I am trying to parse some reStructuredText and want to be able to identify when the indent level has changed. So, I need to be able to see when an indent of 8 spaces has changed to an indent of 4 spaces (for example), so that I can change the color of that text block. Is there a way of using regular expressions to count the number of spaces in the indent and pick out the next line that contains a shallower indent?
Something like this will work:
/
^(\s*)\S.*$ #Find a line with some number of spaces
(?:^\1\S.*$)* #Find more lines with the same starting spaces
^.*$ #This is the line you want here
/xm #x to ignore whitespace in the regex.
#m to have ^and $ match all lines
Related
I have been facing a challenge. I have a text file with the following pattern:
SOME RANDOM TITLE IN CAPS (nnnn)
text text text
more text
...
SOME OTHER RANDOM TITLE IN CAPS (nnnn)
What is for sure is that what I want to extract are lines with a bracket and a date ex: (2015) ; (20008)
After the (nnnn) there is no text, sometimes space and CR LF, sometimes just CR LF
I would like to delete everything else and keep just the TITLE LINE with the brackets
The time I spent I could have done it by hand (there are 100lines) but I like the challenge :)
I thought I could find the issue but I am stuck.
I have tried something along this line:
^.*\(\d\d\d\d\)(?s)(.*)(^.*\(\d\d\d\d\))
But I don't get what I want. I can't seem to stop the (?s)(.*) going all the way to the end of the text instead of stopping at the next occurrence.
I suggest using the Search > Mark feature. Use a pattern like \(\d{4}\) and check the "Bookmark Line" option then click "Mark All". Then use Search > Bookmark > Remove Unmarked Lines. This will remove all lines except the ones that have matched your pattern.
Note: If it's possible to have parentheses with 4 digits within your other lines you could add $ to the end of the expression to ensure that the pattern only matches the end of the line. E.g. more text (1234) and other stuff would be matched by the pattern I gave above but if you use pattern \(\d{4}\)$ it will no longer match.
If you want to be even more specific with your pattern by looking for those lines with only uppercase letters and spaces followed by parentheses with 4 digits inside where the parentheses are at the end of the line, then you could use a pattern like this: [A-Z ]+\(\d{4}\)$
Sample input:
SOME RANDOM TITLE IN CAPS (2008)
text text text
more text
...
SOME OTHER RANDOM TITLE IN CAPS (2010)
Here is how to mark the lines:
After clicking "Mark All" here is what you see:
Now use Search > Bookmark > Remove Unmarked Lines and you get this:
The following RegEx maches the 2 lines with brackets containing 4 numbers:
.*?\(\d{4}\)\s*
It starts matching anything at start zero or more times (non greedy), then it matches a start bracket followed by 4 numbers. Finally ending White Space and new line.
If you want to remove all lines but the ones that end with (4numbers) you may try with this:
^(?!.*\(\d{4}\)\h*$).*(?:\r?\n|\z)
Replace by: (nothing)
See demo
Say you know the starting and ending lines of some section of text, but the chars in some lines and the number of lines between the starting and ending lines are variable, รก la:
aaa
bbbb
cc
...
...
...
xx
yyy
Z
What quantifier do you use, something like:
aaa\nbbbb\ncc\n(.*\n)+xx\nyyy\nZ\n
to parse those sections of text as a group?
You can use the s flag to match multilines texts, you can do it like:
~\w+ ~s.
There is a similar question here:
Javascript regex multiline flag doesn't work
If I understood correctly, you know that your text begins with aaa\nbbbb\ncc and ends with xx\nyyy\nZ\n. You could use aaa.+?bbbb.+?cc(.+?)xx.+?yyy.+?Z so that all operators are not greedy and you don't accidentally capture two groups at once. The text inbetween these groups would be in match group 1. You also need to turn the setting that causes dot to match new line on.
Try this:
aaa( |\n)bbbb( |\n)cc( |\n)( |\n){0,1}(.|\n)*xx( |\n)yyy( |\n)Z
( |\n) matches a space or a newline (so your starting and ending phrases can be split into different lines)
RegExr
At the end of the day what worked for me using Kate was:
( )+aaa\n( )+bbbb\n( )+cc\n(.|\n)*( )+xx\n( )+yyy\n( )+Z\n
using such regexps you can clear pages of quite a bit of junk.
In Notepad++ 6.5.1 I need to replace certain patterns within quote pairs. I want to save the replace as part of a macro, so all replacements need to happen in one step.
For example, in the following string, replace all 'a' characters within quote pairs with a dash, while leaving characters outside the quote pairs untouched:
Input: aa"bbabaavv"kdjhas"bbabaavv"x
Desired result: aa"bb-b--vv"kdjhas"bb-b--vv"x
Note that the quotes are matched up pairwise, such that the 'a' in kdjhas is untouched.
So far I have tried searching for (?:"[^"a]*|\G)\Ka([^"a]*) and replacing with -$1, but that simply replaces all the a's, with the result --"bb-b--vv"kdjh-s"bb-b--vv"x. I'm attempting PCRE regex that will let me recursively replace the quote-delimited text.
