Oracle SQL：Select语句中嵌套REPLACE()的替代方法？

所以我有一个看起来像这样的查询：

SELECT IncidentNumber,ToiletType,ToiletDangers,IncidentDate
FROM Core.LostLawsuits

…返回以下结果：

+----------------+------------+---------------------------+---------------+
| IncidentNumber | ToiletType | ToiletDangers             | Incident Date |
+----------------+------------+---------------------------+---------------+
| 2100           | A          | Spontaneous Combustion    | 2/1/2016      |
+----------------+------------+---------------------------+---------------+
| 2101           | P          | Attracts Bear Stampede    | 2/1/2016      |
+----------------+------------+---------------------------+---------------+

我想做的是获取结果,但在输出时更改ToiletType列结果.目前,我正在使用嵌套的REPLACE(),并想知道是否有更好/不同的方法,同时维护ToiletType的一列结果：

SELECT IncidentNumber,REPLACE(REPLACE(ToiletType,'A','Automatic Standard'),'P','Portable Potty') as ToiletType,IncidentDate
FROM Core.LostLawsuits

解决方法

CASE表达式适用于您的示例：

case ToiletType
   when 'A' then 'Automatic Standard'
   when 'P' then 'Portable Potty'
   end

即

SELECT IncidentNumber,case ToiletType
          when 'A' then 'Automatic Standard'
          when 'P' then 'Portable Potty'
       end as ToiletType,IncidentDate
FROM   Core.LostLawsuits

也许更好的是加入ToiletTypes表？