The carriage return by itself only moves the cursor to the beginning of the line. That’s OK if each new line of output is at least as long as the previous one, but if the new line is shorter, the previous line will not be completely overwritten, e.g.: $ echo -e “abcdefghijklmnopqrstuvwxyz\r0123456789” 0123456789klmnopqrstuvwxyz To actually … Read more
One option is to chain the -replace operations together. The ` at the end of each line escapes the newline, causing PowerShell to continue parsing the expression on the next line: $original_file=”path\filename.abc” $destination_file=”path\filename.abc.new” (Get-Content $original_file) | Foreach-Object { $_ -replace ‘something1’, ‘something1aa’ ` -replace ‘something2’, ‘something2bb’ ` -replace ‘something3’, ‘something3cc’ ` -replace ‘something4’, ‘something4dd’ ` … Read more
In c++11, you can use std::regex_replace: #include <string> #include <regex> std::string test = “abc def abc def”; test = std::regex_replace(test, std::regex(“def”), “klm”); // replace ‘def’ -> ‘klm’ // test = “abc klm abc klm”
The other big option here is simply not to use vim: sed -i ‘s/pattern/replacement/’ <files> or if you have some way of generating a list of files, perhaps something like this: find . -name *.cpp | xargs sed -i ‘s/pattern/replacement/’ grep -rl ‘pattern1’ | xargs sed -i ‘s/pattern2/replacement/’ and so on!
You can just add a space before every uppercase character and trim off the leading and trailing spaces s = s.replace(/([A-Z])/g, ‘ $1’).trim()
Yes, ( ) captures a group. You can use it again with $i where i is the i’th capture group. So: search: (\w+\.someMethod\(\)) replace: ((TypeName)$1) Hint: Ctrl + Space in the textboxes gives you all kinds of suggestions for regular expression writing.
You are already using a range, %, which is short hand for 1,$ meaning the entire file. To go from the current line to the end you use .,$. The period means current line and $ means the last line. So the command would be: :.,$s/\vBEFORE/AFTER/gc But the . or current line can be assumed … Read more
Try ‘\1’ for the replacement (single quotes are important, otherwise you need to escape the \): “foo”.gsub(/(o+)/, ‘\1\1\1’) #=> “foooooo” But since you only seem to be interested in the capture group, note that you can index a string with a regex: “foo”[/oo/] #=> “oo” “Z_123: foobar”[/^Z_.*(?=:)/] #=> “Z_123”
[ ]{2,} SPACE (2 or more) You could also check that before and after those spaces words follow. (not other whitespace like tabs or new lines) \w[ ]{2,}\w the same, but you can also pick (capture) only the spaces for tasks like replacement \w([ ]{2,})\w or see that before and after spaces there is anything, … Read more
var str=”asd-0.testing”; var regex = /(asd-)\d(\.\w+)/; str = str.replace(regex, “$11$2″); console.log(str); Or if you’re sure there won’t be any other digits in the string: var str=”asd-0.testing”; var regex = /\d/; str = str.replace(regex, “1”); console.log(str);