Shifting positional parameter in Linux fedora core ?
Hi I have written following shell script for display value
of positional. But in 11th and 12 field will display without
shifting command. May I know is it advance of Linux of
programming code error?
#!/usr/bin/bash
echo "Bellow is the out of ps command"
echo "`ps`"
echo "The passing Parameter i.e output of \$1,2... value
is:==> $11 "
echo "Total number of passed argument \$# is:==> $#"
echo "Passed argument names (\$*) are:==>$*"
echo "This script PID(\$$) is :=>$$"
echo "The name of executing script(\$0) is :==>$0"
echo "The Parent ID of this script(\$PPID) is:==>$PPID"
And my input to this script is
./scriptname arg1 arg2 arg3 arg4 arg5 arg6 arg7 arg8 arg9
arg10 arg11 arg12
Output Is
Bellow is the out of ps command
PID TTY TIME CMD
2892 pts/0 00:00:00 bash
3172 pts/0 00:00:00 positional_para
3173 pts/0 00:00:00 ps
The passing Parameter i.e output of $1,2... value is:==> arg11
Total number of passed argument $# is:==> 13
Passed argument names ($*) are:==>arg1 arg2 arg3 arg4 arg5
arg6 arg7 arg8 arg9 arg10 arg11 arg12 arg13
This script PID($$) is :=>3172
The name of executing script($0) is :==>./positional_parameter
The Parent ID of this script($PPID) is:==>2892
After the ps output please see the next line I.e $11 value
comes out without shifting the parameter. How is it give me
my friends
No Answer is Posted For this Question
Be the First to Post Answer
How to enable the shell to know what program should be run to interpret the script?
What are different types of shell?
What is shift command in shell script?
How do I read a .sh file?
What is bash command used for?
what is this line in the shell script do ?#!/bin/ksh
Differentiate between ‘ and ” quotes.
What are the various stages of a linux process it passes through?
In a single command how do you run the previous command in the command prompt.
What is difference between shell and bash scripting?
madhar chod unix ke 10 commands dhang se likh nahi sakta hai
Explain about debugging?