oracle 在oracle中使用Decode作为like语句
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Using Decode as a like statement in oracle
提问by Giancarlo Solarino
I need to write a sql statement like this:
我需要写一个这样的sql语句：
SELECT id_segmento AS Segmento, Decode (id_segmento , '1' , 'a', 'b' )
FROM mapchile.segmento
but in this case I will obtain an 'a' when id_segmento is equal to '1', I need it to be 'a' even when the string id_Segmento contains the '1', kind of like and like statment.
但在这种情况下，当 id_segmento 等于 '1' 时，我将获得一个 'a'，即使字符串 id_Segmento 包含 '1'，我需要它是 'a'，有点像和喜欢语句。
There is any other command like Decode that works this way?
还有其他像解码这样的命令吗？
Thanks.
谢谢。
回答by Jim Hudson
I'd use a case statement. Something like
我会使用 case 语句。就像是
case
when id_segmento like '%1%' then 'a'
else 'b'
end
回答by Vincent Malgrat
Use the CASE operator for complex evaluation instead of the DECODE function:
使用 CASE 运算符而不是 DECODE 函数进行复杂计算：
SELECT id_segmento AS Segmento,
CASE
WHEN id_segmento LIKE '%1%' THEN
'a'
ELSE
'b'
END
FROM mapchile.segmento
回答by Vincent Malgrat
if you don't want to use caseyou can still use decodeand instr:
如果您不想使用，case您仍然可以使用decodeand instr：
decode(instr(id_segmento,'1'),0,'b','a')
I'm assuming you want to match on a '1' anywhere in the field. If you want to match on fields that start with a '1' then you could use:
我假设您想在现场的任何地方匹配“1”。如果要匹配以“1”开头的字段，则可以使用：
decode(ascii(id_segmento),49,'a','b')
or
或者
decode(substring(id_segmento,1,1),'1','a','b')
or
或者
decode(instr(id_segmento,'1'),1,'a','b')
