Question

这是我表格REMARKS列中的消息

{"StatusCode":"0","StatusDescription":"","message":"","transactionid":"404897688","enddate":"04/03/2017","formula":"ACCESS"}_SUBS
{"StatusCode":"0","StatusDescription":"","message":"","transactionid":"404894098","enddate":"04/03/2017","formula":"EVASION"}_SUBN
{"StatusCode":"0","StatusDescription":"","message":"","transactionid":"404889188","enddate":"05/03/2017","formula":"LES CHAINES CANAL+"}_SUBS
{"StatusCode":"0","StatusDescription":"","message":"","transactionid":"404880515","enddate":"06/03/2017","formula":"EVASION+"}_SUBS

我在查询中使用它

substr(remarks, (instr(remarks,'formula') + 10), 18) FORMULA

但我也得到了特殊字符}“_因为EVASION+ LES CHAINES CANAL+ EVASION ACCESS的长度不同。

有人可以解释如何排除这些特殊字符，并只获取FORMULA列下显示的名称。

感谢

Answer 1

以下是仅使用标准substr和instr函数（无正则表达式）的解决方案：

select substr( remarks, instr(remarks, '"formula":"') + 11,
               instr(substr(remarks, instr(remarks, '"formula":"') + 11), '"') - 1 )
from   inputs;

Answer 2

如果备注列只是一个字符串，您可以尝试：

with x as (
select '{"StatusCode":"0","StatusDescription":"","message":"","transactionid":"404897688","enddate":"04/03/2017","formula":"ACCESS"}_SUBS' as remarks
from dual
)
select
regexp_substr(remarks, '"formula":"(.*?)"',1,1,'i',1)
from x;

输出：

ACCESS

substr：如何排除像}“_等特殊字符

2 个答案: