I had a problem recently, where a perl script was consuming server resources. I found out it was a perl script by using "top". But it didn't give the path to the script. Nor did ps.Also, ps -ef showed that the process with that pid is /usr/bin/httpd (apache), so it must have been apache serving some perl page I suppose.Is it possible to get the path to currently running perl script, if I only know the process PID of the perl process that is running that script? If so, how? I don't need the path to the perl binary, I need the path to the perl script that binary is currently executing.
I'm the server guy which is why I don't know this, but were staging a new webserver and we use some custom perl scripts and as were moving the site over for testing, apache is blowing a perl error;Can't locate web.pm in @INC (@INC contains: /etc/perl /usr/local/lib/perl/5.10.0 /usr/local/share/perl/5.10.0 /usr/lib/perl5 /usr/share/perl5 /usr/lib/perl/5.10 /usr/share/perl/5.10 /usr/local/lib/site_perl .) at /home/mcp/htdocs/cgi-bin/newgraph.gif line 11.,Now I can copy that file local, but there are a bunch of things, so the question is how do you globally add a folder to the perl search path
my $file = 'test/mybigfile'; my $zip = Archive::Zip->new(); my $zipped = $zip->addFile($file);
[code]....
it simply takes a file (in this case, "mybigfile") and creates a ZIP file containing that file only. Now, my problem is that when I go to decompress the ZIP file, it creates the relative (or absolute, depending on case) path of the original file. To make an example:
Code:
$ pwd /tmp/perl $ ls -R
[code]....
What I need is *not* to store the path information inside of the ZIP file, so that once I decompress the ZIP file I only get the original one without having paths created.
I'm taking here about tins of directories, thousands of files. I'm looking to find a command that makes me able to move the results above to another path, and to create that path once it doesn't exist like below:
I have a program that takes a relative path as input appends it to a some path string to get the actual path.
Now all I can input is the relative path. So if I want to go one level above my input will be ../mypath.
If I know the depth of the path used internally, I can use .. as many times to go to the root directory and then give the absolute path. But suppose I do not know the depth of the directory, can I construct a relative path string such that it considers it as a relative path. One way could be to have enough .. in the path string so that I can force an absolute path for some maximum depth of path.
Is there some path string syntax that I am not aware of but can achieve this?
Experimenting with shell variables, accidentally deleted the path variable how could I return to the original path value. What kinds of problems will I have if I don't have a path variable.
I have a path c:windowsackup I need this string to be changed into /windows/back/up I used the command -bash-3.00$ echo windackup | sed 's/\//g' but the output is windbackup
prefix=user@my-server: find . -depth -type d -name .git -printf '%h�' | while read -d "" path ; do ( cd "$path" || exit $?
[code]....
How shall i go about changing the absolute path to relative path, so that /home/git/mirror/android/adb/ndk.git gets converted to /mirror/android/adb/ndk.git //echo <command> "$prefix$PWD.git" ?? - anything for relative path?
If I use "#!/usr/bin/perl" in the beginning of a perl script the script won't work if the script is at all complicated. Simple scripts like "Hello World" works.
But if I use "#!/usr/bin/perl -w" in the beginning all scripts work?
If I don't use the -w this is whats in the logs:
(2)No such file or directory: exec of '/home/test.net/html/cgi-bin/uh/meny.pl' failed
Premature end of script headers: meny.pl
When I use the -w in the script the error-log shows me this.
meny.pl: Name "main::http_path_cgi" used only once: possible typo at /home/test.net/html/cgi-bin/uh/meny.pl line 24.
I am new here and want to lern CentOS. Current I have installed CentOS 5.5 x64 and Perl 5.8.8. Now i have install Perl 5.12.1 which located to /usr/local/bin/perl. But how I can move it to /usr/bin/perl so root based on Perl 5.12.1?
I recently reinstalled the B module for perl. Now I found out it didn't just do that, it installed a whole new copy of perl into /usr/local/bin/perl. When I type 'which perl' at the terminal, it says '/usr/local/bin/perl' instead of '/usr/bin/perl'. Normally, this wouldn't bother me. But I installed the one in local/bin without threading support. The one that comes with the system already is build with threading support, which is why I want to switch back to the system perl, and possibly remove the one in /usr/local/bin.
So how do I change it so that the result of 'which perl' returns as '/usr/bin/perl'? And what do I have to do to remove /usr/local/bin/perl?
The reason I need threading support is because I'm designing a file copier that copies several chunks of a file simultaneously to speed up the copying process. Guaranteed a useful script if it works.
