Get first n characters of a regex match

Question

I want to get The first n characters of the match from this regex:

(\d+\s*)

Basically I want to right pad with spaces. So in the lines:

12345␢␢␢␢␢␢␢␢123␢␢␢␢␢␢␢
123␢␢␢␢␢␢␢␢␢12345␢␢␢␢␢␢

I want to finish with:

12345␢␢␢␢␢123␢␢␢␢␢␢␢
123␢␢␢␢␢␢␢12345␢␢␢␢␢

There are always two matches on a line and the lines have a constant length.

Match a second time using `.{n}`? Not sure about notepad (never used it) but in [tag:grep] you could simply match again using `grep -o .\{n\}` and you'd get your desired result. — ShellFish, Jul 02 '15 at 13:52
what are you trying to express with your regex? why is something like this `^.{n}` not enough? — bro, Jul 02 '15 at 13:57
@bro Good. It will work for the first column. And the second? Check the updated question — Clodoaldo Neto, Jul 02 '15 at 14:04
And you want to do this in a single search-replace in notepad++? Just asking based on your tags. — NikhilWanpal, Jul 02 '15 at 14:05
Does a global flag exist in notepad? If so try `/\d.{n-1}/g`. — ShellFish, Jul 02 '15 at 14:08
@shell [Can't find it](http://docs.notepad-plus-plus.org/index.php/Regular_Expressions) — Clodoaldo Neto, Jul 02 '15 at 14:25
@ClodoaldoNeto That's a shame, what a fail of notepad there. Can you supply more information. Are there always two matches on a line or any other constant amount? — ShellFish, Jul 02 '15 at 14:28
@Shell There are always two matches on a line and the lines have a constant length. — Clodoaldo Neto, Jul 02 '15 at 14:34

score 2 · Answer 1 · edited May 23 '17 at 11:43

Multiple passes

Based on the extra information about the problem and it's structure I would advise following steps:

Split every line in two, right before the second pattern.
Grab the desired part from every line.
Recombine the lines so matches are on their original line.

This means something like this:

Replace ^(\d*\s*)(\d*\s*)$ with $1\r\n$2. Simply drop the \r if you're not on windows, which I doubt. You should perhaps think of a macro to add at the end of a line. This should be something that is not included in the rest of the document (for instance #). The $1 means, replace the first captured group (stuff inside brackets). So replace it with $1#\r\n$2.
Now grab the desired length of each line: (^.{n}).*(#?) and replace with $1$2. This will capture the first n symbols and insert the macro if it is found.
Remove newlines after macros: #\r\n. Either remove these or replace them with \0.

Notes

You'd have to filter the lines matching (^\d*\s*) first.
If you'd like another macro, sub the occurrences of # in above the answer. It should not be contained in the rest of the file, at least not at the end of a line.
This answer uses backreferences, which should be no problem.

Single pass

A single pass might be possible too here.

^(\d[\d\s]{n-1})[^\d]*(\d[\d\s]{n-1}).*$

Matches these lines, if you extract group one and two, this will filter the desired output from the file. Simply substitute it for $1$2.

Bernhard Barker · Answer 2 · 2015-07-02T15:05:42.803

1

Replace:

(\d[\d\s]{n-1})\s*

With:

$1

This replaces a digit followed by n-1 digits or whitespace characters, followed by any number of whitespace characters by the first n characters of what was matched (thus you should get 2 matches per line).

edited Jul 02 '15 at 15:05

answered Jul 02 '15 at 14:57

Bernhard Barker

54,589
14
104
138

Get first n characters of a regex match

2 Answers2

Multiple passes

Single pass