oracle中的动态列使用sql

8 sql oracle pivot

我有以下表格的例子.Thera可以是无限的分支和客户.我需要对这个分支进行分组并统计他们的客户,然后用不同的列显示它.

BRANCHNAME  CUSTOMERNO
100         1001010
100         1001011
103         1001012
104         1001013
104         1001014
104         1001015
105         1001016
105         1001017
106         1001018
Run Code Online (Sandbox Code Playgroud)

注意,可以有无限的分支和客户,查询必须不仅工作这种情况.

在这种情况下,接受的结果是:

100 103 104 105 106
 2   1   3   2   1
Run Code Online (Sandbox Code Playgroud)

示例SQL DATA

    select '100' BranchName,'1001010' CustomerNo from dual   UNION ALL 
    select '100' BranchName,'1001011' CustomerNo from dual   UNION ALL 
    select '103' BranchName,'1001012' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001013' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001014' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001015' CustomerNo from dual   UNION ALL 
    select '105' BranchName,'1001016' CustomerNo from dual   UNION ALL 
    select '105' BranchName,'1001017' CustomerNo from dual   UNION ALL 
    select '106' BranchName,'1001018' CustomerNo from dual   
Run Code Online (Sandbox Code Playgroud)

Kam*_*dov 6

我认为编写一个返回变量结构流水线表函数是可能的,尽管很复杂.您的管道表函数将使用Oracle Data Cartridge接口和AnyDataSet类型的魔力在运行时返回动态结构.然后,您可以在后续SQL语句中使用它,就好像它是一个表,即

SELECT *
  FROM TABLE( your_pipelined_function( p_1, p_2 ));
Run Code Online (Sandbox Code Playgroud)

还有几个参考文献讨论了相同的示例实现

  • 动态SQL透视
  • Oracle Data Cartridge开发人员指南中的实现接口方法部分
  • 方法4. 下载并安装开源PL/SQL代码后,这是一个完整的实现:

    --Create sample table.
    create table branch_data as
    select '100' BranchName,'1001010' CustomerNo from dual   UNION ALL 
    select '100' BranchName,'1001011' CustomerNo from dual   UNION ALL 
    select '103' BranchName,'1001012' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001013' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001014' CustomerNo from dual   UNION ALL 
    select '104' BranchName,'1001015' CustomerNo from dual   UNION ALL 
    select '105' BranchName,'1001016' CustomerNo from dual   UNION ALL 
    select '105' BranchName,'1001017' CustomerNo from dual   UNION ALL 
    select '106' BranchName,'1001018' CustomerNo from dual;
    
    --Create a dynamic pivot in SQL.
    select *
    from table(method4.dynamic_query(
        q'[
            --Create a select statement
            select
                --The SELECT:
                'select'||chr(10)||
                --The column list:
                listagg(
                    replace(q'!sum(case when BranchName = '#BRANCH_NAME#' then 1 else 0 end) "#BRANCH_NAME#"!', '#BRANCH_NAME#', BranchName)
                    , ','||chr(10)) within group (order by BranchName)||chr(10)||
                --The FROM:
                'from branch_data' v_sql
            from
            (
                --Distinct BranchNames.
                select distinct BranchName
                from branch_data
            )
        ]'
    ));
    
    Run Code Online (Sandbox Code Playgroud)

  • 这将需要更多的时间,当我有空闲时间,我会为你做:)所以懒惰的开发人员;) (4认同)