I´m trying to get out some value inside a text with regex.
我试图用正则表达式在文本中找出一些值。
Text example:
COMISION 002...................50.00........15.060000...............753.00 IVA 21 %
COMISION 002 ................... 50.00 ........ 15.060000 ............... 753.00 IVA 21%
I would like to get: 753.00
我想得到:753.00
I´m using this regex:
我正在使用这个正则表达式:
Pattern pattern = Pattern.compile("(?<=\\.\\.)(.*)(?=IVA 21 %)");
The problem is that this regex is outputting:
问题是这个正则表达式正在输出:
.................50.00........15.060000...............753.00
So I assume that the first time the engine finds the two dots (..) sets a limit.
所以我假设引擎第一次发现两个点(..)设置了一个限制。
What I want and can´t resolve is something like: "find the words "IVA 21 %", then look back and bring me all of the data till you see two dots together"
我想要和不能解决的是:“找到”IVA 21%“的字样,然后回头看看我带来的所有数据,直到你看到两个点在一起”
I´m new in the regex world so any help is appreciate.
我是正则表达世界的新手,所以任何帮助都是值得欣赏的。
2 个解决方案
#1
2
You can use this regex to capture your number:
您可以使用此正则表达式来捕获您的号码:
\\d+(?:\\.\\d+)?(?= IVA 21 %)
In your regex negative lookbehind (?<=\.\.) will assert first two dots in the input that are right after 002.
在正则表达式中,负向后观(?<= \。\。)将在输入中断言前两个点,这些点位于002之后。
#2
#1
2
You can use this regex to capture your number:
您可以使用此正则表达式来捕获您的号码:
\\d+(?:\\.\\d+)?(?= IVA 21 %)
In your regex negative lookbehind (?<=\.\.) will assert first two dots in the input that are right after 002.
在正则表达式中,负向后观(?<= \。\。)将在输入中断言前两个点,这些点位于002之后。