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write a scwipt that
a) takes exactly one argument, a directory name.
b) if the number of argument is more or less than one,print
a usage message
c) if the argument is not adirectory, print another message
d) for the given directory, print the five biggest files
and the five files that were most recently modified.
e) save the output to a file called q2output.
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Answer / sudhir
for var in `ls -1S` === -S sort the file primary key as
size of the file and -1 column o/p
do
if [[ $i -eq 5 ]]
then
break;
fi
print $var
let i=i+1;
done
Same code can be used for time and ls command option will be
ls -1t
| Is This Answer Correct ? | 3 Yes | 3 No |
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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
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