Oracle:使用动态列名称更新 table

Oracle: update table using dynamic column names

我正在使用 Oracle 11g。我的 tables 包括名称和 l_name(名称列的小写)等列。我正在尝试遍历 table space 中的所有列,以将 l_ 列设置为其各自大写列的小写。这是我尝试过的:

for i in (select table_name from user_tables) loop
    SELECT SUBSTR(column_name,3) bulk collect into my_temp_storage FROM user_tab_columns WHERE table_name = i.table_name and column_name like 'L\_%' escape '\';
    for j in (select column_name from user_tab_columns where table_name = i.table_name) loop
        for k in 1..my_temp_storage.count
        loop
            if(j.column_name like 'L\_%' escape '\' and SUBSTR(j.column_name,3) = my_temp_storage(k)) then
                DBMS_OUTPUT.PUT_LINE( 'update ' || i.table_name || ' set ' || j.column_name || ' = LOWER(' ||my_temp_storage(k)|| ') where ' || j.column_name || ' is not null');
                execute immediate 'update ' || i.table_name || ' set ' || j.column_name || ' = LOWER(' ||my_temp_storage(k)|| ') where ' || j.column_name || ' is not null';
            end if;
        end loop;
    end loop;
end loop;

我将所有列的名称以大写形式存储在 my_temp_storage 中,并使用 my_temp_storage 中列的较低值更新 table。这给了我一条错误消息:

Error report -
ORA-00900: invalid SQL statement
ORA-06512: at line 8
00900. 00000 -  "invalid SQL statement"
*Cause:    
*Action: 

但 DBMS 输出似乎没问题:

`update EMPLOYEE set L_NAME = LOWER(NAME) where L_NAME is not null` 

你能帮我用我做的方式或任何其他方式吗?

程序当然可以简化:

begin
    for i in (select table_name, column_name from user_tab_columns 
              where column_name like 'L\_%' escape '\') 
    loop
        l_sql := 'update ' || i.table_name || ' set ' || i.column_name 
                  || ' = LOWER(' ||substr(i.columm_name,3)
                  || ') where ' || i.column_name || ' is not null';
        execute immediate l_sql;
    end loop;          
end;

不过这似乎是一个奇怪的数据库设计。您是否考虑过虚拟列、and/or 基于函数的索引,而不是手动维护的列?