使用逗号分隔值的SQL查询

我有一个要求，其中我有3张桌子

联系信息

联系人角色

IdContactRoleLookup

我需要一个能给出这样结果的查询

idcontactrolename            Contacts
--------------------------------------------------------
Admin                    Blah1,Blah1;Blah2,Blah2,Blah3,Blah3
Secretary                Blah4,Blah4

这就是我现在所拥有的

SELECT l.sValue
    , SUBSTRING((
                        SELECT C.sLast + ',' + C.sFirst + ';'
                        FROM contactinfo c
                        inner join contactrole crole
                        ON crole.idcontactinfo = c.idcontactinfo
                        WHERE crole.idcompany = '<<blah>>'
                        and c.idcompany = '<<blah>>'
                        FOR XML PATH('')

                    ), 1, 1000000)
FROM CONTACTROLE CR
inner join contactinfo c on 
cr.idcontactinfo = c.idcontactinfo 
inner join lookupdata l on cr.idlookuprole = l.idlookupdata
where c.idcompany = '<<blah>>'

---

我会这样做，在子查询中使用主查询的ID

SELECT     
l.sValue
    , SUBSTRING((
                        SELECT C.sLast + ',' + C.sFirst + ';'
                        FROM contactinfo c
                        inner join contactrole crole
                        ON crole.idcontactinfo = c.idcontactinfo
                        WHERE c.idcompany = '<<blah>>'
AND c.idcontactrole = CR.idcontactrole
                        FOR XML PATH('')

                    ), 1, 1000000)

FROM CONTACTROLE CR
inner join lookupdata l on cr.idlookuprole = l.idlookupdata

---

这将为您提供一个公司的所有角色。如果角色不存在，您将有一个空值。

我会这样做，在子查询中使用主查询的ID

SELECT     
l.sValue
    , SUBSTRING((
                        SELECT C.sLast + ',' + C.sFirst + ';'
                        FROM contactinfo c
                        inner join contactrole crole
                        ON crole.idcontactinfo = c.idcontactinfo
                        WHERE c.idcompany = '<<blah>>'
AND c.idcontactrole = CR.idcontactrole
                        FOR XML PATH('')

                    ), 1, 1000000)

FROM CONTACTROLE CR
inner join lookupdata l on cr.idlookuprole = l.idlookupdata

SELECT
     RoleName,
     STUFF(
         (
            SELECT  ';' + b.FirstName + ', ' + b.LastName
            FROM    ContactRole a
            INNER JOIN ContactInfo b
                  ON a.idContactInfo = b.idContactinfo
            WHERE   a.IdContactRoleName = r.idcontactRole
            FOR XML PATH (''))
            , 1, 1, '')  AS NamesList
FROM  IdContactRoleLookup r
GROUP BY idcontactRole, RoleName

这将为您提供一个公司的所有角色。如果角色不存在，您将拥有空值

SELECT
     RoleName,
     STUFF(
         (
            SELECT  ';' + b.FirstName + ', ' + b.LastName
            FROM    ContactRole a
            INNER JOIN ContactInfo b
                  ON a.idContactInfo = b.idContactinfo
            WHERE   a.IdContactRoleName = r.idcontactRole
            FOR XML PATH (''))
            , 1, 1, '')  AS NamesList
FROM  IdContactRoleLookup r
GROUP BY idcontactRole, RoleName

您可以使用交叉应用来获得结果：

select distinct l.RoleName,
  left(list, len(list) -1) list
FROM CONTACTROLE CR
inner join IdContactRoleLookup l 
  on cr.IdContactRoleName = l.idcontactRole
cross apply
(
  select C1.Lastname + ',' + C1.FirstName + ';'
  from CONTACTROLE CR1
  inner join contactinfo c1
    on cr1.idContactInfo = c1.idContactinfo
  where cr.IdContactRoleName = cr1.IdContactRoleName
  FOR XML PATH('')
) t2 (list)

请参见

您可以使用交叉应用来获得结果：

select distinct l.RoleName,
  left(list, len(list) -1) list
FROM CONTACTROLE CR
inner join IdContactRoleLookup l 
  on cr.IdContactRoleName = l.idcontactRole
cross apply
(
  select C1.Lastname + ',' + C1.FirstName + ';'
  from CONTACTROLE CR1
  inner join contactinfo c1
    on cr1.idContactInfo = c1.idContactinfo
  where cr.IdContactRoleName = cr1.IdContactRoleName
  FOR XML PATH('')
) t2 (list)

---

请参见您的查询和表定义不匹配？哪个是正确的？您的查询和表定义不匹配？哪个是正确的？