Question

我在Oracle上做了很多查询，现在我正在使用SQL Server。我在sql server（stuff）中看到了使用类似函数的方式，如来自oracle的listagg。

    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，但结果却没有。

目标是例如2个顶行，结果为1 - RH，Planta Brasilia

问题出在哪里？

Answer 1

您应该先distinct id_question，inner join到tb_departament这样

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

Answer 2

id_Question错过群组...

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

Answer 3

CREATE TABLE tblSample (
EMPNAME VARCHAR(10)
,DEPTNAME VARCHAR(10)
,LEAVETYPE VARCHAR(20)
,TOTALLEAVE INT
);

    INSERT INTO tblSample (EMPNAME, DEPTNAME, LEAVETYPE, TOTALLEAVE)
        VALUES ('ANDREW','CSE','SICKLEAVE',3)
            ,('GEORGE','IT','CASUALLEAVE',1)
            ,('ANDREW','CSE','CASUALLEAVE',2)
            ,('GEORGE','IT','SICKLEAVE',2);

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
-----------------------------------------------

Answer 4

感谢帮助人员，这里最终的sql工作正常：

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

SQL STUFF无法正常工作，为什么？

4 个答案: