CODEWITHSUNDEEP
bashshellglob
Panfeng Li
Panfeng Li

Reputation: 3616

Shell, copy files with similar names

I would like to copy a series of similar files from the current directory to the target directory, the files under the current directory are:

prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0001_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0001_uz.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0002_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0002_uz.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0003_ux.hst
prod07_sim0500-W31-0.2_velocity-models-2D_t80_f0003_uz.hst

Where sim is from sim0001 to sim0500 and f is from f0001 to f0009. I only need f0002, f0005 and f0008. I write the following code:

target_dir="projects/data"

for i in {0001..0500}; do
    for s in f000{2,5,8}; do
        files="[*]$i[*]$s[*]"
        cp $files target_dir
    done
done

I am very new to Shell, and wondering how to write the $files="[*]$i[*]$s[*]"$, so that it could match only the f0002, f0005 and f0008. The reason why I also use for i in {0001..0500}; do is that the files are too large and I would like to make sure I could access some completed ones (for example, including all sim0001) in the beginning.

Edit: changed for s in f0002 f0005 f0008; do to f000{2,5,8}.

Upvotes: 1

Views: 200

Answers (1)

l0b0
l0b0

Reputation: 58808

What you need is globbing and a bit different quoting:

cp *"$i"*"$s"* "$target_dir"

Not storing this in a variable is intentional - it's faster and it's safe. If you end up with such a large list of files that you start running into system limits you'll have to look into xargs.

Upvotes: 3

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