULL,
    AddressID integer NOT NULL,
    DriversLicenseNumber varchar(35), 
    LastUpdatedBy varchar(20) NOT NULL,
    LastUpdated date NOT NULL,
    CONSTRAINT PK_Insured PRIMARY KEY (InsuredID),
    CONSTRAINT FK_InsuredAddress FOREIGN KEY (AddressID) references  Address);

    CREATE TABLE Policy(
    PolicyID integer NOT NULL,
    EffectiveDate date NOT NULL,
    TerminationDate date NOT NULL,
    Amount Numeric (8,2) NOT NULL,
    PolicyYear integer NOT NULL,
    PolicyType char(1) NOT NULL,
    InsuredID integer NOT NULL,
    AddressID integer NOT NULL,
    LastUpdatedBy varchar(20) NOT NULL,
    LastUpdated date NOT NULL,
    CONSTRAINT PK_Policy PRIMARY KEY (PolicyID),
    CONSTRAINT FK_PolicyAddress FOREIGN KEY (AddressID) references Address,
    CONSTRAINT FK_PolicyInsured FOREIGN KEY (InsuredID) references Insured);

    CREATE TABLE Home(
    PolicyID integer NOT NULL,
    ExteriorType varchar(30) NOT NULL,
    Alarm char(3) NOT NULL,
    DistanceToFireStation integer NOT NULL,
    LastUpdatedBy varchar(20) NOT NULL,
    LastUpdated date NOT NULL,
    CONSTRAINT PK_Home PRIMARY KEY (PolicyID),
    CONSTRAINT FK_HomePolicy FOREIGN KEY (PolicyID) references Policy);

    CREATE TABLE Auto(
    PolicyID integer NOT NULL,
    VinNumber varchar(30) NOT NULL,
    Make varchar(15) NOT NULL,
    Model varchar(20) NOT NULL,
    MilesPerYear integer NOT NULL,
    LastUpdatedBy varchar(20) NOT NULL,
    LastUpdated date NOT NULL,
    CONSTRAINT PK_Auto PRIMARY KEY (PolicyID),
    CONSTRAINT FK_AutoPolicy FOREIGN KEY (PolicyID) references Policy);

我认为查询需要表地址，保险，策略和外部右侧或左侧连接，但我不能让SQL服务器识别它，因为它不断形成内部连接和交叉连接。对于返回包含住房政策的被保险人及其地址和没有保单及其地址的被保险人的查询，我需要什么？

到目前为止我已尝试过：

SELECT Insured.InsuredID, Insured.FirstName, 
       Insured.LastName, Address.HouseNumber, 
       Policy.PolicyID 
FROM Address RIGHT JOIN Policy 
ON Address.AddressID = Policy.AddressID 
RIGHT JOIN Insured ON Policy.AddressID = Insured.AddressID 
ORDER BY Insured.InsuredID

这是最新的查询，它返回了我对所有受保人的家庭政策的要求，但对于没有政策的被保险人，我在地址中获得了空白。

SELECT i.InsuredID, i.FirstName, i.MI, i.LastName,            
a.HouseNumber, a.Street, a.CityCounty, a.StateAbb, a.CountryAbb, a.Zip 
FROM INSURED i 
LEFT JOIN (SELECT * FROM Policy WHERE PolicyType = 'H') HomePolicy on    
i.InsuredID = HomePolicy.InsuredID 
LEFT JOIN Address a on HomePolicy.AddressID = a.AddressID;

Answer 1

您可以尝试此查询：

SELECT i.InsuredID,
i.FirstName,
i.LastName,
a.HouseNumber,
p.PolicyID
FROM insured i
LEFT JOIN policy p ON i.AddressID = p.AddressID AND p.PolicyType = 'H'
LEFT JOIN address a ON i.AddressID = a.AddressID
ORDER BY i.InsuredID;

我认为联接的顺序错误。这能满足您的需求吗？更新：将Insured表连接到Address表将显示地址，无论他们是否有策略。

Answer 2

数据库设计似乎很好。我认为我们对＆＃34; PolicyType＆＃34;柱。它拥有什么价值以及本专栏的目的是什么。

说PolicyType='H'这意味着它是家庭政策。或者查找相同查询的其他方法是检查policyid表中是否存在home。

这是对的吗？

主要查询，

对于使用家庭政策返回被保险人的查询，我需要什么？他们的地址和被保险人没有政策和地址？

检查此脚本，

--insureds with home policies and their addresses 
select i.InsuredID,
i.FirstName,
i.LastName
A.HouseNumber
,1 INDICATE
from Insured i
INNER JOIN policy p ON i.InsuredID = p.InsuredID 
INNER JOIN [Address] A ON A.ADDRESSID=I.ADDRESSID
WHERE  EXISTS(SELECT PolicyID FROM Home H WHERE h.PolicyID=P.PolicyID)
AND NOT EXISTS(SELECT PolicyID FROM [Auto] a WHERE A.PolicyID=P.PolicyID)

UNION ALL
--insureds with no policy and their addresses
select i.InsuredID,
i.FirstName,
i.LastName
,A.HouseNumber
,0 INDICATE
from Insured i
INNER JOIN [Address] A ON A.ADDRESSID=I.ADDRESSID
WHERE  EXISTS(SELECT InsuredID FROM   policy p WHERE i.InsuredID = p.InsuredID )

我使用＆＃34; EXISTS条款＆＃34;因为输出中不需要该表列。

SQL外部联接 - 加入需要3个表

2 个答案: