scripting bash for loop awk output in argument

Question

im writing some bash exercises, this in particular wants to check all the files named file-a(sonenumber) and move it to directory named dirA But there's some problem with for loop i can do that without for loop just simply writing

mv $(ls -la | awk '$9 - /-a[0-9]+/ {print $9}') dirA

but i wanna do it with a for and i cant do that there's the script

FILES=(ls -la | awk '$9 - /-a[0-9]+/ {print $9}')

for arg in "$FILES"
do
       mv $arg dirA
done

Answer 1

No need a for loop, just a shell glob:

mv -- *-a[0-9]* dirA

If you insist for a loop:

files=( *-a[0-9]* )

for f in "${files[@]}"; do
    mv "$f" dirA/
done

or

for f in *-a[0-9]*; do
    mv "$f" dirA/
done

But please, avoid parsing ls output.

ls is a tool for interactively looking at directory metadata. Any attempts at parsing ls output with code are broken. Globs are much more simple AND correct: for file in *.txt. Read http://mywiki.wooledge.org/ParsingLs

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Don't use UPPER case variables

Answer 2

You're not setting FILES to the output of the command. For that, you need to use $(command). To make it an array, wrap another set of () around that.

Then to loop over the array contents, use "${FILES[@]}.

I'm not sure why you're using ls -la when you could just use ls -a. Then each filename will be a single line, you don't have to extract $9.

FILES=($(ls -a | awk '/-a[0-9]+/'))

for arg in "${FILES[@]}"
do
       mv "$arg" dirA
done

However, it's best not to parse the output of ls.

Answer 3

I modified your code to work in the for loop.

FILES=$(ls -la | awk  '/-a[0-9]+/ {print $9}') 
for arg in $FILES 
do 
  mv $arg dirA
done