I'm very new to regex, what I'm trying to do is to match a line only if the next line is an empty line.
For example:
This is some text
( empty line )
This is some text
This is some text
This is some text
( empty line )
This is some text
( empty line )
In the previous example, I would like to be able to select only line 1,5,7.
Is this possible with regex in notepad++?
Thanks in advance.
You can use this regex,
(.*)\n\s*\n
and replace with
\1
Working Demo
It uses a concept of group capture, so here you can use \1 to use captured group, that is line before newline
You could try the below positive lookahead based regex.
^.*?\S.*(?=\n[ \t]*$)
\S matches any non-space character. So .*?\S.* matches the line which has at-least one non-space character and the following (?=\n[ \t]*$) asserts that the match must be followed by a newline character and then followed by zero or more space or tab characters.
OR
^.*?\S.*(?=\n\n)
If you mean empty line as line which don't have any single space or tab characters, then you could use the above regex. (?=\n\n) asserts that the match must be followed by a blank line.
DEMO 1
DEMO 2
This should do the trick:
/(.)*\n\n/
If you're looking for an easy way to test / verify regex rubular is pretty good.
Related
How would you use the regex in Notepad++ to format replacing a single character that it finds in every line excepts for the duplicate ones in the certain line further?
test1:_|TEST:-TEST.|
test2:_|TEST:-TEST.|
test3:_|TEST:-TEST.|
As shown in the test code, there are two colons; I'm trying to replace the first colon with each line to a ; and NOT the second one found; the result of me doing the regex should equal to this:
test1;_|TEST:-TEST.|
test2;_|TEST:-TEST.|
test3;_|TEST:-TEST.|
Ctrl+H
Find what: ^.+?\K:
Replace with: ;
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
.+? # 1 or more any character but newline, not greedy
\K # forget all we have seen until this position
: # colon
Screen capture (before):
Screen capture (after):
I'm guessing that maybe this expression,
(\w+)\s*(?::)(\s*_\s*\|\s*\w+\s*:\s*-\w+\.\|)
with a replacement of $1;$2 might work.
DEMO 1
Or with less boundaries, this expression:
([^:]+):(.*)
with the same replace.
DEMO 2
It's done like this
Find (?m)^[^:\r\n]*\K:
Replace ;
https://regex101.com/r/rT1vG9/1
How can I use regex in sublime to target the end of every third line, so that I can insert a semicolon.
I know I can target/wrap every third line like this:
(.*\n){3}
And target the end of each line like this: $
But how can I target the END of every THIRD line so that I can insert a semicolon?
You shouldn't match the third newline character. Try the following regex:
^.*(?:\R.*){2}\K
See live demo here
In above regex \R means any kind of newline character, \K means reset match output and ^ matches at start of each line by default in Sublime Text (so no need for (?m)).
Put the cursor at the beginning of file content then search for the given regex and replace with ;.
I need to add text to first line of all my JSP's in eclipse, this is the regex I a using \A.* but some how it selects the first line, I just want to prepend text to the start of the file. any help will be very much appreciated.
The .* pattern matches any 0+ chars other than line break characters, so it matches the first line.
It seems that Eclipse Find/Replace regex feature does not match entirely zero-width patterns (e.g. (?=,) will not find and insert a text before commas).
A workaround is to match and capture some text with (...) (where ... stand for a consuming pattern) capturing group and use $1 in the replacement pattern to reinsert the matched text.
Use
\A(.*)
Replace with MY_NEW_TEXT_HERE_AT_THE_START_OF_FILE$1.
How to get (or remove) all comment lines from a matlab file?
Lines may start with no or an arbitrary number of whitespaces followed by one or more %, followed by the comment.
Using
only_comments = regexp(raw_string, '(?m)^[ ]*[%].*?$', 'match');
fails. Also, how to make sure tabs will be catched?
As I understand this its
(?m) line mode
^ beginning of line
[ ]* none or any number of white spaces
[%].*?$ followed by a % and then any charachter until the line end is reached.
Whats wrong?
Seems like you want something like this,
only_comments = regexp(raw_string, '(?m)^[ ]*[%]+.*?$', 'match');
OR
only_comments = regexp(raw_string, '(?m)^ *%+.*$', 'match');
Explanation:
^ Asserts that we are at the start.
<space>* Matches zero or more spaces.
%+ Matches one or more %
.* Matches any character but not of line breaks.
$ Asserts that we are at the end.
(?m)^[ ]*%+.*$
Think you need this.your regex (?m)^[ ]*[%].*?$ does not quantify %.It will match only 1 %.You need to use %+ to match one or more of it.
How we can remove in Notepad++ with regular expressions the not needed text around a specific string? The string with numbers don't has to be removed. The numbers (string) we need is surrounded always by "onRemoveVariable([0-9]*)".
Source:
<table>
<tr><td style="css">
del
edit
</td></tr>
<tr><td style="css">
del
edit
</td></tr>
Result:
12354
1231584
Does anybody has an idea?
Beste regards
Mario
You could use this regex to delete everything except the numbers between the onRemoveVariable parts:
^.*?onRemoveVariable\((\d+)\).*$|.*
This will attempt to get the numbers first, and if not found, match the whole line.
Replacement string:
$1
If the number was matched, the replacement string will thus put only the number back. If not, then $1 will be null and the result will be an empty line.
regex101 demo
If you now want to remove the multiple blank lines, you can use something like:
\R+
And replace with:
\r\n
Then remove manually any remaining empty lines (there can be at most 2 with this replace, one at the beginning and one at the end). \R matches any line break and \R+ thus matches multiple line breaks. The above thus replaces multiple line breaks with single line breaks.
^ # Beginning of line
.*? # Match everything until...
onRemoveVariable\( # Literal string oneRemoveVariable( is matched
(\d+) # Store the digits
\) # Match literal )
.* # Match any remaining characters
$ # End of line
| # OR if no 'onRemoveVariable(` is found with digits and )...
.* # Match the whole line
You need find all digits \d+ with onRemoveVariable( before it and ) after it.
Use lookahead and lookbehind assertions.
(?<=onRemoveVariable\()(\d+)(?=\))
You can use this regex to match just numbers you want :
/onRemoveVariable\((\d+)\)/g
DEMO (Look at the match information on the right panel)
Hope it helps.