Edit: Quote marks within a quoted string are escaped with an extra quote, e.g. "". However, assume I will have already replaced these in a previous pass with a special character. Therefore a regex solution to this problem will not have to deal with escaped quotes.
It is hard to tell if this is possible as you've only provided one line of input text.
But assuming that input follows this pattern:
BOL|any text|string with two groups of a's|any text|string with two groups of a's|any text|EOL
aa "bbabaavv" kdjhas "bbabaavv" x
I was able to create this regexp search string:
^(.+?\".+?)([a]+)(.+?)([a]+)(.*?\")(.+?\".+?)([a]+)(.+?)([a]+)(.*?\".*)$
With this replace string:
\1-\3-\5\6-\8-\A
and it turn your input string from this:
aa"bbabaavv"kdjhas"bbabaavv"x
into this:
aa"bb-b-vv"kdjhas"bb-b-vv"x
Now naturally the search an replace will fail if the input varies from that pattern described as the search is looking for those four groups of a's inside the two groups of quoted strings.
Also I tested that regexp using Zeus which can create a regexp with more than 9 groups.
As you can see the regexp requires 10 groups.
I'm not familar with Notpad++ so I don't know if it supports that many groups.
If your data have variable number of occurrences of quoted strings, then it is not possible to perform replacements only via regex at least in its form offered by Notepad++.
To replace using regex, you would need to perform regex find in existing regex match. As far as I know such a functionality is not available in Notepad++ regexes.
Self-answer
I may have been reaching for the stars in trying to get Notepad++ to do this regex replace, but I think I found a workaround.
The actual task I was attempting involved creating a SQL Server VALUES list from an Excel spreadsheet, where I was copying and pasting selected cells into Notepad++. The delimiters are \t and \r\n. But, cells can have linefeeds too, which are delimited by ". So, I was going to replace these linefeeds with <br> (or something like it), so that
"line1
line2"
would become "line1<br>line2", before processing the actual end-of-row line feeds.
Having such parsing work reliably, especially when more than two lines were in a single cell, may have been too much to ask of Notepad++'s regex capability.
So I came up with a workaround that seems to be working:) Basically it starts with selecting a blank "dummy" column to the right of my column selection (which I can insert if I'm partially selecting from the middle). This will leave a trailing \t at the end of each row, which effectively sets these EOL's apart from ones that might exist with a text cell, freeing me from having to parse line feeds from a "..." field.
So I compiled a macro from the following steps, which seems to be working well:
replace ' with ''
replace \t\r\n with '\)\r\n, \('
replace \t with ', '
replace "" with ''
replace " with <blank>
replace ^ with \(' (cleanup - first row only)
replace ^, \('$ with <blank> (cleanup - last row only)
Example transformation:
from
line1 line 2
"line3
line3b
line3c" line 4
to
('line1', 'line 2')
, ('line3
line3b
line3c', 'line 4')
which can now be easily modified into a SELECT statement:
SELECT *
FROM (VALUES('line1', 'line 2')
, ('line3
line3b
line3c', 'line 4')
) t(a,b)
I need to edit a large EDI message, which basically is a textfile of thousands of short lines. The reason is that it must comply to the standard specification and doesn't because in some of the segments there are an extra QTY+220 line that must be removed. It is in those segments that has 4 QTY lines where QTY+220 must be deleted. Here is a correct segment:
SEQ++79'
MOA+9:1.87945:NOK'
QTY+58:0'
QTY+136:5'
QTY+260:5'
Here is an incorrect segment:
SEQ++365'
MOA+9:1.31896:NOK'
QTY+58:0'
QTY+136:4'
QTY+220:0' <---- this line must be removed
QTY+260:4'
The complete textfile is about 75.000 lines and there are more than 2200 of these validation errors in the xml schema. I tried to make a seach and replace with notepad++ and regular expressions, but I can't make it match over multiple lines. Here is a single line:
^QTY.*'
But I want it to find matches of 4 QTY-lines and remove the 3rd line. How can I do that?
Use \n to match linebreaks.
In your example, replace
(QTY[^\n]+)\n(QTY[^\n]+)\n(QTY[^\n]+)\n(QTY[^\n]+)
with
$1\n$2\n$4
to remove the third line
I am trying to use grep -v -e '' to exclude comments (lines with # as the first non-whitespace character) from a file.
The # can appear either at the begining of the line or there could be a combination of several blank space and tabs in any combination before the first # is encountered.
Assume the file np4 contains this:
# hash at the begining of the line
## two hashes at the begining of the line
#### four hashes at the begining of the line
# two white spaces then a hash
a good line
another good line starting with a few spaces
a good line starting with a combination of spaces and tabs
# two white spaces, two tabes and then a hash
## two tabs, two white spaces and then two hashes
# tab, ws, tab, ws, tab then hash
I tried using the command below, but it does not work as I thought it would. I should only get three lines as the output.
grep -v -e '^\s*#.*$' np4
I believe that what you are missing is the plus sign to match one or more of the pound symbol. I didn't grep as i tested, but this looked good. i've supplied a permalink to my test.
^\s*#+.*$
Here is a permalink to the test on regexpal.com.
not sure if \s works well for grep.
Could you try "^[ ]#." ?
In the [] there is a space and a tab, two characters.