When I add this to my vimrc:
set iskeyword-=#
It removes some syntax. I've only tested this in viml and ruby code. What's the reason for this? I've included images to show the effect in viml code.
EDIT: How would I go about removing it without affecting syntax? The reason why I removed it was to make it easier to select "$#" as a word. Anyone can offer a solution to this problem or an alternative way of doing that is greatly appreciated.
From 'isfname', which 'iskeyword' points to:
If the character is '#', all characters where isalpha() returns TRUE are included. … To include '#' itself use "#-#".
# (64) isn’t actually in the character set to begin with for me:
iskeyword=#,48-57,_,192-255
but if it somehow got in there for you, you can add a part starting with a caret to exclude it:
set iskeyword+=^#-#
Related
I've search both SO and Rstudio's community pages and failed to find aquestion, much less and answer to this annoyance I have experienced with Rstudio. (The Rstudio help pages won't let me post a second question within 12 hours of my first, which was explained as a bug.)
If I type:
(test)
... and then realize that test should be quoted, then putting the cursor at the end to test and entering a double-quote will give me two double-quotes "". It will not do this if I first enter a quote between ( and t and then it will also not give me doubling of the double-quote character at the end of `test. Why should it matter whether I first correct my error at the end of the symbol or at the beginning? Is there anything I can do to modify this quirk> It seems that a syntax aware console editor out to be able to tell when a doubling of quotes does not make sense. It's obviously making that "decision" when the quotes are entered between an open-paren and a character. Why not suppress the unhelpful behavior when it is between a character and a close-paren?
I have many text files, and I need to add some text (e.g. MNP) to the beginning of the first line in each file.
How can I do this in Notepad++?
(I'm using v6.6.9)
Make sure to backup your work beforehand, and set proper extension of files to affect and folder to search through before you do this.
You can use regular expressions. Several places around the internet claim that the regex \A works, but it wasn't working for me, it was cycling byte by byte through. I found that \A^ sticks to 0 position of the file.
Oddly, I additionally found that I couldn't replace \A or \A^ and have it take effect. This is what worked for me.
Find: \A^(.*?)
Replace MNP\1
Truthfully, the \1 in Replace isn't even necessary since I'm cheating and basically telling notepad to look for 0 characters.
This should work just as well.
Find: \A^.*?
Replace MNP
Please backup your work beforehand though.
Alternatively, this also seems to work.
Find: .{0}(.*)
Replace: MNP\1
It effectively looks for 0 characters followed by the whole document/line (depending on whether . matches newline is checked, this choice won't matter for the outcome however).
Notepad++ obviously recognizes all comments as such. Is there a way to simply delete all?
Edit: Stat-R's bookmark method has helped greatly, not only for removing comments but for conditionally removing lines in general.
For a general file, first of all you need to know the comment operator of the language you are writing the file in. For example, in java script the comment operator is //.
For the following code...
In NP++, you need to
Mark the lines that contains '//'. Make sure the bookmark option is enabled.
Then, choose from NP++ menu Search>Bookmark>Remove Bookmarked lines
EDIT:
Another solution after #Chris Mirno 's suggestion is as follows:
Use regular expression. See the image below. It is self explanatory
To understand it better, refer to these
In the Find & Replace Dialog, put the following regex and adjust the search options as depicted.
/\*.*?\*/
Replace with: (empty)
Select Mode: Regular Expression AND .(dot) matches newline
This should remove all your C style comments spanned across lines.
Star-R and Chris Mirno Answer are also Correct and Good.
But For Line Comment:
//.*?(?=\r?$)
Explanation:
// will be the Starting Position
.*? Will be any character
(?=\r?$) will search to the end of the line (as it is required in line comment)
Note:
But Still check each of the line because for example if your code contains soap format like
//www.w3.org/2001/XMLSchema-instance\x2......");
it will capture this line because the starting is // and it goes to end of the line so watch out for this :)
Warning to all using Stat-R's solution:
This method will remove lines of code if formatted like this:
echo "hello"; //This comment will be detected
Following his method, the entire line will be removed.
Therefore make sure to go through and make these comments, their own line before doing this method.
I have had some luck running a macro for the above. Basically:
search for // (F3)
select to end of line (shift+end)
delete (delete)
Put // into the search dialog by just searching for it once. Then record the three steps in a macro, then play it back until EOF.
The first time I did it I had a problem, but then it worked, not sure what I did differently.
Anton Largiader's answer was the most reliable one, including complex inline comments.
However, it will leave many empty lines, including ones with empty characters (space, tabs...) so I would just add another step to make it almost perfect:
After running the macro, just do:
Edit > Line Operations > Remove Empty Lines
OR
Edit > Line Operations > Remove Empty Lines (Containing Blank Characters)
1st option is good if you wish to remove only really empty lines
2nd options will remove every empty line even containing space etc. so there will be no more actual spacing left between code blocks. 1st option might be the safest with some manual cleanup afterwards.
