Shell Programming and Scripting

Hi All,

"Please read the below information carefully."

i have tried the below code for counting the number of lines present in text file ignoring blank lines

Code:

#! /bin/bash
clear
rdCount=0;
while read myline
do
  if [ -z "${myline}" ]; then
    echo "line is empty"
  else
   echo $myline
   let rdCount=$rdCount+1
  fi
done < ps5.txt
echo "COUNT=$rdCount"

and my ps5.txt contains data-

Code:

a
 
s
 
c

its giving me the correct output i.e

Code:

a
line is empty
line is empty
line is empty
s
line is empty
line is empty
line is empty
c
COUNT=3

now i have edited this file ,nw ps5.txt contains

Code:

hkj
h
h
 
 
h

when i have run the above script its giving me the wrong output i.e

Code:

hkj
h
h
 
h
COUNT=6

i guess their might b a problm of new line charecter but i dont know how to solve this.

please help me to solve the above problm


Thanks

1. Please post the output of cat -A ps5.txt
2. Or open ps5.txt in vi and check if you can see ^M at the end of each line. If so, do this in vi :%s/<ctrl+v+m>//g

Hi Balajesuri,

See the below output

Code:

[user01@rk03 exercise]$ cat -A ps5.txt
hkj^M$
h^M$
h^M$
^M$
^M$
h^M$

As I suspected, try what I mentioned in point #2 of post #2.

Alternatively, you can also use dos2unix or the following too:
sed -i 's/<ctrl+v+m>//g' inputfile

Hi balajesuri,

tried below code

Code:

:%s/<ctrl+v+m>//g

getting error "patteren not fount :<ctrl+v+m>/


can you please tell me in details

Do not type <ctrl+v+m> literally. I meant, press "ctrl, v, m" keys together so that you get ^M.

dos2unix not working?

--ahamed