I'm having problems about a regexp.
I'm trying to implement a regex to select just the tab indent blocks, but i cant find a way of make it work:
Example:
INDENT(1)
INDENT(2)
CONTENT(a)
CONTENT(b)
INDENT(3)
CONTENT(c)
So I need blocks like:
INDENT(2)
CONTENT(a)
CONTENT(b)
AND
INDENT(3)
CONTENT(c)
How I can do this?
really tks, its almost that, here is my original need:
table
tr
td
"joao"
"joao"
td
"marcos"
I need separated "td" blocks, could i adapt your example to that?
It depends on exactly what you are trying to do, but maybe something like this:
^(\t+)(\S.*)\n(?:\1\t.*\n)*
Working example: http://www.rubular.com/r/qj3WSWK9JR
The pattern searches for:
^(\t+)(\S.*)\n - a line that begins with a tab (I've also captured the first line in a group, just to see the effect), followed by
(?:\1\t.*\n)* - lines with more tabs.
Similarly, you can use ^( +)(\S.*)\n(?:\1 .*\n)* for spaces (example). Mixing spaces and tabs may be a little problematic though.
For the updated question, consider using ^(\t{2,})(\S.*)\n(?:\1\t.*\n)*, for at least 2 tabs at the beginning of the line.
You could use the following regex to get the groups...
[^\s]*.*\r\n(?:\s+.*\r*\n*)*
this requires that your lines not begin with white space for the beginning of the blocks.
Related
Sometimes I find myself editing a C source file which sees both use of tab as four spaces, and regular tab.
Is there any tool that attempts to parse the file and "normalize" this, i.e. convert all occurrences of four spaces to regular tab, or all occurrences of tab to four spaces, to keep it consistent?
I assume something like this can be done even with just a simple vim one-liner?
There's :retab and :retab! which can help, but there are caveats.
It's easier if you're using spaces for indentation, then just set 'expandtab' and execute :retab, then all your tabs will be converted to spaces at the appropriate tab stops (which default to 8.) That's easy and there are no traps in this method!
If you want to use 4 space indentation, then keep 'expandtab' enabled and set 'softtabstop' to 4. (Avoid modifying the 'tabstop' option, it should always stay at 8.)
If you want to do the inverse and convert to tabs instead, you could set 'noexpandtab' and then use :retab! (which will also look at sequences of spaces and try to convert them back to tabs.) The main problem with this approach is that it won't just consider indentation for conversion, but also sequences of spaces in the middle of lines, which can cause the operation to affect strings inside your code, which would be highly undesirable.
Perhaps a better approach for replacing spaces with tabs for indentation is to use the following substitute command:
:%s#^\s\+#\=repeat("\t", indent('.') / &tabstop).repeat(" ", indent('.') % &tabstop)#
Yeah it's a mouthful... It's matching whitespace at the beginning of the lines, then using the indent() function to find the total indentation (that function calculates indentation taking tab stops in consideration), then dividing that by the 'tabstop' to decide how many tabs and how many spaces a specific line needs.
If this command works for you, you might want to consider adding a mapping or :command for it, to keep it handy. For example:
command! -range=% Retab <line1>,<line2>s#^\s\+#\=repeat("\t", indent('.') / &tabstop).repeat(" ", indent('.') % &tabstop)
This also allows you to "Retab" a range of the file, including one you select with a visual selection.
Finally, one last alternative to :retab is that to ask Vim to "reformat" your code completely, using the = command, which will use the current 'indentexpr' or other indentation configurations such as 'cindent' to completely reindent the block. That typically respects your 'noexpandtab' and 'smarttabstop' options, so it use tabs and spaces for indentation consistently. The downside of this approach is that it will completely reformat your code, including changing indentation in places. The upside is that it typically has a semantic understanding of the language and will be able to take that in consideration when reindenting the code block.
I know that you can use this to remove blank lines
sed /^$/d
and this to remove comments starting with #
sed /^#/d
but how to you do delete all the comments starting with // ?
You just need to "escape" the slashes with the backslash.
/\/\//
the ^ operator binds it to the front of the line, so your example will only affect comments starting in the first column. You could try adding spaces and tabs in there, too, and then use the alternation operator | to choose between two comment identifiers.
