Hi folks,
I am wondering how to output awk back to a variable.
I am new to Unix/Linux.
I am trying to get rid of a decimal number and put the output back in a variable for further use in the script. here is how I used awk:
var=$1
echo $var |awk '{print $1 *100}' | $var
echo $var
this... (4 Replies)
I have a file which I am processing using awk to spit out the following:
export CLIENT=1 ; export USER=1 ; export METABASE=1 ; export TASK=1 ; export TOTAL=3
What i want to do now is execute that within the script so those variables are available to other commands. I've tried piping the... (3 Replies)
Hi,
I need to get the pid of a process and have to store the pid in a variable and i want to use this value(pid) of the variable for some process. Please can anyone tell me how to get the pid of a process and store it in a variable. please help me on this.
Thanks in advance,
Amudha (7 Replies)
Hi
I am trying to store the output of awk into a variable in a shell script. I can run it successfully from the command line but not from a ksh shell script.
ls -al test.txt | grep -v grep | awk '{print $1}'
returns -rw-r--r--
#!/bin/ksh
perm=$(`ls -al test.txt | grep -v grep | awk... (2 Replies)
Not sure why it is not working the following :
set -- $@
stype ="a"
for shell_args in "$@"
do
$stype=` awk '{print substr ("'"$shell_args"'", 0, 3)}' `
echo $stype
done
Thank you (5 Replies)
Hi all,
Hope someone can help me out here.
I have this BASH script (see below)
My problem lies with the variable path.
The output of the command find will give me several fields. The 9th field is the path. I want to captured that and the I want to filter this to a specific level.
The... (6 Replies)
hi i want to find the size of a folder and assign it to a variable and then compare if it is greater than 1 gb.
i am doin this script, but it is throwing error....
#!/bin/ksh
cd . | du -s | size = awk '{print $1}'
if size >= 112000
then
echo size high
fi
ERROR : (4 Replies)
Hi Experts,
I am trying to get system output to capture inside awk , but not working:
Please advise if this is possible :
I am trying something like this but not working, the output is coming wrong:
echo "" | awk '{d=system ("date") ; print "Current date is:" , d }'
Thanks, (5 Replies)
I am reading an xml file with date tag as <Date>Default</Date> using the below command.
Dt=$(awk -F'' '/<Date>/{print $3}' /home/test/try.xml
and getting the value from the xml file stored in this variable "Dt"
echo $Dt gives me a value. Dt=Default.
Now according to my requirement, If... (2 Replies)
Hi,
I am trying to use variable output in awk to append a string to a word in a line. But that is not happening. Could you please help me on this.
YouTube Video Tutorial: How to Use Code Tags and Format Posts @UNIX.com
The below is the code
#!/bin/ksh
set -x
src=/users/oracle/Temp... (2 Replies)
Discussion started by: pvmanikandan
2 Replies
LEARN ABOUT LINUX
pldd
PLDD(1) Linux User Manual PLDD(1)NAME
pldd - display dynamic shared objects linked into a process
SYNOPSIS
pldd PID
pldd OPTION
DESCRIPTION
The pldd command displays a list of the dynamic shared objects that are linked into the process with the specified process ID. The list
includes the libraries that have been dynamically loaded using dlopen(3).
OPTIONS
-?, --help
Display program help message.
--usage
Display a short usage message.
-V, --version
Display the program version.
VERSIONS
pldd is available since glibc 2.15.
CONFORMING TO
The pldd command is not specified by POSIX.1. Some other systems have a similar command.
EXIT STATUS
On success, pldd exits with the status 0. If the specified process does not exist, the user does not have permission to access its dynamic
shared object list, or no command-line arguments are supplied, pldd exists with a status of 1. If given an invalid option, it exits with
the status 64.
EXAMPLE
$ echo $$ # Display PID of shell
1143
$ pldd $$ # Display DSOs linked into the shell
1143: /usr/bin/bash
linux-vdso.so.1
/lib64/libtinfo.so.5
/lib64/libdl.so.2
/lib64/libc.so.6
/lib64/ld-linux-x86-64.so.2
/lib64/libnss_files.so.2
NOTES
The command
lsof -p PID
also shows output that includes the dynamic shared objects that are linked into a process.
SEE ALSO ldd(1), lsof(1), dlopen(3), ld.so(8)GNU 2014-09-27 PLDD(1)