AIX

Hello all, could somebody help..?
I have following 6 files (with white spaces in their names)

Code:

This is file
This is file1
This is file2
This is file3
This is file4
This is file5
This is file6

I tried to run the below script, and it did not give me desired ouput..

Code:

$ for i in `ls -1`
> do
> echo $i
> done

This
is
file
This
is
file1
This
is
file2
This
is
file3
This
is
file4 and so on..

Can somebody help on how to get the output as the filename itself..?

A few things: first, it's recommended to use $() instead of ``, since you can nest it.
Second, whenever the contents of a variable or a command output might contain a whitespace (which is also a parameter separator) that has to be preserved, surround it with double quotes.
Third, (not applicable in this case) whenever you list files, and the number of files might in extreme cases be very high, use something else than ls, or you'll fail on an "argument list too long" error.

Considering those:

Code:

for i in "$( ls )"
do
    echo "$i"
done

Pludi is right .

edit by bakunin: ahem... which added insight is your post supposed to transport? If you don't have anything substantial to say please don't say it here.

Thank you very much for the quick help.. that works..

Code:

while read line
do
        echo $line
done < `ls -l`

try the above script also

The requester said, there are white space in the file names.
So this will not give the desired output.

Aside from that it will make the issue even worse as it's a long listing, including the owner, group, permissions, ... , cluttering to output even more.