i have multiple document which has lines which starts with blank space. those lines should be removed.
how can one remove a line which contains a blank space in the beginning and no other chars?
With Notepad++, you can use regular expression search and replace.
Go to Replace, select "Regular Expression" in the Search Mode box, and enter the following search expression:
\r\n \r\n
This means: newline, space, newline.
Replace by nothing. That should do the trick!
This is in Notepad++
Search
Replace
Find what: ^ +$
Replace with: (blank)
Search Mode: Regular expression
I my case regEx not solve the problem, I use plugin TextFX and I solve like this:
delete all spaces:
delete lines:
Find and Replace (Ctrl + h) than on Search Mode settings check Regular expression. After you sett the box to search with regex patterns enter this into the Find what field: ^([ ]+)$ and leave the replace field empty. Than click Replace All. That's it.
This will find all lines starting with a space character (may contain more that one space) but that have NO other character after the space/s.
Go to Edit-> Line Operations->Remove Empty Lines(Containing Blank Characters)
Related
I have a log file, and I want to get rid of the third column that start with "external", this column is not always in the third place so I need to find the word "external" and then delete it with the string that follows the colon.
I was thinking in using -replace for that, but does "-replace" accept some regex to delete the rest of the string (after the semicolons) that is always changing?
or maybe there is a better way to do this?
02/02/2020 name:VAL_NATURE external:af2045b2-5992-432e-b790-c1ad4743038 status:good
cat mylog.log | %{$_ -replace "external???",""}
With any delimited file, the first thought I have is to break it at the delimiters (in your case, the white space) and treat it like an object. Deleting a column is trivial if you do that, and it lets you have easy access to the data for other purposes.
If, however, your only task is to remove that column with 'external' + colon + all text up to the next bit of white space, that is an easy thing to do with a regex replace.
$line = '02/02/2020 name:VAL_NATURE external:af2045b2-5992-432e-b790-c1ad4743038 status:good'
$line -replace 'external:.*\s',''
EDIT: Tested the code above, and got this output:
02/02/2020 name:VAL_NATURE status:good
The . is any character, and .* says "any character zero or more times" it continues matching until it gets to whitespace, which is represented by the \s. So this regex matches the word 'external' followed by a ':' followed by zero or more other characters followed by whitespace (space/tab/etc).
This is related to cleaning files before parsing them elsewhere, namely, malformed/ugly CSV. I see plenty of examples for removing/matching all characters between certain strings/characters/delimiters, but I cannot find any for specific strings. Example portion of line would look something like:
","Should now be allowed by rule above "Server - Access" added by Rich"\r
To be clear, this is not the entire line, but the entire line is enclosed in quotes and separated by "," and ends in ^M (Windows newline/carriage return).The 'columns' preceding this would be enclosed at each side by ",". I would probably use this too to remove cruft that appears earlier in the line.
What I am trying to get to is the removal of all double quotes between "," and "\r ("Server - Access" - these ones) without removing the delimiters. Alternatively, I may just find and replace them with \" to delimit them for the Ruby CSV library. So far I have this:
(?<=",").*?(?="\\r)
Which basically matches everything between the delimiters. If I replace .*? with anything, be that a letter, double quotes etc, I get zero matches. What am I doing wrong?
Note: This should be Ruby compatible please.
If I understand you correctly, you can use negative lookahead and lookbehind:
text = '","Should now be allowed by rule above "Server - Access" added by Rich"\r'
puts text.gsub(/(?<!,)"(?![,\\r])/, '\"')
# ","Should now be allowed by rule above \"Server - Access\" added by Rich"\r
Of course, this won't work if the values themselves can contain comas and new lines...
I am new to writing in bash and I just finished this long script but I made the mistake of not adding quotation marks to all the variables beginning with the unary operator $. Adding all the quotation marks by hand is going to take a while. Is there a short cut I can use so all the words in the text file beginning with $ get quotation marks around them? So if a line in the file looks like:
python myProgram.py $car1 $car2 $speed1 $speed2
Then after the shortcut it will appear as
python myProgram.py "$car1" "$car2" "$speed1" "$speed2"
I am writing the script using nano.
Use global search and replace with the expression (\$\w+).
Switch to search and replace mode with C-\.
Switch to regex mode with Alt-R.
Type the expression (\$\w+). Hit Enter.
Type in the replacement expression "\1" replace the captured expression with quotations. Hit Enter.
On the match, hit A for All.
Given your need, it doesn't seem mandatory to provide a solution based on that editor. If you have access to a shell you might try this simple sed command:
sed -i.bak -r 's/\$\w+/"&"/g' my-script.sh
This is far from being perfect but should do the job in your particular case. If the above command:
-i.bak will perform the replacement "in place" -- that is modifying the original file, making a backup with the .bak extension
s/..../..../g is the usual sed command to search and replace using a pattern. The search pattern is between the first two \. The replacement is between the last two /
\$\w+ this pattern correspond to a $ followed by one or more letters (\w+). The backslash before $ is needed because that character normally has special meaning in a search pattern.
"&" is the replacement string. In there, the & is replaced by the string found in the search pattern. Broadly speaking this put quotes arround any string matching the search pattern.
I want to count the characters without whitespace of a visual selection.
Intuitively, I tried the following
:'<,'>w !tr -d [:blank:] | wc -m
But vim does not like it.
This is possible with the following substitute command:
:'<,'>s/\%V\S//gn
The two magical ingredients are
the n flag of the substitute command. What it does is
Report the number of matches, do not actually substitute. (...) Useful to count items.
See :h :s_flags, and check out :h count-items, too.
the zero-width atom \%V. It matches only inside the Visual selection. As a zero-width match it makes an assertion about the following atom \S "non-space", which is to match only when inside the Visual selection. See :h /\%V.
The whole command thus substitutes :s nothing // for every non-whitespace character \S inside the Visual selection \%V, globally g – only that it doesn't actually carry out any substitutions but instead reports how many times it would have!
In order to count the non-whitespace characters within a visual selection in vim, you could do a
:'<,'>s/\S/&/g
Vim will then tell how many times it replaced non-whitespace characters (\S) with itself (&), that is without actually changing the buffer.
You must escape special character for the shell, and use [:space:] better because it will delete also the newline character. It should be:
:'<,'>w !tr -d '[:space:]' | wc -m
I'm using a script to remove trailing spaces and then save the file.
The problem is that all my code foldings expand when I use it. How do I change the command so it will keep the code foldings?
You can use foldingStartMarker & foldingStopMarker to indicate the folds to TextMate.
To define a block that starts with { as the last non-space character on the line and stops with } as the first non-space character on the line, we can use the following patterns:
foldingStartMarker = '\{\s*$';
foldingStopMarker = '^\s*\}';
pressing the F1 key will fold any code folds present.
Reference: http://manual.macromates.com/en/navigation_overview#collapsing_text_blocks_foldings