Shell，复制具有相似名称的文件

Question

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

Answer 1

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 .