Sql server SQL的东西不起作用，为什么？

我在Oracle上做了很多查询，现在我正在使用SQL Server。 我在SQLServer中看到了使用类似于oracle的Listag函数的方法

    Select
    sqd.id_question,
    STUFF((Select ',' + nm_departament from tb_departament where sqd.id_departament = id_departament for xml path('')),1,1,'') nm_departements
from
    tb_survey_question_departament sqd  

sintax是正确的，但结果不正确

我们的目标是拥有例如2个顶排，结果是1-RH，巴西足底

---

问题在哪里？

缺少按id分组的问题

Select
sqd.id_question,
STUFF((Select ',' + nm_departament from tb_departament where sqd.id_departament = id_departament for xml path('')),1,1,'') nm_departements
    from
tb_survey_question_departament sqd  
group by sqd.id_question

你应该先明确身份证问题，然后像这样加入tb部门

Select
    sqd.id_question,
    STUFF(( 
             SELECT ',' + td.nm_departament 
             from tb_departament td
             INNER JOIN tb_survey_question_departament  sqd1 ON sqd1.id_departament = td.id_departament 
             WHERE  sqd1.id_question = sqd.id_question
             FOR XML PATH('')
         )
         ,1,1,'') AS nm_departements
from
    (
       SELECT DISTINCT sqd.id_question 
       FROM tb_survey_question_departament sqd 
    ) sqd

---

感谢大家的帮助，这里的最终sql运行良好：

Select
    sqd.id_question,
    que.ds_question,
    STUFF(( 
             SELECT ',' + td.nm_departament 
             from tb_departament td
             INNER JOIN tb_survey_question_departament  sqd1 ON sqd1.id_departament = td.id_departament 
             WHERE  sqd1.id_question = sqd.id_question
             FOR XML PATH('')
         )
         ,1,1,'') AS nm_departements
from
    (
       SELECT DISTINCT sqd.id_question 
       FROM tb_survey_question_departament sqd 
    ) sqd
    inner join tb_survey_question que on sqd.id_question = que.id_question

---

不，这行不通。这只会引发一个错误，因为GROUPBY SQL将无法正确地对内容进行分组。问题在于STUFF语句本身，因为它没有正确地连接到外部查询，但是在没有看到示例数据的情况下，很难推荐如何修复它。是的，如果我放入group，您是对的。问题是：列“tb\u survey\u question\u department.id\u department”在选择列表中无效，因为它不包含在聚合函数或group BY子句中。数据存在我在前两行显示的问题，即一个id_问题有一个或多个id_部门关联，因此我只需要按id_问题显示一行并聚合“n”departements@MCunha98您的资料应该如下所示：STUFFSELECT'，“+d.nm_从tb_出发作为d加入tb_调查_问题_出发作为sqd2上的sqd2.id_出发=d.id_出发，其中sqd2.id_出发=sqd.id_出发作为XML路径1，1，您可以尝试我的答案@MCunha98

SELECT EMPNAME
,DEPTNAME
,STUFF((
        SELECT ',' + LEAVETYPE + '-' + CAST(TOTALLEAVE AS VARCHAR(5))
        FROM tblSample
        WHERE EMPNAME = T.EMPNAME
            AND DEPTNAME = T.DEPTNAME
        FOR XML PATH('')
        ), 1, 1, '') AS LEAVETYPE
,SUM(TOTALLEAVE) AS TOTALLEAVE
FROM tblSample T
GROUP BY EMPNAME
,DEPTNAME
-----------------------------------------------

Select
    sqd.id_question,
    STUFF(( 
             SELECT ',' + td.nm_departament 
             from tb_departament td
             INNER JOIN tb_survey_question_departament  sqd1 ON sqd1.id_departament = td.id_departament 
             WHERE  sqd1.id_question = sqd.id_question
             FOR XML PATH('')
         )
         ,1,1,'') AS nm_departements
from
    (
       SELECT DISTINCT sqd.id_question 
       FROM tb_survey_question_departament sqd 
    ) sqd

Select
    sqd.id_question,
    que.ds_question,
    STUFF(( 
             SELECT ',' + td.nm_departament 
             from tb_departament td
             INNER JOIN tb_survey_question_departament  sqd1 ON sqd1.id_departament = td.id_departament 
             WHERE  sqd1.id_question = sqd.id_question
             FOR XML PATH('')
         )
         ,1,1,'') AS nm_departements
from
    (
       SELECT DISTINCT sqd.id_question 
       FROM tb_survey_question_departament sqd 
    ) sqd
    inner join tb_survey_question que on sqd.id_question = que.id_question