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>
Related
I'm currently using the 'sed' command (below) which searches for the word {{Title}} and replaces it with the current working directory ${PWD##*/}/.
sed -i "s/{{Title}}/${PWD##*/}/g" filename.x
How do I modify the above 'sed' command to work for the full path of the current working directory?
Any help will be much appreciated!
Change the s command to use a separator that is not part of the path. Generally, : is not allowed in path names so it's safe to use here instead of /, like instead of:
s/expression/$PWD/
use
s:expression:$PWD:
I wrote a perl script that prints a path to a very specific file. I want to define a personal environment variables (by using setenv in .aliases file) that gives the output of this script.
For example, let's say that the file "myscript.pl" prints the path "/home/files/reports/file". Let's call the variable (that I want to define in .aliases file) 'myoutput'. I want that when I type "most $myoutput" in Unix, this file will be opened by most, and when I type "echo $myoutput", Unix will print the path.
How can I define a personal variable which value is determined by a script?
If you use bash, you can put the following to your .bashrc:
export myoutput=$(perl /path/to/myscript.pl)
For tcsh, use .cshrc instead, and modify the line to
setenv myoutput `perl /path/to/myscript.pl`
You need to start a new session to make the variable exist.
myoutput=$(perl myscript.pl)
When your script prints more, select the correct line:
myoutput=$(perl myscript.pl | grep /home)
I have a script in one of my application folders.Usually I just cd into that locatin in Unix box and run the script e.g.
UNIX> cd My\Folder\
My\Folder> MyScript
This prints the required result.
I am not sure how do I do this in Bash script.I have done the following
#!/bin/bash
mydir=My\Folder\
cd $mydir
echo $(pwd)
This basically puts me in the right folder to run the required script . But I am not sure how to run the script in the code?
If you can call MyScript (as opposed to ./MyScript), obviously the current directory (".") is part of your PATH. (Which, by the way, isn't a good idea.)
That means you can call MyScript in your script just like that:
#!/bin/bash
mydir=My/Folder/
cd $mydir
echo $(pwd)
MyScript
As I said, ./MyScript would be better (not as ambiguous). See Michael Wild's comment about directory separators.
Generally speaking, Bash considers everything that does not resolve to a builtin keyword (like if, while, do etc.) as a call to an executable or script (*) located somewhere in your PATH. It will check each directory in the PATH, in turn, for a so-named executable / script, and execute the first one it finds (which might or might not be the MyScript you are intending to run). That's why specifying that you mean the very MyScript in this directory (./) is the better choice.
(*): Unless, of course, there is a function of that name defined.
#!/bin/bash
mydir=My/Folder/
cd $mydir
echo $(pwd)
MyScript
I would rather put the name in quotes. This makes it easier to read and save against mistakes.
#!/bin/bash
mydir="My Folder"
cd "$mydir"
echo $(pwd)
./MyScript
Your nickname says it all ;-)
When a command is entered at the prompt that doesn't contain a /, Bash first checks whether it is a alias or a function. Then it checks whether it is a built-in command, and only then it starts searching on the PATH. This is a shell variable that contains a list of directories to search for commands. It appears that in your case . (i.e. the current directory) is in the PATH, which is generally considered to be a pretty bad idea.
If the command contains a /, no look-up in the PATH is performed. Instead an exact match is required. If starting with a / it is an absolute path, and the file must exist. Otherwise it is a relative path, and the file must exist relative to the current working directory.
So, you have two acceptable options:
Put your script in some directory that is on your PATH. Alternatively, add the directory containing the script to the PATH variable.
Use an absolute or relative path to invoke your script.
My colleague has created an JAVA_HOME variable somewhere but he could not remember.
I have check /etc/profile, /etc/bashrc, /root/.bash_profile, /root/.bashrc. All these files does not have a line to set JAVA_HOME, but it still keep coming back with old values.
So, is there a way to find out where is it coming from?
Here's one crude but effective way:
cd /
grep -r "JAVA_HOME" .
This will recursively search all subdirectories under the starting point (the UNIX root directory, in this example) for any file containing the string 'JAVA_HOME'.
You can use grep -r --include="*.ext" "JAVA_HOME" . if you want to restrict your searches to files having the extension .ext.
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