i'm new to shell can handle this script for me plz


 
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# 1  
Old 02-17-2010
i'm new to shell can handle this script for me plz

Write a shell script named displayargs that prints FOUR lines. The first line tells the name that was used to invoke the script, the second line tells how many parameters there were, the third line tells what the last parameter was, and the fourth line tells what the first parameter was.
For example:
Code:
  $ display args  a b c
  My name is  displayargs.
  There were 3  parameters.
  The last one  was c.
  The first one  was a.


Last edited by zxmaus; 02-18-2010 at 05:35 AM.. Reason: code tags added
# 2  
Old 02-17-2010
quick fix
Code:
first="$1"
let n=$#-1
shift $n
last="$1"

echo "
My name is $0
There were $# parameters
The last one was $last
The first one was $first
"

# 3  
Old 02-18-2010
simpler :
Code:
echo "My name is $0"
echo "There were $# parameters"
echo "The last is ${!#}"
echo "The first is $1"

# 4  
Old 02-19-2010
thank u but thats it?
# 5  
Old 02-19-2010
What do you mean "that's it"? Of course that's it, you don't need any more code to display some command line arguments.
# 6  
Old 02-19-2010
i'm confuse bcoz i got this output which is not the same output in the question

my name is ./displayargs
there were 0 parameteres
the last is ./displayargs
the first is
# 7  
Old 02-19-2010
that's because you did not give any argument.
I'm sur you are able to modify the script so that it take care of that possibility. The syntax :
Code:
if ...
then ...
else ...
fi ...

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