I want to assign the result of an expression (i.e., the output from a command) to a variable and then manipulate it – for example, concatenate it with a string, then echo it.  Here's what I've got:

cd ~/Desktop;
thefile= ls -t -U | grep -m 1 "Screen Shot";
echo "Most recent screenshot is: "$thefile;

But that outputs:

Screen Shot 2011-07-03 at 1.55.43 PM.png
Most recent screenshot is: 

So, it looks like that isn't getting assigned to $thefile, and is being printed as it's executed.

What am I missing?


A shell assignment is a single word, with no space after the equal sign. So what you wrote assigns an empty value to thefile; furthermore, since the assignment is grouped with a command, it makes thefile an environment variable and the assignment is local to that particular command, i.e. only the call to ls sees the assigned value.

You want to capture the output of a command, so you need to use command substitution:

thefile=$(ls -t -U | grep -m 1 "Screen Shot")

(Some literature shows an alternate syntax thefile=`ls …` ; the backquote syntax is equivalent to the dollar-parentheses syntax except that quoting inside backquotes is weird sometimes, so just use $(…).)

Other remarks about your script:

  • Combining -t (sort by time) with -U (don't sort with GNU ls) doesn't make sense; just use -t.

  • Rather than using grep to match screenshots, it's clearer to pass a wildcard to ls and use head to capture the first file:

      thefile=$(ls -td -- *"Screen Shot"* | head -n 1)
  • It's generally a bad idea to parse the output of ls. This could fail quite badly if you have file names with nonprintable characters. However, sorting files by date is difficult without ls, so it's an acceptable solution if you know you won't have unprintable characters or backslashes in file names.

  • Always use double quotes around variable substitutions, i.e. here write

      echo "Most recent screenshot is: $thefile"

    Without double quotes, the value of the variable is reexpanded, which will cause trouble if it contains whitespace or other special characters.

  • You don't need semicolons at the end of a line. They're redundant but harmless.

  • In a shell script, it's often a good idea to include set -e. This tells the shell to exit if any command fails (by returning a nonzero status).

If you have GNU find and sort (in particular if you're running non-embedded Linux or Cygwin), there's another approach to finding the most recent file: have find list the files and their dates, and use sort and read (here assuming bash or zsh for -d '' to read a NUL-delimited record) to extract the youngest file.

IFS=/ read -rd '' ignored thefile < <(
  find -maxdepth 1 -type f -name "*Screen Shot*" -printf "%T@/%p\0" |
    sort -rnz)

If you're willing to write this script in zsh instead of bash, there's a much easier way to catch the newest file, because zsh has glob qualifiers that permit wildcard matches not only on names but also on file metadata. The (om[1]) part after the pattern is the glob qualifiers; om sorts matches by increasing age (i.e. by modification time, newest first) and [1] extracts the first match only. The whole match needs to be in parentheses because it's technically an array, since globbing returns a list of files, even if the [1] means that in this particular case the list contains (at most) one file.

set -e
cd ~/Desktop
thefile=(*"Screen Shot"*(om[1]))
print -r "Most recent screenshot is: $thefile"
  • 8
    Wow! More info than I possibly could have hoped for. Thank you many times for your answer; I really appreciate all the effort you put into this! You've just shown me that I really, really have a lot to learn. I'm off to buy a book on bash :P. – Nathan G. Jul 4 '11 at 0:10
  • 3
    That's what I would call the definitive answer! :) – alex Jul 4 '11 at 6:19

If you want to do it with multiline/multiple command/s then you can do this:

output=$( bash <<EOF
#multiline/multiple command/s


#multiline/multiple command/s


output="$( bash <<EOF
echo first
echo second
echo third
echo "$output"


  • This is a nice answer. is there a way that I can specify a variable as part of the command? for example output=$(echo $someVariable) – Zach Smith Apr 4 '19 at 12:32

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