Question

7

Batch files renaming w/ filename's shift

rated 0 times [ 7] [ 0] / answers: 1 / hits: 2590 / 1 Year ago, thu, january 5, 2023, 3:18:56

I have a ton of files named this way: [name]_[phonenumber]_HH-mm-ss_dd-MM-yyyy.mp3.
How can I shift [name] and [phonenumber] to the end of file name, and put the date in the beginning in form of yyyy-MM-dd_HH-mm-ss so I get yyyy-MM-dd_HH-mm-ss_[name]_[phonenumber]?

Here is an actual file name: [Unknown]_[+74999519075]_18-01-36_17-01-2014.mp3

I've tried rename, but due to lack of regexp knowledge I didn't come up with a working solution.

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2 Years ago

answered 1 Year ago singerrupt · Accepted Answer

This should work:

for i in *mp3; do rename 's/(.+?)_(.+?)_(.+?)-(.+?)-(.+?)_(.+?)-(.+?)-(.+?).mp3/$8-$7-$6_$3-$4-$5_$1_$2.mp3/' "$i"; done

The parentheses capture patterns. The 1st captured match is $1, the 2nd $2 etc. So, the command above:

will look for everything up to the first _

.+? means match the shortest pattern possible because of the ?,

then everything up to the 2nd _ etc.

and renames accordingly.

I tested it with:

$ touch [name]_[phonenumber]_HH-mm-ss_dd-MM-yyyy.mp3

$ ls

[name]_[phonenumber]_HH-mm-ss_dd-MM-yyyy.mp3

$ for i in *mp3; do rename 's/(.+?)_(.+?)_(.+?)-(.+?)-(.+?)_(.+?)-(.+?)-(.+?).mp3/$8-$7-$6_$3-$4-$5_$1_$2.mp3/' "$i"; done

$ ls

yyyy-MM-dd_HH-mm-ss_[name]_[phonenumber].mp3