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
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
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
我认为写一个
pipelined table function that returns a variable structure是可能的,虽然很复杂.你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一个动态结构.然后,您可以在后续sql语句中使用它,就好像它是一个表,即
原文链接:https://www.f2er.com/oracle/204995.htmlSELECT * FROM TABLE( your_pipelined_function( p_1,p_2 ));
还有几个参考文献讨论了相同的示例实现
> Dynamic SQL Pivoting
> Oracle Data Cartridge开发人员指南的Implementing the Interface Approach部分
> Method4.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 ) ]' ));