如何有条件地选择 Oracle 查询中的列
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How to conditionally select a column in an Oracle query
提问by Travis
I want to do something like:
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
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.
不幸的是，Oracle 似乎不喜欢这种语法，而且我没有发现原因或找到我应该使用的其他函数。
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.
基本上，如果lookup值为8，我想得到08，否则我想要lookup的值。我确定我只是愚蠢。
回答by Gordon Linoff
You want a case statement:
你想要一个 case 语句：
select (case when lookup = 8 then 8 else lookup end) as lookup
If lookupis a character string, you probably want:
如果lookup是字符串，您可能想要：
select (case when lookup = '08' then '08' else lookup end) as lookup
If lookupis an integer and you want to convert it to a string, then:
如果lookup是整数并且您想将其转换为字符串，则：
select (case when lookup = 8 then to_char(lookup, '00') else to_char(lookup, '00') end) as lookup
However, that would seem redundant to me.
然而，这对我来说似乎是多余的。
