I'm trying to replace a string, inside of a file, with perl from inside a Makefile.
InstallTo = $(PWD)/WebTest
BuildApache:
mkdir -p WebTest
cd Source/httpd; ./configure --prefix=$(InstallTo) --exec-prefix=$(InstallTo)
cd Source/httpd; make; make install
cd $(InstallTo)/conf; perl -pi -e 's/ServerRoot \"$(InstallTo)\"/ServerRoot/g' httpd.conf
cd $(InstallTo)/conf; cp -f httpd.conf httpd.conf.orig
I'm not sure exactly what I'm doing though, I've just tried to modify the perl line from something I found on the net. I think its the \" thats messing things up but I don't know enough about Perl to fix it.
You might want to try:
s|ServerRoot "$(InstallTo)"|ServerRoot|g
You're pasting a value with a slash in it as part of the search expression. It ends up as:
s/ServerRoot \"PWD/WebTest\"/ServerRoot/g
(Where PWD stands for any literal directory spec.) Since you can't escape the slash, that's always going to be a problem unless you use an alternative delimiter.
Since your variable contains '/' you need to use a different character for regular expressions, also you may want to use quotemeta or \Q..\E in regular expressions having variables which can contain special characters
s#\QServerRoot "$(InstallTo)"\E#ServerRoot#g
Refer to this post for more details how-do-i-handle-special-characters-in-a-perl-regex
Related
I want to delete all files in a folder, which contain he word TRAR in their filename.. I hav etried the following :
CONFIG_DIR=`pwd`
VENDOR=ericsson-msc
RELEASE=v1
BASE_DIR=/appl/virtuo/gways
system ("cd /appl/virtuo/gways/config/ericsson-msc/v1/spool/input_d; rm-rf *TRAR");
remove all your config lines ( are they even perl? )
CONFIG_DIR=`pwd`
VENDOR=ericsson-msc
RELEASE=v1
BASE_DIR=/appl/virtuo/gways
and
system ("cd /appl/virtuo/gways/config/ericsson-msc/v1/spool/input_d; rm -rf *TRAR")
should work but you should really be using perl code (unlink, etc)
I suspect you are confusing the usage of perl with how you will use awk in bash scripts.
As #Steffen Ullrich said, that isn't Perl or Shell. But I'll try to make it a little more Perlish for you:
First, note that
variables in Perl start with a $
strings need "quotes around them"
statements end with a ;
spaces around = are ok and make it all easier to read
so
$CONFIG_DIR = `pwd`;
$VENDOR = "ericsson-msc";
$RELEASE = "v1";
$BASE_DIR = "/appl/virtuo/gways";
Next, see how you can combine these into a single string like this (I'm guessing that's what you want to do)
$DIR_FOR_CLEANING = "$BASE_DIR/config/$VENDOR/$RELEASE/spool/input_d";
Lastly, you should be really careful whenever using the -r command to rm along with a wildcard like *. Look up the man page for rm and see if -r is something you want to do. I don't think you need it here, unless you have directories named *TRAR that you want to recurse into to remove. I'll bet you only have files named *TRAR in that input_d directory.
Also, the command the way you wrote it could fail the cd if that directory doesn't exist, and would then proceed to recursively remove *TRAR from whatever directory you're running the script from. But you don't need to change directories at all. Try something like this
system ("echo rm -f $DIR_FOR_CLEANING/*TRAR");
If the echo command lists the files you do in fact want it to remove, then remove the "echo" and the rm will start deleting stuff.
I'm using zsh, and am trying to write a function to operate on a URL and a pathname:
function my-function
{
somecommand --url $1 $(readlink -f $2)
}
(to complicate things somewhat, the function actually uses sh syntax, as it is sourced from my ~/.zshrc using a trick like this). The readlink is there to expand symlinks and ensure directories such as . are evaluated correctly (the directory name is stored for later use by somecommand).
When I type a command from the command-line like this:
my-function http://example.org/example /tmp/myexampledirectory
... it works fine, even if I autocomplete the directory name. However, if the directory name contains spaces, zsh completes it like this:
my-function http://example.org/example /tmp/My\ Example\ Directory
For most "normal" commands (cp, mv, etc.) that never seems to cause a problem. However, in my case, somecommand sees $2 as only being /tmp/My - presumably the rest is seen as another argument.
How can I avoid this situation? I would prefer not to alter the standard zsh autocompletion, but rather find a way for my function to handle this.
