正则表达式可在Notepad ++中反转搜索

Question

I have a string

2012-02-19 00:11:12,128|DEBUG|Thread-1|@@@ Time taken is 18 ms 

Below regex allows me to search for 18 ms

\d\d\s[m][s]

What I want to do is search for string prior to 18 ms in Notepad++ and then delete it. So that out of thousands rows I have, I can just extract out timings.

Also, I need regex mentioned above to work with timings which are in 3 digits as well as 2 digits. For example it should be able to search for 18 ms as well as 999 ms .

Please help.

Answer 1

You may put your regex into a positive lookahead:

^.*(?=\d{2,3}\sms\s*$)

In case you have some text after 18 ms , you need to use a word boundary \\b :

\\b allows you to perform a "whole words only" search using a regular expression in the form of \\bword\\b

^.*(?=\d{2,3}\sms\b)

See demo

{2,3} is a limiting quantifier that lets you match 2 or 3 preceding subpattern.

There's an additional quantifier that allows you to specify how many times a token can be repeated. The syntax is {min,max} , where min is zero or a positive integer number indicating the minimum number of matches, and max is an integer equal to or greater than min indicating the maximum number of matches. If the comma is present but max is omitted, the maximum number of matches is infinite.

You can replace with empty string and 18 ms will stay on the line.

Note you can use \\d+ to allow 1 or more digits to be matched (without restrictions on the digit number).

Note 2 : if your number is the first of many on the line you need to use lazy matching , ie use .*? instead of .* in the beginning of pattern.

Answer 2

Also, I need regex mentioned above to work with timings which are in 3 digits as well as 2 digits.

.*?(?=\d{2,3}\sms\b)

Use the above regex and then replace the match with empty string.

Answer 3

You can use capturing group :

Find:

^.*(\d{2,}\s[m][s])$

Replace with:

\1