just encountered the code for doing tab expansion in perl, here is the code:
1 while $string =~ s/\t+/' ' x (length($&) * 8 - length($`) % 8)/e;
I tested it to be working, but I am too much a rookie to understand this, anyone care to explain a bit about why it works? or any pointer for related material that could help me understand this would be appreciated, Thanks a lot.
Perl lets you embed arbitrary code as replacement expressions in regexes.
$& is the string matched by the last pattern match—in this case, some number of tab characters.
$` is the string preceding whatever was matched by the last pattern match—this lets you know how long the previous text was, so you can align things to columns properly.
For example, running this against the string "Something\t\t\tsomething else", $& is "\t\t\t", and $` is "Something". length($&) is 3, so there are at most 24 spaces needed, but length($`)%8 is 1, so to make it align to columns every eight it adds 23 spaces.
The e flag on the regex means to treat the replacement string (' ' x (...etc...) as perl code and interpret/execute it for each match. So, basically look for any place there's 1 or more (+) tab characters (\t), then execute the small perl snippet to convert those tabs into spaces.
The snippet calculates how many tabs were matched, multiplies that number by 8 to get the number of spaces required, but also accounts for anything which may have come before the matched tabs.
Related
I'm using the vscode vimplugin. I have a bunch of lines that look like:
Terry,169,80,,,47,,,22,,,6,,
I want to remove all the alphanumeric characters after the first comma so I get:
Terry,,,,,,,,,,,,,
In command mode I tried:
s/^.+\,[a-zA-Z0-9-]\+//g
But this does not appear to do anything. How can I get this working?
edit:
s/^[^,]\+,[a-zA-Z0-9-]\+//g
\+ is greedy; ^.\+, eats the entire line up to the last ,.
Instead of the dot (which means "any character") use [^,] which means "any but a comma". Then ^[^,]\+, means "any characters up to the first comma".
The problem with your requirement is that you want to anchor at the beginning using ^ so you cannot use flag g — with the anchor any substitution will be done once. The only way I can solve the puzzle is to use expressions: match and preserve the anchored text and then use function substitute() with flag g.
I managed with the following expression:
:s/\(^[^,]\+\)\(,\+\)\(.\+\)$/\=submatch(1) . submatch(2) . substitute(submatch(3), '[^,]', '', 'g')/
Let me split it in parts. Searching:
\(^[^,]\+\) — first, match any non-commas
\(,\+\) — any number of commas
\(.\+\)$ — all chars to the end of the string
Substituting:
\= — the substitution is an expression
See http://vimdoc.sourceforge.net/htmldoc/change.html#sub-replace-expression
submatch(1) — replace with the first match (non-commas anchored with ^)
submatch(2) — replace with the second match (commas)
substitute(submatch(3), '[^,]', '', 'g') — replace in the rest of the string
The last call to substitute() is simple, it replaces all non-commas with empty strings.
PS. Tested in real vim, not vscode.
My program contains a string like
$abc= "mojo logo sfdgsdj2123 *** mojo **";
I want to change it to
$abc= "mojo *** mojo **";
How can i do this?
Also the characters between "logo" and the first " * " can be anything other than " * " (ie not necessarily sfdgsd2123).
So basically the question is to remove till the first occurrence of "*" after first "mojo".
Please help...
I'll give you the "here's how you would do it answer" rather than write the substitution code, as you did ask "How can I do this?"
So here's how:
Make the regex for "a sequence of zero or more non-asterisk characters that are preceded by the word mojo"
Substitute the first occurrence of the substring that matches that regex with the empty string.
That's all there is to it. It's a little one-liner in Perl and most languages with sophisticated enough regex engines to support positive lookbehind.
If all that sounded crazy, feel free to walk through the string character by character. Find where "mojo" appears first. Then continue walking through the string, removing all the non-asterisk characters you encounter.
I am trying to get that Perl split working for more than 2 hours. I don't see an error. Maybe some other eyes can look at it and see the issue. I am sure its a silly one:
#versionsplit=split('.',"15.0.3");
print $versionsplit[0];
print $versionsplit[1];
print $versionsplit[2];
I just get an empty array. Any idea why?
