oracle11g regexp_替换json

我选择：

select regexp_replace(regexp_substr('[{"date": "01_2016", "val":"100_22"},{"date": "02_2016","val": "200.10"}]'
,'"val":\s*("(\w| )*")', 1, level)
,'"val":\s*"((\w| )*)"', '\1', 1, 1) val
from dual
connect by regexp_substr('[{"date": "01_2016", "val":"100_22"},{"date": "02_2016","val": "200.10"}]', '"val":\s*("(\w| )*")', 1, level) is not null
;

如果我的值的格式为100_10，则可以。但是我想要100.10，这个选择不支持这个。如何编写regexp\u替换

使用

（\d+）\uuz（\d+）

仅匹配由下划线分隔的数值：

SELECT REGEXP_REPLACE(
         '[{"date": "01_2016", "val":"100_22"},{"date": "02_2016","val": "200.10"}]',
         '"val":"(\d+)_(\d+)"',
         '"val":"\1.\2"'
       )
FROM   DUAL;

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谢谢大家。我找到了解决办法

 select regexp_replace(regexp_substr('[{"date": "01-2016", "val":"100.22"},{"date": "02-2016","val": "200.10"},{"date": "03-2016","val": "200.15"}]','"val":\s*("(\w|[..])*")', 1, level),'"val":\s*"((\w|[..])*)"', '\1', 1, 1) val, regexp_replace(regexp_substr('[{"date": "01-2016", "val":"100.22"},{"date": "02-2016","val": "200.10"},{"date": "03-2016","val": "200.15"}]' ,'"date":\s*("(\w|[-])*")', 1, level) ,'"date":\s*"((\w|[-])*)"', '\1', 1, 1) date_period from dual connect by regexp_substr('[{"date": "01-2016", "val":"100.22"},{"date": "02-2016","val": "200.10"},{"date": "03-2016","val": "200.15"}]', '"val":\s*("(\w|[..])*")', 1, level) is not null

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你的目的是什么，输入和输出？你能确定结果在执行时应该是什么样子吗？