Linux bash script: using ls with glob and getting output in variable not working

Question

On my CentOS 7 machine, I have a folder which contains 2 files, with names like:

myapp-nightly__v3.0-SNAPSHOT__b59__lite.zip
myapp-nightly__v3.0-SNAPSHOT__b59.zip

These files are erased and generated every night, and the number after the "b" (in the example above, 59), is incremented.

I want to have a bash script file, in which I can get, in a variable, the name of the file that doesn't contain "_lite" in it (in the example above, that would be "myapp-nightly__v3.0-SNAPSHOT__b59.zip")

So, I first tried "ls" with globbing in a terminal, to see how that works:

ls myapp-nightly__v3.0-SNAPSHOT__b+([0-9]).zip

This correctly outputs "myapp-nightly__v3.0-SNAPSHOT__b59.zip".

Then I tried making a simple bash script, that runs the above command, and stores its output in a variable:

#!/bin/bash
MYAPP_KIT_FULL_PATH=$(ls myapp-nightly__v3.0-SNAPSHOT__b+([0-9]).zip)
echo "MYAPP_KIT_FULL_PATH=$MYAPP_KIT_FULL_PATH"

This gives me the following error:

./test.sh: command substitution: line 3: syntax error near unexpected token `('
./test.sh: command substitution: line 3: `ls myapp-nightly__v3.0-SNAPSHOT__b+([0-9]).zip)'

So my question is, how do I correctly execute such a "ls" command in a bash script, and store its output in a variable? Does this not work in bash because of the glob characters?

Extra remarks:

I know the error message says "line 3" but I'm pretty sure it refers to line 2 (line 3 just does an echo). Even if I completely erase line 3 from the script (so that the script only has 2 lines), it still gives the error about line 3

what do you want to do after you get those files? btw, using ls to get file in script is not recommended — Kent
– Kent, Commented Dec 13, 2018 at 10:20
after i get the filename, i want to unzip it. So i want to do "unzip $MYAPP_KIT_FULL_PATH" — Shivan Dragon
– Shivan Dragon, Commented Dec 13, 2018 at 10:23
That looks like an extended glob (wildcard) expression, but the error indicates you don't have extended globs enabled. You can add shopt -s extglob at the beginning. But also, you really shouldn't parse the output of ls. — Gordon Davisson
– Gordon Davisson, Commented Dec 13, 2018 at 16:55

Aaron · Accepted Answer · 2018-12-13 17:19:40Z

3

You could use find :

find dir -maxdepth 1 -name 'myapp-nightly__v3.0-SNAPSHOT__b*.zip' -not -name '*lite.zip'

To unzip the target file in your current directory :

find dir -maxdepth 1 -name 'myapp-nightly__v3.0-SNAPSHOT__b*.zip' -not -name '*lite.zip' -exec unzip {} \;

edited Dec 13, 2018 at 17:19

answered Dec 13, 2018 at 10:27

Aaron

24.9k2 gold badges41 silver badges61 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Shivan Dragon Over a year ago

Thanks, this works great. So instead of "dir" i specify the actual folder where these files are located.

Aaron Over a year ago

Exactly ! You can use . if you're already in the directory that contain the files.

Gordon Davisson Over a year ago

Careful; if there are subdirectories, this will search all of them for matching files as well (and try to operate on any it finds). I'd add a -maxdepth 1 restriction in there.

Gordon Davisson · Accepted Answer · 2018-12-13 17:06:47Z

You're using an extended glob (wildcard) expression, but the error indicates you don't have extended globs enabled. You can add shopt -s extglob at the beginning. But also, you really shouldn't parse the output of ls.

There are several options for doing this "right" in bash. @Aaron's already shown how to use find, so I'll give a couple of pure-bash options. First, you can just use a bare wildcard expression (no ls) and capture the result to an array:

#!/bin/bash
shopt -s extglob
myapp_kit_full_paths=(myapp-nightly__v3.0-SNAPSHOT__b+([0-9]).zip)
# Operate on just the first match:
unzip "${myapp_kit_full_paths[0]}"
# Operate on *all* matches (in case there are several):
for file in "${myapp_kit_full_paths[@]}"
    unzip "$file"
done

(BTW, note that I used a lowercase variable name here. There are a bunch of all-caps names with special meanings, and using one of those by accident can cause trouble; therefore, lower- and mixed-case variable names are safer for your own purposes.)

Another option is to just expand the glob expression directly in a for statement:

#!/bin/bash
shopt -s extglob
for myapp_kit_full_path in myapp-nightly__v3.0-SNAPSHOT__b+([0-9]).zip
    unzip "$myapp_kit_full_path"
done

Collectives™ on Stack Overflow

Linux bash script: using ls with glob and getting output in variable not working

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related