I want to print certain columns only from ps output that is PID, PPID, command, memory utilization and CPU utilization columns.
when I run ps command I get the following output.
Now I only want some columns from this output so I use -o flag as mentioned in this tutorial.
But I am getting this error.
I don't understand where is the problem. I have also tried usin --help and it is not showing -o flag. So I am confused here.
I am using the windows operating system. And using Git Bash terminal to run all these Linux commands.
Git Bash is a terminal for Windows that emulates the Linux bash (shell) functionality. It is not 100% compatible to a "real" bash shell. As you've empirically seen, its ps executable doesn't support all the flags you're used to from Linux. The --help option will show you what flags are supported.
Hello
Maybe put 2 things together, ps and grep? Then try this...
ps | grep -o -E "^[ 0-9]{1,9}"
...and is this working on your system?
( The Space in [ ] is important )
Related
I am trying to locate a file used by a binary file during its execution. Using strace helps but its way too convoluted, macroed with grep is good enough, but does there exist an utility which can help me dump only files used by a binary?
you can try using:
lsof -p PID of the running process
lsof -c ssh would show all files opened by processes starting with the letter
Or try ltrace or maybe fuser
I've seen strace be used with some complex grep piping.. but it all depends on what exactly the end goal is.
You can also utilize the -e options in strace to filter, example is:
sudo strace -t -e trace=open,close,read,getdents,write,connect,accept whoami >/dev/null
and grep from there..
the -L flag provided in kill does not work in zsh.
When I run the command kill -L Using zsh the result is:
kill: unknown signal: SIGL
kill: type kill -l for a list of signals
Running kill -L Using bash gives the list of signal names as expected.
-L, --table
List signal names in a nice table.
Please help me understand why this inconsistency, and can it be "fixed"?
kill is a shell builtin for both zsh and bash, with different implementations and options on each. The zsh builtin does support the POSIX -l option for listing signals, but not the GNU -L extension.
You can always use /bin/kill to run the freestanding program version if you desire. On OSes with a GNU runtime, that'll also support -L.
As per project requirement, i need to check the content of zip file generated which been generated on remote machine.This entire activity is done using automation framework suites. which has been written in shell scripts. I am performing above activity using ssh command abd execute unzip command with -l and -q switches. But this command is getting failed. and shows below error messages.
[SOMEUSER#MACHINE IP Function]$ ./TESTS.sh
ssh SOMEUSER#MACHINE IP unzip -l -q SOME_PATH/20130409060734*.zip | grep -i XML |wc -l
unzip: cannot find or open SOME_PATH/20130409060734*.zip, SOME_PATH/20130409060734*.zip.zip or SOME_PATH/20130409060734*.zip.ZIP.
No zipfiles found.
0
the same command i had written manually but that works properly. I really have no idea.Why this is getting failed whenever i executed via shell scripts.
[SOMEUSER#MACHINE IP Function]$ ssh SOMEUSER#MACHINE IP unzip -l -q SOME_PATH/20130409060734*.zip | grep -i XML |wc -l
2
Kindly help me to resolve that issue.
Thanks in Advance,
Priyank Shah
when you run the command from your local machine, the asterisk character is being expanded on your local machine before it is passed on to your remote ssh command. So your command is expecting to find SOME_PATH/20130409060734*.zip files on your machine and insert them into your ssh command to be passed to the other machine, whereas you (I'm assuming) mean, SOME_PATH/20130409060734*.zip files on the remote machine.
for that, precede the * character by a backslash ( \ ) and see if it helps you. In some shells escape character might be defined differently and if yours is one of them you need to find the escape character and use that one instead. Also, use quotes around the commands being passed to other server. Your command line should look something like this in my opinion:
ssh SOMEUSER#MACHINE_IP "/usr/bin/unzip -l -q SOME_PATH/20130409060734\*.zip | grep -i XML |wc -l"
Hope this helps
I am writing a portable shell script to get system process information, I need process id, command, pwdx (linux). On linux I am able to get this information as follows.. but it fails on all other unix flavours.
$ ps -awwwwwww -u <userid> -o pid,cmd|grep -i <filter_term> | egrep -v grep
$ pwdx <pid>
what I should use on AIX, HPUX and Solaris to get the similar information, or there any cross platform command
On Solaris I have tried /usr/ucb/ps but that support formatted output and lsof for pwdx equivalent but that also doesn't show what I need
On Solaris I have tried /usr/ucb/ps but that support formatted output:
What is wrong with formatted output ?
and lsof for pwdx equivalent but that also doesn't show what I need.
That doesn't make sense. pwdx is a Solaris native command and was even originally implemented on that OS.
Linux != Unix. And in the same hand, the commands are not always going to be the same, for instance GNU ps is not like Solaris ps or HP-UX ps etc. In some cases the Vendor Unix flavors offer a "compatibility binary" like those stashed in /usr/ucb on solaris. But ultimately you need to look at the man page for each version and review the output format options.
Edit. That is for in general all commands. Including grep, egrep etc.
To show the full command name, use this
ps -eo comm
This will show the command that was run. (ps is from /usr/bin on my Solaris system 5.11)
When I do
find /
on a terminal and then do on another terminal
lsof -a -d 0-2 -c fin
I see o/p listed from execution of lsof command.
But when I do
echo hi ; read -t 30 hello
hi
on the same terminal ( as find) and do (on different terminal)
lsof -a -d 0-2 -c read
I don't get any output from lsof command
Why ? Is it because read is bash built in ? Whats happening here ?
You got it right. "read" is a shell built-in. The process name remains sh (or bash, or zsh, or whatever else is your shell of choice).
Moreover, though for some shell built-ins there are binary alternatives, there isn't one for read. Really because of its syntax, it takes in the name of a shell variable that gets assigned the result of reading from the stdin. If it was an external program, it could never set the variable in the calling shell.