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I've got this directory full of images, and I can do this:

echo *.jpg image1.jpg image2.jpg image3.jpg # and so on

How can I get the output in a plain text file in this format?

image1.jpg image2.jpg image3.jpg
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2 Answers 2

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Avoid using ls, bash globs can do it better

printf '%s\n' *.jpg >output_file
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  • Why is bash globbing better than using ls? Commented Jan 26, 2012 at 8:18
  • If you just want to print the list, ls is good. But if you want to process the files, glob is better: for file in * will work for files with spaces in their names, while ls | while read file will not. Commented Jan 26, 2012 at 8:58
  • @choroba. I am completely in favour of using bash globbing here, but you can easily get ls to handle multiple spaces in a filename by using IFS=$'\n'. Yes, it is more of a hassle, but it isn't a show-stopper for ls. The main restriction with output from ls is that it can't easily handle \n in filenames other than via the -b option (print octal escapes for nongraphic characters) which makes it rather impractical when compared to using globbing... Come to think of it: when a filename does have \n in it, then perhaps ls -b is the easiset way to hanlde it in an output file. Commented Jan 26, 2012 at 11:01
  • Ok, ls writes non-printable characters 'as is' when the output is sent to a pipe (I was misled from the terminal output). However, ls is a further process to start, and I try to avoid it whenever possible. Commented Jan 26, 2012 at 11:28
  • Can this be done recursively? Commented Jan 26, 2012 at 16:35
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I'd prefer find over ls. With find you can specify a lot of options, such as directory depth (min and max), access/modified time, use regular expressions and a lot more. Have a look at the find manpage.

runejuhl@lapaz:~/Pictures$ find . -iname '*.jpg'|head ./2011/03/09/CIMG9447.JPG ./2011/03/09/CIMG9445.JPG ./2011/03/09/CIMG9444.JPG ./2011/03/09/CIMG9442.JPG ./2011/03/09/CIMG9443.JPG ./2011/03/09/CIMG9440.JPG ./2011/03/09/CIMG9441.JPG ./2011/03/09/CIMG9446.JPG ./2011/06/26/CIMG9512.JPG ./2011/06/26/CIMG9585.JPG runejuhl@lapaz:~/Pictures$ find . -maxdepth 2 -iname '*.jpg' |head ./Webcam/2012-01-10-120822.jpg ./Webcam/2010-11-08-192524.jpg ./Webcam/2012-01-10-170146.jpg ./Webcam/2012-01-10-120928_4.jpg ./Webcam/2012-01-10-120835_4.jpg ./Webcam/2012-01-10-120928_3.jpg ./Webcam/2012-01-10-120828.jpg ./Webcam/2012-01-10-120928_2.jpg ./Webcam/2012-01-10-120945_1.jpg ./Webcam/2012-01-10-120945_2.jpg

To get only the filename, you can use the -printf argument:

-printf format

True; print format on the standard output, interpreting \' escapes and%' directives. Field widths and precisions can be specified as with the printf' C function. Please note that many of the fields are printed as %s rather than %d, and this may mean that flags don't work as you might expect. This also means that the-' flag does work (it forces fields to be left-aligned). Unlike -print, -printf does not add a newline at the end of the string.

(look at the man page for more information)

To print only the filename, you'd use printf with "%f":

runejuhl@lapaz:~/Pictures$ find . -maxdepth 2 -iname '*.jpg' -printf '%f\n' |head 2012-01-10-120822.jpg 2010-11-08-192524.jpg 2012-01-10-170146.jpg 2012-01-10-120928_4.jpg 2012-01-10-120835_4.jpg 2012-01-10-120928_3.jpg 2012-01-10-120828.jpg 2012-01-10-120928_2.jpg 2012-01-10-120945_1.jpg 2012-01-10-120945_2.jpg

If using xargs, and passing the filename in this way, it might be a good idea to use "%f\0" and xargs -0 to use null delimiters.

When you feel like it, have a look at the -print0 argument and a real good look at xargs or parallel

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  • Is it possible to only retrieve and save the names of the files minus the directory? Commented Jan 27, 2012 at 7:38
  • Yes, no problem. You can use the <code>-type</code> argument to specify which types. (I just read the question again, this isn't what you're looking for. I'll update my reply later today.) Commented Jan 28, 2012 at 9:42

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