New answers tagged replace
0
This may actually be easier on the command line with perl than within emacs:
perl -i.bak -pe 's/import org.apache.log4j.Logger;/import org.apache.log4j.Logger;/;' * */*
(Add more */*/* if you have deeper subfolders.)
Added advantage: backup copies of your original files are created with the suffix ".bak".
0
If you use library Dired+ then you can use command diredp-do-eval-in-marked-recursive to do what you want.
If you use it directly it asks you for a sexp to evaluate at the beginning of each file to be acted on. In this case, the sexp you want is just a replace-string sexp.
So this simple command does the job:
(defun bar ()
(interactive)
(diredp-do-eval-...
1
This is trivial in a Dired buffer for the directory possibly containing (possibly with subdirs, recursively) matches for some search pattern.
Just use Q, which is bound to dired-do-find-regexp-and-replace:
Q runs the command dired-do-find-regexp-and-replace (found in
dired-mode-map), which is an interactive autoloaded compiled Lisp
function in dired-aux.el.
...
1
You could do this with wgrep, which you can install from Melpa.
The workflow would be:
Call grep: M-x grep
Fill in an invocation that will match your files and patterns:
grep --color=auto -nH --null -e 'import org.apache.log4j.Logger;' ./*
(note that the first part is the default, you only need to enter 'import org.apache.log4f.Logger;'./*
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