Hive3 中case when的一个乱码问题与实践

现象描述： 

select a.apply_no,
       (case
         when coalesce(a.final_repay_date, '2999-12-31') > '2020-05-01'  then
          a.uid
         when coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01'  then
          a.uid
         else
          'null'
       end) as aa
  from dwm_asset_user_detail_comm_day a
 where apply_no = '20191125004334'
或者 
 select a.apply_no,
       (case
         when coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01' then
          a.uid
         when coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01' then
          a.uid
         else
          'null'
       end) as aa
  from dwm_asset_user_detail_comm_day a
 where apply_no = '20191125004334'

以上代码如果  coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01' 和  coalesce(a.final_repay_date, '2999-12-31') > '2020-05-01' 条件同时被满足即输出乱码

但是在以下情况下不会出现乱码，虽然when的两个条件同时被满足了： 

 select a.apply_no,
       (case
         when 1=1 then
          a.uid
         when 2=2 then
          a.uid
         else
          'null'
       end) as aa
  from dwm_asset_user_detail_comm_day a
 where apply_no = '20191125004334'
或者 
 select a.apply_no,
       (case
         when coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01' then
          a.uid
         when 1=1 then
          a.uid
         else
          'null'
       end) as aa
  from dwm_asset_user_detail_comm_day a
 where apply_no = '20191125004334'

解决方案：

如果then的输出字段一样，则合并两个when的条件改成：

 select a.apply_no,
       (case
         when coalesce(a.final_repay_date, '2999-12-31') > '2020-05-01' or
              coalesce(a.user_final_repay_date, '2999-12-31') > '2020-05-01'  then
          a.uid 
         else
          'null'
       end) as aa
  from dwm_asset_user_detail_comm_day a
 where apply_no = '20191125004334'

如果then的输出字段不一样，则考虑从逻辑进行调整，保持各个条件的满足保持互斥关系。