You need:
#versionsplit=split(/\./,"15.0.3");
The first argument to split is a regular expression, not a string. And . is the regex symbol which means ‘match any character’. So all the characters in your input string were being treated as separators, and split wasn't finding anything between them to return.
the "." represents any character.You need to escape it for split function to recognise as a field separator.
change your line to
#versionsplit=split('\.',"15.0.3");
This question is nearly identical to this question except that I have to go to three spaces (company coding guidelines) rather than four and the accepted solution will only double the matched pattern. Here was my first attempt:
:%s/^\(\s\s\)\+/\1 /gc
But this does not work because four spaces get replaced by three. So I think that what I need is some way to get the count of how many times the pattern matched "+" and use that number to create the other side of the substitution but I feel this functionality is probably not available in Vim's regex (Let me know if you think it might be possible).
I also tried doing the substitution manually by replacing the largest indents first and then the next smaller indent until I got it all converted but this was hard to keep track of the spaces:
:%s/^ \(\S\)/ \1/gc
I could send it through Perl as it seems like Perl might have the ability to do it with its Extended Patterns. But I could not get it to work with my version of Perl. Here was my attempt with trying to count a's:
:%!perl -pe 'm<(?{ $cnt = 0 })(a(?{ local $cnt = $cnt + 1; }))*aaaa(?{ $res = $cnt })>x; print $res'
My last resort will be to write a Perl script to do the conversion but I was hoping for a more general solution in Vim so that I could reuse the idea to solve other issues in the future.
Let vim do it for you?
:set sw=3<CR>
gg=G
The first command sets the shiftwidth option, which is how much you indent by. The second line says: go to the top of the file (gg), and reindent (=) until the end of the file (G).
Of course, this depends on vim having a good formatter for the language you're using. Something might get messed up if not.
Regexp way... Safer, but less understandable:
:%s#^\(\s\s\)\+#\=repeat(' ',strlen(submatch(0))*3/2)#g
(I had to do some experimentation.)
Two points:
If the replacement starts with \=, it is evaluated as an expression.
You can use many things instead of /, so / is available for division.
The perl version you asked for...
From the command line (edits in-place, no backup):
bash$ perl -pi -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
(in-place, original backed up to "YOUR_FILE.bak"):
bash$ perl -pi.bak -e 's{^((?: )+)}{" " x (length($1)/2)}e' YOUR_FILE
From vim while editing YOUR_FILE:
:%!perl -pe 's{^((?: )+)}{" " x (length($1)/2)}e'
The regex matches the beginning of the line, followed by (the captured set of) one or more "two space" groups. The substitution pattern is a perl expression (hence the 'e' modifier) which counts the number of "two space" groups that were captured and creates a string of that same number of "three space" groups. If an "extra" space was present in the original it is preserved after the substitution. So if you had three spaces before, you'll have four after, five before will turn into seven after, etc.
My usual 'x' usage was :
print("#" x 78, "\n");
Which concatenates 78 times the string "#". But recently I came across this code:
while (<>) { print if m{^a}x }
Which prints every line of input starting with an 'a'. I understand the regexp matching part (m{^a}), but I really don't see what that 'x' is doing here.
Any explanation would be appreciated.
It's a modifier for the regex. The x modifier tells perl to ignore whitespace and comments inside the regex.
In your example code it does not make a difference because there are no whitespace or comments in the regex.
The "x" in your first case, is a repetition operator, which takes the string as the left argument and the number of times to repeat as the right argument. Perl6 can replicate lists using the "xx" repetition operator.
Your second example uses the regular expression m{^a}x. While you may use many different types of delimiters, neophytes may like to use the familiar notation, which uses a forward slash: m/^a/x
The "x" in a regex is called a modifier or a flag and is but one of many optional flags that may be used. It is used to ignore whitespace in the regex pattern, but it also allows the use of normal comments inside. Because regex patterns can get really long and confusing, using whitespace and comments are very helpful.
Your example is very short (all it says is if the first letter of the line starts with "a"), so you probably wouldn't need whitespace or comments, but you could if you wanted to.
Example:
m/^a # first letter is an 'a'
# <-- you can put more regex on this line because whitespace is ignored
# <-- and more here if you want
/x
In this use case 'x' is a regex modifier which "Extends your pattern's legibility by permitting whitespace and comments." according to the perl documentation. However it seems redundant here