As someone suggested in another post, the simplest and most reliable is maybe to export the all text in .RTF format using Menu Plugin-->NppExport-->Export to RTF and then:
-Open the newly created file in Word
-Select any part of any comment
-On the top-right side of Word clic Select--> Select all texts with similar formatting
-Remove the selected comments all at once (del or cut if doesn't work)
To remove Powershell comments if someone find it handy:
Removing Comment in a Powershell using Notepad ++
To find just lines beginning with # (and not with # elsewhere in the line).
Notepad++ SEARCH Menu > Find
‘Mark‘ Tab – fill in as below.
Select ‘Mark All’ (clear all marks if used previously).
Regex ^[#}
enter image description here
SEARCH Menu > bookmark > Remove (or do anything on the list with
them)
Clear all marks to reset
You can select no comments just code by doing the following:
Regex ^[^#}
enter image description here
Enter ctrl+shift+K to remove comment
can any body tell me how to use regex for negation of string?
I wanna find all line that start with public class and then any thing except first,second and finally any thing else.
for example in the result i expect to see public class base but not public class myfirst:base
can any body help me please??
Use a negative lookahead:
public\s+class\s+(?!first|second).+
If Peter is correct and you're using Visual Studio's Find feature, this should work:
^:b*public:b+class:b+~(first|second):i.*$
:b matches a space or tab
~(...) is how VS does a negative lookahead
:i matches a C/C++ identifier
The rest is standard regex syntax:
^ for beginning of line
$ for end of line
. for any character
* for zero or more
+ for one or more
| for alternation
Both the other two answers come close, but probably fail for different reasons.
public\s+class\s+(?:(?!first|second).)+
Note how there is a (non-capturing) group around the negative lookahead, to ensure it applies to more than just the first position.
And that group is less restrictive - since . excludes newline, it's using that instead of \S, and the $ is not necessary - this will exclude the specified words and match others.
No slashes wrapping the expression since those aren't required in everything and may confuse people that have only encountered string-based regex use.
If this still fails, post the exact content that is wrongly matched or missed, and what language/ide you are using.
Update:
Turns out you're using Visual Studio, which has it's own special regex implementation, for some unfathomable reason. So, you'll be wanting to try this instead:
public:b+class:b+~(first|second)+$
I have no way of testing that - if it doesn't work, try dropping the $, but otherwise you'll have to find a VS user. Or better still, the VS engineer(s) responsible for this stupid non-standard regex.
Here is something that should work for you
/public\sclass\s(?:[^fs\s]+|(?!first|second)\S)+(?=\s|$)/
The second look a head could be changed to a $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)
I've been trying to construct a ruby regex which matches trailing spaces - but not indentation placeholders - so I can gsub them out.
I had this /\b[\t ]+$/ and it was working a treat until I realised it only works when the line ends are [a-zA-Z]. :-( So I evolved it into this /(?!^[\t ]+)[\t ]+$/ and it seems like it's getting better, but it still doesn't work properly. I've spent hours trying to get this to work to no avail. Please help.
Here's some text test so it's easy to throw into Rubular, but the indent lines are getting stripped so it'll need a few spaces and/or tabs. Once lines 3 & 4 have spaces back in, it shouldn't match on lines 3-5, 7, 9.
some test test
some test test
some other test (text)
some other test (text)
likely here{ dfdf }
likely here{ dfdf }
and this ;
and this ;
Alternatively, is there an simpler / more elegant way to do this?
If you're using 1.9, you can use look-behind:
/(?<=\S)[\t ]+$/
but unfortunately, it's not supported in older versions of ruby, so you'll have to handle the captured character:
str.gsub(/(\S)[\t ]+$/) { $1 }
Your first expression is close, and you just need to change the \b to a negated character class. This should work better:
/([^\t ])[\t ]+$
In plain words, this matches all tabs and spaces on lines that follow a character that is not a tab or a space.
Wouldn't this help?
/([^\t ])([\t ]+)$/
You need to do something with the matched last non-space character, though.
edit: oh, you meant non blank lines. Then you would need something like /([^\s])\s+/ and sub them with the first part
I'm not entirely sure what you are asking for, but wouldn't something like this work if you just want to capture the trailing whitespaces?
([\s]+)$
or if you only wanted to capture tabs
([ \t]+)$
Since regexes are greedy, they'll capture as much as they can. You don't really need to give them context beforehand if you know what you want to capture.
I still am not sure what you mean by trailing indentation placeholders, so I'm sorry if I'm misunderstanding.
perhaps this...
[\t|\s]+?$
or
[ ]+$