Leading blanks

laki47 · 05-12-2010, 01:55 AM

Can you please help me with the following:

# num_count=`ls /tmp/countdir |wc -l`
# echo "Number of files in /tmp/countdir is ($num_count)"
Number of files in /tmp/countdir is ( 5)

Why do I have 7 leading blanks before number 5?

How could I avoid that?

Thanks!

vikas027 · 05-12-2010, 02:01 AM

Quote:

Originally Posted by laki47

Can you please help me with the following:

# num_count=`ls /tmp/countdir |wc -l`
# echo "Number of files in /tmp/countdir is ($num_count)"
Number of files in /tmp/countdir is ( 5)

Why do I have 7 leading blanks before number 5?

How could I avoid that?

Thanks!

Use this code

Code:

num_count=`ls /tmp/countdir |wc -l`
num_count1=`echo $num_count | sed -e 's/^[ \t]*//'`
echo "Number of files in /tmp/countdir is ($num_count1)"

laki47 · 05-12-2010, 02:04 AM

Quote:

Originally Posted by laki47

Can you please help me with the following:

# num_count=`ls /tmp/countdir |wc -l`
# echo "Number of files in /tmp/countdir is ($num_count)"
Number of files in /tmp/countdir is ( 5)

Why do I have 7 leading blanks before number 5?

How could I avoid that?

Thanks!

Can I use (KORN SHELL):

typeset -i num_count
...

catkin · 05-12-2010, 03:08 AM

Quote:

Originally Posted by laki47

Why do I have 7 leading blanks before number 5?

How could I avoid that?

Use my system! I could not reproduce the problem:

Code:

c@CW8:~$ num_count=`ls /tmp | wc -l`
c@CW8:~$ echo "Number of files in /tmp is ($num_count)"
Number of files in /tmp is (4)
c@CW8:~$ ls /tmp | wc -l
4

What is the output of ls /tmp/countdir |wc -l on your system?

Which shell are you using? If ksh, how have you declared num_count or is it implicitly declared as an ordinary variable?

EDIT: in ksh, to print the attributes of variable num_count try typeset -p | grep num_count (not tested; no ksh to test on).