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Question

我有5张桌子- 定义-

ID | defName |
--------------
1  | def1    |
2  | def2    |
3  | def3    |
4  | def4    |

def_treat_relations

defID | treatID|
---------------
3     | 5      |
3     | 2      |
2     | 3      |
4     | 3      |

def_specialty_relations

defID | specID |
----------------
3     | 2      |
3     | 3      |
2     | 4      |
4     | 1      |

特色

ID | specName |
---------------
1  | spec1    |
2  | spec2    |
3  | spec3    |
4  | spec4    |

苦恼

ID | treatName |
----------------
1  | treat1    |
2  | treat2    |
3  | treat3    |
4  | treat4    |

我需要一个查询，向我显示所有定义以及一系列处理方法和特色

结果应如下所示：

defname | speces    | treat       |
-------------------------------
def1    |           |             |
def2    |spec4      |treat3       |
def3    |spec2,spec3|treat5,treat2|
def4    |spec1      |treat3       |

我尝试了一些尝试，但遇到了麻烦...在此先感谢

Answer 1

您需要使用某种字符串聚合来实现此目的。较新版本的SQL Server具有string_agg可以为您完成此操作，但是由于您尚未指定版本，因此这里是使用for xml的解决方案，而stuff则向后兼容：

declare @definition table(ID int, defName varchar(10));
insert into @definition values(1,'def1'),(2,'def2'),(3,'def3'),(4,'def4');

declare @def_treat_relations table(ID int, treatID int);
insert into @def_treat_relations values (3,5),(3,2),(2,3),(4,3);

declare @def_specialty_relations table(ID int, specID int);
insert into @def_specialty_relations values (3,2),(3,3),(2,4),(4,1);

declare @specialty table(ID int, specName varchar(10));
insert into @specialty values (1,'spec1'),(2,'spec2'),(3,'spec3'),(4,'spec4');

declare @tretments table(ID int, treatName varchar(10));
insert into @tretments values (1,'treat1'),(2,'treat2'),(3,'treat3'),(4,'treat4'),(5,'treat5');


select d.defName
      ,isnull(
          stuff((select ', ' + s.specName
                 from @def_specialty_relations as sr
                     left join @specialty as s
                         on sr.specID = s.ID
                 where d.ID = sr.ID
                 order by s.specName
                 for xml path('')
                 )
                 ,1,2,'')
        ,'')
        as specs

      ,isnull(
            stuff((select ', ' + t.treatName
                   from @def_treat_relations as tr
                       left join @tretments as t
                          on tr.treatID = t.ID
                   where d.ID = tr.ID
                   order by t.treatName
                   for xml path('')
                   )
                   ,1,2,'')
          ,'')
          as treat
from @definition as d
    left join @def_treat_relations as tr
        on d.ID = tr.ID
    left join @tretments as t
        on tr.treatID = t.ID
order by d.defName;

输出

+---------+--------------+----------------+
| defName |    specs     |     treat      |
+---------+--------------+----------------+
| def1    |              |                |
| def2    | spec4        | treat3         |
| def3    | spec2, spec3 | treat2, treat5 |
| def3    | spec2, spec3 | treat2, treat5 |
| def4    | spec1        | treat3         |
+---------+--------------+----------------+

从树表中选择并根据ID合并

1 个答案:

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