Java applet not loading image with relative path(e.g. images/1.jpg) but loads image with absolute path(i.e. from /root/user/images/1.jpg) . This is a problem when i want to host the applet on web server
I'm new in perl programming and linux OS. What is the difference between perl and perl-devel? What does mean devel? Iwant to install Catalyst and before install as required I have to check if make, gcc and perl-devel are installed in my system. make and gcc are installed. But I have to install perl-devel. First I searched for make, gcc and perl-devel in YAST Software Management and search did not find perl-devel. I visited the software.opensuse.org and wrote "perl-devel" and searched. The result was many similar zips with a prefix perl-devel and I can not choose one for needed perl-devel.
how to add a path to PATH variable permanently so that it remains persisent even after closing shell and rebooting the system when i added a path, to variable it remained there as long as i didn't closed the shell. but when i reopened it ,changed were undone.
I am trying to figure out how i can add the path /usr/sbin/ into the $PATH variable. I want this to be used from the normal account. I am bored settinh this manualy each time my computer starts.
After saving above changes, I enter the command: source ~/.bashrc Now if I do echo $PATH, the path shows both the old PLAY_HOME and new PLAY_HOME. This is really bad and messes up a lot of things in my project. This problem only goes away if I logout or reboot, a rather very long process. What is happening is that the old path is added to new path element and the old path includes the old path element you want to remove.
Does anyone know how to get the path with a inode number by C programming? Or can I get the absolute path without giving a "path" but a inode number by C?
like this: get_path(unsigned inode); not such this function: getcwd(".", xxx); taowuwen@gmail.com
I Installed perl-doc in ubuntu 10.04 using sudo apt-get install perl-doc,in the same way I installed doc for postgreSQL. How to use perl-doc so that I can get from perl-doc.
I want the output of this file to be in a column, not next to each ohter. I tired putting a a newline escape character in a few places, but it breaks the script. It is easy in awk, just ls -ltr | awk '{print $8 }'
I'm using my Linux (SLES 10) server as a File Server at this point. I need to set File Permissions to nested folders differently to different groups. For example:
homesharedengineering* should be read only for groupA homesharedengineeringadmin should be read & write for groupB Plus read only for groupA homesharedengineeringautocad should be read & write for groupC Plus read only for groupA
I've been using Webmin and Putty to set permissions but Putty only allows me the Default Group, it won't allow me to set several groups on the same directory. Webmin seems to allow me to add multiple groups (Webmin --> Others --> File Manager --> Info & ACL tab will provide extended abilities) but when I add multiple groups, they don't seem to take effect? I'm wondering if my setup at the 'Share' level or at the hierarchy of my folder structure (unix based) needs to be set specifically?
running Windows 7 64bit with VMware Workstation 7.01-build 227600. I have some knowledge of Linux, I have installed f12 and have updated the system as of 03/22/2010. All updates completed successfully.
1) How do I install VMTools on the f12 (after mounting the CD/DVD tool package)
2) How do I update the gcc files it says are dependencies?
Here's what I get on installation:Before you can compile modules, you need to have the following installed...
make gcc kernel headers of the running kernel
and then I am prompted for this input from the install script:
Searching for GCC... The path "" is not valid path to the gcc binary. Would you like to change it? [yes]
and this is where I get stuck. How do I get around this or satisfy the requirements for the install?
I've read one explanation saying it can allow malicious executables that are in the cwd to run in place of similarly named built-ins and core utilities.For something to wreak true havoc, wouldn't this only apply to root? And in any case, if '.' were appended to PATH instead of prepended, shouldn't that circumvent this "replacement" attack, since PATH's directories are scanned in order? (Aren't they scanned in order?)Are there any other issues besides this one?
I am making some scripts to automaticly install linux-software.gcc,kernel,binutils, etc.) I have made a command to extract the packages, but how can I check if I can re-use my already extracted source-archives?
I'm trying to simply pull a line (semi-static entry) from an xml file online with a shell script, or perl script. Basically, I've got a URL that loads an XML file. From this file I want to pull a text entry:
<blah xml code blah 'http://static.address.com/static_directory/dynamic_filename.123' blah xml code blah>
I would like to do this from a shell or perl script. I would like to parse the xml file so that it just pulls the http - .123 information and ends there, that's the only information I need. it will then run a local command on this file.
I want to compare two files in perl, I have two files file1.txt & file2.txt. if column1 on file2.txt match column1 on file1.txt then I want my result on file3.txt (column1 column2 file1.txt + column1 column2 column3 file2.txt). this problem was solved with "awk" but I want to do in perl.