/^[ \t]*(\/\/|$)/
Edit:
If you simply want to remove comments from the file, then you can do something like:
/(\/\/|$).*/
I don't know what the 'd' operator at the end does, but the above expression should match for you modulo having to escape the parentheses or the alternation operator (the '|' character)
Edit 2:
I just realized that using a Mac you may be "shelling" that command and using the system sed. In that case, you could try putting quotation marks around the search pattern so that the shell doesn't do anything crazy to all of your magic characters. :) In this case, 'd' means "delete the pattern space," so just stick a 'd' after the last example I gave and you should be set.
Edit 3:
Oh I just realized, you'll want to beware that if you don't catch things inside of quotes (i.e. you don't want to delete from # to end of line if it's in a string!). The regexp becomes quite a bit more complicated in that case, unfortunately, unless you just forgo checking lines with strings for comments. ...but then you'd need to use the substitution operation to sed rather than search-and-delete-match. ...and you'd need to put in more escapes, and it becomes madness. I suggest searching for an online sed helper (there are good regex testers out there, maybe there's one for sed?).
Sorry to sort of abandon the project at this point. This "problem" is one that sed can do but it becomes substantially more complex at every stage, as opposed to just whipping up a bit of Python to do it.
I have been looking at regular expressions to try and do this, but the most I can do is find the start of a line with ^, but not replace it.
I can then find the first characters on a line to replace, but can not do it in such a way with keeping it intact.
Unfortunately I don´t have access to a tool like cut since I am on a windows machine...so is there any way to do what I want with just regexp?
Use notepad++. It offers a way to record an sequence of actions which then can be repeated for all lines in the file.
Did you try replacing the regular expression ^ with the text you want to put at the start of each line? Also you should use the multiline option (also called m in some regex dialects) if you want ^ to match the start of every line in your input rather than just the first.
string s = "test test\ntest2 test2";
s = Regex.Replace(s, "^", "foo", RegexOptions.Multiline);
Console.WriteLine(s);
Result:
footest test
footest2 test2
I used to program on the mainframe and got used to SPF panels. I was thrilled to find a Windows version of the same editor at Command Technology. Makes problems like this drop-dead simple. You can use expressions to exclude or include lines, then apply transforms on just the excluded or included lines and do so inside of column boundaries. You can even take the contents of one set of lines and overlay the contents of another set of lines entirely or within column boundaries which makes it very easy to generate mass assignments of values to variables and similar tasks. I use Notepad++ for most stuff but keep a copy of SPFSE around for special-purpose editing like this. It's not cheap but once you figure out how to use it, it pays for itself in time saved.
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
[ ]+$
Alright, this one's interesting. I have a solution, but I don't like it.
The goal is to be able to find a set of lines that start with 3 periods - not an individual line, mind you, but a collection of all the lines in a row that match. For example, here's some matches (each match is separated by a blank line):
...
...hello
...
...hello
...world
...
...wazzup?
...
My solution is as follows:
^\.\.\..*(\n\.\.\..*)*$
It matches all those, so it's what I'm using for now - however, it looks kinda silly to repeat the \.\.\..* pattern. Is there a simpler way?
Please test your regex before submitting it, rather than submit what "should work." For example, I tried the following first:
(^\.\.\..*$)+
which only returned individual lines, even though in my mind it looks like it would do the trick - I guess I just don't understand regex internals. (And no, I didn't need to set any flags to get ^ and $ to match line boundaries, since I'm implementing this in Ruby.)
So I'm not totally sure there's a good answer, but one would be much appreciated - thanks in advance!
In most regex implementations you can shorten \.\.\. using \.{3} so your solution would turn into \.{3}.*(\n\.{3}.*)*.
What you already have is already simple and understandable. Keep in mind that more "clever" RegExps may very well be slower and undoubtedly less readable.
Assuming lines are terminated by a \n:
((^|\n)\.{3}[^\n]*)+
I am not familiar with Ruby, so depending on how it returns matches you might need to "nonmatch" groups:
((?:(?:^|\n)\.{3}[^\n]*)+)
^([.]{3}.*$\n?)+
This doesn't really need $ in there.
You are pretty close to a solution with (^\.\.\..*$)+, but because the + modifier is on the outside of the group, it is getting overwritten each time and you are only left with the last line. Try wrapping it in an outer group: ((^\.\.\..*$)+) and looking at the first submatch and ignoring the inner one.
Combined with the other suggestion: ((^\.{3}.*$)+)