小编典典

使用SQL在Oracle中创建动态列

sql

我有以下表格示例。Thera可以是无限的分支机构和客户。我需要对该分支进行分组并计算他们的客户,然后用不同的列来显示它。

BRANCHNAME  CUSTOMERNO
100         1001010
100         1001011
103         1001012
104         1001013
104         1001014
104         1001015
105         1001016
105         1001017
106         1001018

请注意,可以有无限的分支和客户,查询不仅必须在这种情况下有效。

在这种情况下,可接受的结果是:

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

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2021-04-19

共1个答案

小编典典

我认为写一个返回变量结构流水线表函数虽然很复杂,但却是可能的。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,您可以在后续的SQL语句中使用它,就像它是一个表一样,即

SELECT *
  FROM TABLE( your_pipelined_function( p_1, p_2 ));

还有更多参考文献讨论了相同的示例实现

  • 动态SQL透视
  • 《Oracle数据盒带开发人员指南》的“实现接口方法”部分
  • 方法4。 下载并安装开源PL / SQL代码后,这里是一个完整的实现:

    --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
    )
    ]’
    ));

2021-04-19