Question

我有四个表员工，associations_employees，association，association_items 。下面的选择查询会生成连接的行。

注意：我为SQL和Coldfusion语言制作了标签，这是因为我使用coldfusion作为我的脚本语言。我不确定我是应该依赖SQL还是使用我的脚本语言。

查询

SELECT  AE.userid, E.firstname,
    A.title, AI.itemvalue
FROM  associations_employees AE INNER JOIN 
    employees E on E.userid = AE.useridFK INNER JOIN 
    associations A on A.associationid = AE.associationidFK INNER JOIN
    association_items AI on AI.associationidFK = AE.associationidFK

当前选择输出

userID  firstname  title           itemvalue
------  ---------  -----           ---------
5603    Jesh       Learner Type    Family Literacy   
5603    Jesh       Learner Type    Elementary School 
5603    Jesh       Learner Type    Academic             
5603    Jesh       Personnel Type  Staff             
5605    jennone    Personnel Type  Site Supervisor   
5605    jennone    Personnel Type  Rops member       
5607    Sharon     Personnel Type  Rops member       
5607    Sharon     Personnel Type  Site Supervisor
5607    Sharon     Mentor Type     High School
5607    Sharon     Mentor Type     Op. Read
5607    Sharon     Mentor Type     Enrichment
5607    Sharon     Mentor Type     General

正如您所注意到的，除了'itemvalue'列之外，还有多行类似。我需要组合这些行以产生以下结果。

需要的输出

userID  firstname  title           itemvalue
------  ---------  ------          ---------
5603    Jesh       Learner Type    Family Literacy;Elementary School;Academic
5603    Jesh       Personnel Type  Staff
5605    jennone    Personnel Type  Rops member;Site Supervisor;Staff
5607    Sharon     Personnel Type  Rops member;Site Supervisor
5607    Sharon     Mentor Type     Enrichment;General;High School;Op. Read

Answer 1

您可以使用STUFF方法来实现此目的：

SELECT AE.userid,
       E.firstname,
       A.title,
       STUFF((SELECT ',' + [AI.itemvalue]
              FROM association_items AI
              WHERE AI.associationidFK = AE.associationidFK
              FOR XML PATH('')), 1, 1, '') AS itemvalue
FROM associations_employees AE
INNER JOIN employees E ON E.userid = AE.useridFK
INNER JOIN associations A ON A.associationid = AE.associationidFK
GROUP BY AE.userid, E.firstname, A.title,

这尚未经过测试，因此可能需要进行一些微调。

Answer 2

如果您需要ColdFusion解决方案，cfoutput的group属性将起作用。第1步是在查询中添加order by子句。

order by userid, title

接下来，你的cfoutput标签。

<cfoutput query="yourquery" group = "userid">
  <cfoutput group = "title">
    #userid # #firstname# #lastname# #title#
    <cfset items = ''>
    <cfoutput>
      <cfset items = listappend(items,itemvalue,';')>
    </cfoutput>
    #items#
  </cfoutput>
</cfoutput>

这是基本方法。您必须添加格式以及如何从itemvalue列表中排除尾部分号。

Answer 3

你可以使用类似于Dan的数组方法。数组可能更快

<cfoutput query="yourquery" group = "userid">
  <cfoutput group = "title">
    #userid # #firstname# #lastname# #title#
    <cfset items = []>
    <cfoutput>
        <cfset ArrayAppend(items, itemvalue)>
    </cfoutput>
    #ArrayToList(items, ";")#
  </cfoutput>
</cfoutput>

通过将列填充到列表中来组合多个行。

3 个答案: