Find files with similar names in a folder using shell script

Question

I have a requirement of selecting xml files from a folder of same ID's. E.g. There are files with name

1. S-000001-0-0.xml
2. S-000001-0-1.xml
3. S-000001-0-2.xml
4. S-000001-0-3.xml
5. S-000001-0-4.xml
6. S-000002-0-0.xml
7. S-000003-0-0.xml

From 1 to 5 Xml belongs to one ID and 6,7 belongs to different ID. My requirement is to select files of type 1 to 5 together and move to a new folder.

Answer 1

You can use a glob like so:

mv S-000001-*.xml dir-to-move-to

Answer 2

for x in `ls S*.xml | egrep -o '[0-9]{6}' | sort | uniq`; do mkdir -p /tmp/$x; cp S-$x-* /tmp/$x/.; done

This will create a folder ID (for each ID) in /tmp and then will copy there all files with the same ID.

You can replace cp with mv if tests are ok.

I assumed that ID has 6 digits and all other numbers are smaller.

Answer 3

find /path/to/old/directory -name 'S-??????-?-?.xml' -type f -print0 | \
    xargs -r0 mv -vt /path/to/new/directory