using command subsitition in a bash for loop

Question

I am probably making an obvious mistake but I could use a strong hint:-)

for file in *; do ls $file | var=$(ls $file | grep -Eo '.{18}$' | cut -d '.' -f1 |sed 's/.\{12\}/&./'); echo "var is $var. File is $file";done

RESULT: $var is empty. $file is not. Why is $var empty?

When I manually put my test string in $file and run as follows, $var contains what I want:

var=$(ls $file | grep -Eo '.{18}$' | cut -d '.' -f1 |sed 's/.\{12\}/&./')

So there is something about the for loop which is causing this. It seems the $var variable has a scope limitation preventing me from using in following statements.

Have a great Easter:-)

full example:

As asked, here is an example to hopefully clarify my query. The file's inventory_a_test-20190605161153.txt1 date must match the date in its file name. First, I set the file to the wrong date:

$ touch inventory_a_test-20190605161153.txt
$ ls -l --full-time inventory_a_test-20190605161153.txt
-rw-rw-r-- 1 root root 0 **2021-04-05** 15:02:05.000000000 +0200 inventory_a_test-20190605161153.txt

Now I want to write a command flow like the one below as a for loop in bash to correct the date:

$ file=inventory_a_test-20190605161153.txt
$ ls $file | grep -Eo '.{18}$' | cut -d '.' -f1 |sed 's/.\{12\}/&./'
201906051611.53
$ touch -a -m -t 201906051611.53 $file
$ ls -l --full-time inventory_a_test-20190605161153.txt
-rw-rw-r-- 1 root root 0 **2019-06-05** 16:11:53.000000000 +0200   inventory_a_test-20190605161153.txt

Answer 1

This part doesn't make sense:

ls $file | $(something)

The $(command) format means "run command and treat its output as a variable". This isn't something that can accept piped input. I think what you meant to write was:

for file in *; do 
    var=$(ls "$file" | grep -Eo '.{18}$' | cut -d '.' -f1 |sed 's/.\{12\}/&./'); 
    echo "var is $var. File is $file"
done

However, this will fail quite badly for strange file names. You really don't want to be parsing the output of ls. More importantly, the ls is absolutely not needed here! You already have the file name so the ls is adding nothing but a complication and possible source of error. This should do what you want:

for file in *; do
    timestamp=$(sed -E 's/.*(.{12})(..)\.[^.]*$/\1.\2/' <<<"$file")
    touch -a -m -t "$timestamp" -- "$file"
done

The sed will print only the last 14 characters of the file that come before the final extension (\.[^.]*$/ will match a . and then 0 or more non-. until the end of the file name), but separates the last 2 from the previous 12 with a . effectively generating your timestamp.

Answer 2

@terdon no thank you for your kind solution. For the other linux beginners out there. I had some time zone issue as well so the final result is:

for file in *; do
     timestamp=$(sed -E 's/.*(.{12})(..)\.[^.]*$/\1.\2/') <<<"$file"
     TZ=UTC-2 touch -a -m -t "$timestamp" -- "$file"
done

ps: I use this site to analyse a regex: https://regex101.com/