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我有一个从小型数据库中提取联系人列表的 SQL 查询；它可以很好地提取数据。然而，SQL 的问题在于它会根据电子邮件数量和与联系人关联的地址多次列出同一个联系人。

它应该只列出一次。这是我编写的 SQL：

SELECT DISTINCT 
                Contact_Main.contactid, first_name, last_name, p_title,
                prefix, suffix, email, address1, address2, 
                city, state, zip, Contact_Address_Type.addresstype, 
                Contact_Email_Types.emailtype, 
                addressid, emailid, Contact_Address_Type.addresstypeid, 
                Contact_Email_Types.emailtypeid,
                (SELECT COUNT(addressid) 
                       FROM Contact_Address 
                       WHERE contactid = Contact_Main.contactid) as num_addresses,
                (SELECT COUNT(emailid) 
                        FROM Contact_Email 
                        WHERE contactid = Contact_Main.contactid) as num_emails
        FROM Contact_Main
        LEFT JOIN Contact_Email ON Contact_Main.contactid = 
                                                    Contact_Email.contactid
        LEFT JOIN Contact_Email_Types ON Contact_Email_Types.emailtypeid =
                                                    Contact_Email.emailtypeid
        LEFT JOIN Contact_Address ON Contact_Address.contactid = 
                                                    Contact_Main.contactid
        LEFT JOIN Contact_Address_Type ON Contact_Address_Type.addresstypeid  = 
                                                    Contact_Address.addresstypeid 
        ORDER BY last_name, first_name;

我确定这是一个简单的修复程序，我需要将其中的某些部分作为子查询，但由于某种原因我无法弄清楚。

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2 回答 2

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所以这很好用吗？

SELECT DISTINCT 
                cm.contactid
                , first_name
                , last_name
                , p_title
                , prefix
                , suffix
                , email
                , address1
                , address2
                , city
                , state
                , zip
                , cat.addresstype
                , cet.emailtype
                , addressid
                , emailid
                , cat.addresstypeid
                , cet.emailtypeid
                --, (SELECT COUNT(addressid) 
                --     FROM Contact_Address 
                --     WHERE contactid = cm.contactid) as num_addresses
                --, (SELECT COUNT(emailid) 
                --      FROM Contact_Email 
                --      WHERE contactid = cm.contactid) as num_emails
        FROM Contact_Main cm
        LEFT JOIN Contact_Email ce ON cm.contactid = 
                                                    ce.contactid
        LEFT JOIN Contact_Email_Types cet ON cet.emailtypeid =
                                                    ce.emailtypeid
        LEFT JOIN Contact_Address ON ca.contactid = 
                                                    cm.contactid
        LEFT JOIN Contact_Address_Type ON cat.addresstypeid  = 
                                                    ca.addresstypeid 
        ORDER BY cm.last_name, cm.first_name;
于 2012-07-27T03:19:15.957 回答
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好的，所以当我在工作时，我问我的老板 SQL 代码是怎么回事。我们都看了看，他确定它正在返回所有可能的结果。在 SQL Access 中进行了模拟之后，我们必须执行 LEFT OUTER JOIN 而不仅仅是 LEFT JOIN。

    SELECT DISTINCT Contact_Main.contactid, first_name, last_name, p_title,
                    prefix, suffix, email, address1, address2, 
                    city, state, zip, Contact_Address_Type.addresstype, 
                    Contact_Email_Types.emailtype, 
                    addressid, emailid, Contact_Address_Type.addresstypeid, 
                    Contact_Email_Types.emailtypeid,
  (SELECT COUNT(addressid) FROM Contact_Address 
   WHERE contactid = Contact_Main.contactid) as num_addresses,
  (SELECT COUNT(emailid) FROM Contact_Email 
  WHERE contactid = Contact_Main.contactid) as num_emails
  FROM Contact_Main
  LEFT OUTER JOIN Contact_Email 
  ON Contact_Main.contactid = Contact_Email.contactid
  LEFT OUTER JOIN Contact_Email_Types
  ON Contact_Email_Types.emailtypeid = Contact_Email.emailtypeid
  LEFT OUTER JOIN Contact_Address 
  ON Contact_Address.contactid = Contact_Main.contactid
  LEFT OUTER JOIN Contact_Address_Type 
  ON Contact_Address_Type.addresstypeid  = Contact_Address.addresstypeid 
  ORDER BY last_name, first_name;

这就是诀窍。感谢所有的建议！

于 2012-07-29T02:40:37.550 回答