bash: variable is a filename not a list

ezekieldas · 02-15-2014, 11:46 AM

I have a list of three filenames:

Code:

this_is_a_filename
this
this is a filename

If i do:

Code:

for i in `cat /tmp/l`; do echo $i; done

I get:

Code:

this_is_a_filename
this
this
is
a
filename

Which is 6 elements and I want 3. How can I preserve the name of the file "this is a filename" ? It seems bash is interpreting this filename as a list.

dwhitney67 · 02-15-2014, 12:42 PM

There's probably several approaches that can be used to resolve the "space in file name" issue. Here's one simple solution (but perhaps not the best):

Code:

IFS=$'\n'
for i in `ls`; do echo $i; done
unset IFS

Edit:

I just realized I slapped an 'ls' in the example above. I should have copied/pasted the OP's code. Doh!

ezekieldas · 02-15-2014, 12:58 PM

Thanks! I think this will do it!

NevemTeve · 02-15-2014, 12:59 PM

Code:

for i in `cat /tmp/l`; do echo $i; done

This method is over-used by beginners, but it has serious flaws. Try this:

Code:

while read i; do echo "$i"; done </tmp/l

Firerat · 02-15-2014, 01:07 PM

Code:

while read fn;do
    echo "${fn}"
done </tmp/1

don't use `ls` in a for loop

Code:

for fn in *;do
    echo "${fn}"
done

http://mywiki.wooledge.org/BashGuide/Patterns