如何在Oracle查询中有条件地选择列 [英] How to conditionally select a column in an Oracle query

问题描述

我想做类似的事情:

select (if lookup = 8 then 08 else lookup) lookup
  , //more columns
from lookup_table
order by lookup

不幸的是，Oracle似乎不喜欢这种语法，而且我无法找到原因或找到我应该使用的其他功能.

Unfortunately, Oracle doesn't seem to like this syntax, and I am failing to discover why or find what other function I should be using.

基本上，如果查找值为8，我想得到08，否则，我想要查找的值.我确定我只是愚蠢.

Basically, if the lookup value is 8, I want to get 08, otherwise I want the value of lookup. I'm sure I'm just being stupid.

推荐答案

您需要个案说明:

select (case when lookup = 8 then 8 else lookup end) as lookup

如果lookup是字符串，则可能需要:

If lookup is a character string, you probably want:

select (case when lookup = '08' then '08' else lookup end) as lookup

如果lookup是整数，并且您想将其转换为字符串，则:

If lookup is an integer and you want to convert it to a string, then:

select (case when lookup = 8 then to_char(lookup, '00') else to_char(lookup, '00') end) as lookup

但是，这对我来说似乎是多余的.

However, that would seem redundant to me.

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