The zsh completion system works very well here, and the solution is very simple, just put double-quotes around the readlink argument in the script:
somecommand --url $1 $(readlink -f "$2")
The point is that without quotes readlink removes backslashes which escape whitespaces. Compare three results:
1. Without backslashes and quotes readlink -f assumes that there are three different files/directories (with default path in current directory) and produces
$ readlink -f /tmp/My Example Directory
/tmp/My
/home/jimmij/Example
/home/jimmij/Directory
2. With escaping backslashes but without quotes readlink -f understands that there is only one directory, but removes backslashes from output, so that somecommand takes three separate arguments
$ readlink -f /tmp/My\ Example\ Directory
/tmp/My Example Directory
3. With backslashes and with double-quotes readlink -f gives the output with backslashes what is (most probably) expected by somecommand
$ readlink -f "/tmp/My\ Example\ Directory"
/tmp/My\ Example\ Directory
BTW, as a rule of thumb: if there are any problems with whitespaces in the shell-like scripts (bash, zsh, whatever) the first thing to play with is different quotation marks around variables.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Renaming lots of files in Linux according to a pattern
I have multiple files in this format:
file_1.pdf
file_2.pdf
...
file_100.pdf
My question is how can I rename all files, that look like this:
file_001.pdf
file_002.pdf
...
file_100.pdf
I know you can rename multiple files with 'rename', but I don't know how to do this in this case.
You can do this using the Perl tool rename from the shell prompt. (There are other tools with the same name which may or may not be able to do this, so be careful.)
rename 's/(\d+)/sprintf("%03d", $1)/e' *.pdf
If you want to do a dry run to make sure you don't clobber any files, add the -n switch to the command.
note
If you run the following command (linux)
$ file $(readlink -f $(type -p rename))
and you have a result like
.../rename: Perl script, ASCII text executable
then this seems to be the right tool =)
This seems to be the default rename command on Ubuntu.
To make it the default on Debian and derivative like Ubuntu :
sudo update-alternatives --set rename /path/to/rename
Explanations
s/// is the base substitution expression : s/to_replace/replaced/, check perldoc perlre
(\d+) capture with () at least one integer : \d or more : + in $1
sprintf("%03d", $1) sprintf is like printf, but not used to print but to format a string with the same syntax. %03d is for zero padding, and $1 is the captured string. Check perldoc -f sprintf
the later perl's function is permited because of the e modifier at the end of the expression
If you want to do it with pure bash:
for f in file_*.pdf; do x="${f##*_}"; echo mv "$f" "${f%_*}$(printf '_%03d.pdf' "${x%.pdf}")"; done
(note the debugging echo)
I have around 230 files which are *.pl , *.txt and some are *.conf files which has a default path set to the current environment say /home/AD/USR/perl/5.8.0/bin/perl. I need to replace "/home/AD/USR" with an environment variable ${USR_PATH}. The files I want to modify are in subdirectories. Which means my script should find e.g find .|xargs grep -l "/home/AD/USR" all the files and then replace the string.
OLD: /home/AD/USR/perl/5.8.0/bin/perl
New : ${USR_PATH}/perl/5.8.0/bin/perl
Can some one give me a clue how do I do that?
Shell : /bin/bash
Env : Linux x86_64
If you replace part of a string with ${USR_PATH} you will refer to the perl variable $USR_PATH, not the environment variable, which is in perl referred to as $ENV{USR_PATH}.
perl -pi.bak -we 's#/home/AD/USR(?=/perl/5.8.0/bin/perl)#\$ENV{USR_PATH}#g'
*.pl *.txt *.conf
Using the lookahead will save you the trouble of replacing the rest of the path afterwards.
I assume you want to replace it with the literal value. If you want to replace it with the actual value in the environment variable, just remove the backslash in front of $ENV.
While using an environment variable seems handy and all, it will reduce your scripts portability. Why not use a configuration file? If you had done that from the start, you wouldn't be having this trouble. Search CPAN for a nice module.
perl -i -pe 's|/home/AD/USR/perl/5.8.0/bin/perl|\${USR_PATH}/perl/5.8.0/bin/perl|' <your files>
I couldn't find an answer for this exact problem, so I'll ask it.
I'm working in Cygwin and want to reference previous commands using !n notation, e.g., if command 5 was which ls, then !5 runs the same command.
The problem is when trying to do substitution, so running:
!5:s/which \([a-z]\)/\1/
should just run ls, or whatever the argument was for which for command number 5.
I've tried several ways of doing this kind of substitution and get the same error:
bash: :s/which \([a-z]*\)/\1/: substitution failed
As far as I can tell the s/old/new/ history substitution syntax only does simple string substitution; it does not support full regexes. Here's what man bash has to say:
s/old/new/
Substitute new for the first occurrence of old in the event line. Any delimiter can be used in place of /. The final delimiter is optional if it is the last character of the event line. The delimiter may be quoted in old and new with a single backslash. If & appears in new, it is replaced by old. A single backslash will quote the &. If old is null, it is set to the last old substituted, or, if no previous history substitutions took place, the last string in a !?string[?] search.
Never fear, though. There are in fact easier ways to accomplish what you are trying to do:
!$ evaluates to the last argument of the previous command:
# ls /etc/passwd
/etc/passwd
# vim !$
vim /etc/passwd
!5:$ evaluates to the last argument of command #5:
# history
...
5: which ls
...
# !5:$
ls
You can also use Alt+. to perform an immediate substitution equivalent to !$. Alt+. is one of the best bash tricks I know.
This worked for me using Bash in Cygwin (note that my which ls command was number 501 in my history list; not 5 like yours):
$(!501 | sed 's/which \([a-z]\)/\1/')
You could also do it this way (which is shorter/cleaner):
$(!501 | sed 's/which //')