I have symbols that mean end of line
\r\n
Example of string:
D:\INSTALL\_SysinternalsSuite\processhacker-2.39-bin\x86\r\n
My regex looks like this
([a-zA-Z]:)(\\.*\\r\\n)
PS. log looks like this:
blablabla\r\n D:\INSTALL\_SysinternalsSuite\processhacker-2.39-bin\x86\r\n blablabla\r\n
I need catch only line like this
D:\INSTALL_SysinternalsSuite\processhacker-2.39-bin\x86\r\n
I win it with adding ? , it makes it non-greedy. regex: (?([a-zA-Z]:)(\.*?\r\n))
I can math only lines, like in the middle
blablabla\r\n D:\INSTALL_SysinternalsSuite\processhacker-2.39-bin\x86\r\n blablabla\r\n
I win it with adding ? , it makes it non-greedy. regex: (?([a-zA-Z]:)(\.*?\r\n))
I can math only lines, like in the middle
blablabla\r\n D:\INSTALL_SysinternalsSuite\processhacker-2.39-bin\x86\r\n blablabla\r\n
What about the following regex:
(\w\:[^\r\n]+)
Regards,
J
It's catch all log file, but i should rich and of the line in strings like this:
D:\INSTALL_SysinternalsSuite\processhacker-2.39-bin\x86\r\n
(?([a-zA-Z]:)(\.*\r\n))
So that the following lines with \r\n are not caught
Need to stop regular expression at first match \r\n in line like this
D:\INSTALL_SysinternalsSuite\processhacker-2.39-bin\x86\r\n
its's better , but it catch other lines with \r\n
Full log
Jun 5 14:39:40 blabla-pc.blabla.bla bla|10.3.0.0 Результат: Помещено на карантин: not-a-virus:HEUR:AdWare.Script.Generic\r\nПользователь: BLA\i.blablabla (Активный пользователь)\r\nОбъект: C:\users\i.blablabla\appdata\local\Google\Chrome\User Data\Default\Cache\f_000244\r\n
By default .
doesn't match newline characters. You can try using [\r\n.]*
instead of .*
.