oracle 用一個表欄位更新另一個表欄位三種方法____oracle

1. update （select .....） set column1 = column2;

update (select iy.company_name company_name1, cc.company_name_jc company_name2
        from income_year_item iy , city_company cc
       where iy.company_code = cc.code
       )
   set company_name1 = company_name2;

 

2.只能單行子查詢s

update  CITY_PROJECT_SCALE_INFO c  set
(c.value) = (
       select d.value from CITY_PROJECT_SCALE_INFO @test d where d.project_id = '7d7fd580a06240b2a9137dc2bbe831e9'
          and d.project_id = c.project_id and c.company_code = d.company_code
)
 where exists (
  select 1 from  CITY_PROJECT_SCALE_INFO @test d where d.project_id = '7d7fd580a06240b2a9137dc2bbe831e9'
          and d.project_id = c.project_id and c.company_code = d.company_code
)

 

請你尤其注意這裡的where子句，你可以嘗試不寫where子句來執行以下這句話，你將會使得CITY_PROJECT_SCALE_INFO中的很多值變成空。

這是因為在oracle的update語句中如果不寫where子句,oracle將會預設的把所有的值全部更新，即使你這裡使用了子查詢並且某在值並不能在子查詢裡找到，你就會想當然的以為,oracle或許將會跳過這些值吧，你錯了，oracle將會把該行的值更新為空白。

 

3.使用merg inot 語句

--更新生產基礎欄位
merge into city_cfg_data_column_common cf1 using
      city_cfg_data_column_common2 cf2 on
      (cf1.resourceid = cf2.resourceid)
   when matched then
        update set   cf1.template_type = cf2.template_type,
                     cf1.chinese_name = cf2.chinese_name,
                     cf1.column_name = cf2.column_name,
                     cf1.column_type = cf2.column_type,
                     cf1.column_size = cf2.column_size
  when not matched then
       insert (cf1.resourceid,cf1.template_type,cf1.chinese_name,cf1.column_name,cf1.column_type,cf1.column_size
              ,cf1.is_can_edit,cf1.is_unique,cf1.is_can_cover,cf1.show_order)
       values (cf2.resourceid,cf2.template_type,cf2.chinese_name,cf2.column_name,cf2.column_type,cf2.column_size
              ,cf2.is_can_edit,cf2.is_unique,cf2.is_can_cover,cf2.show_order)