Renaming a batch of files

[alfirdaous]

Good day to you,

I would like to rename a batch of files using command line:

00001-fqAQHelS.mp4
00002-cKIaSnn1.mp4
00003-d6lha9ys.mp4
00004-eUyAzS7x.mp4
00005-nVFSjsiW.mp4
..
00010-ffdsfselS.mp4
00011-fqAQHelS.mp4
00012-cKIaSnn1.mp4
00013-d6lha9ys.mp4
00014-eUyAzS7x.mp4
00015-nVFSjsiW.mp4

Some files have 4 zeros at the begining (0001) and some 3 (00010), then 2 (00100),.. my command line is:

N=1; for i in *.mp4; do echo $i title_$[N+1].mp4; N=$[N+1]; done

The output is something like:

title_2.mp4
title_3.mp4
..
title_10.mp4
title_11.mp4
title_12.mp4

How can I keep the same number of zeros as the first ones:

title_00002.mp4
title_00003.mp4
..
title_00010.mp4
title_00011.mp4
title_00012.mp4

so number of digits (5) will remain the same??

Thanks in advance

[steeldriver]

If you are set on using your current loop solution, you could use printf to create the new name with a zero-padded number i.e. (simple illustration)

$ n=1; for i in {1..10}; do printf -v newname "title_%05d" $((i++)); echo "$newname"; done
title_00001
title_00002
title_00003
title_00004
title_00005
title_00006
title_00007
title_00008
title_00009
title_00010

However you could just use the perl-based 'rename' command to substitute the original numbering e.g.

$ rename -vn 's/(\d+)-[^.]*/title_$1/' *.mp4
00010-ffdsfselS.mp4 renamed as title_00010.mp4
00011-fqAQHelS.mp4 renamed as title_00011.mp4
00012-cKIaSnn1.mp4 renamed as title_00012.mp4
00013-d6lha9ys.mp4 renamed as title_00013.mp4
00014-eUyAzS7x.mp4 renamed as title_00014.mp4
00015-nVFSjsiW.mp4 renamed as title_00015.mp4

[TheFu]

There is a regex rename tool already ... called rename.
rename 's/[-].*$/.mp4/g' *mp4
will remove the - to the end.

Then rename 's/^/title_/g' *mp4
will insert the beginning.

Clear as mud? BTW, I didn't test this, but I use rename ALL_THE_TIME.

Now, if you WANT to retain your method, we need to know which scripting language are you using.

I like steeldrivers answer better. I didn't remember the use of substitutions off the top of my head. Pretty slick, huh?

[papibe]

EDIT: this solves your sequence output, but not your renaming problem.

Hi alfirdaous.

You can use 'seq' to generate your number's sequence. For instance:

$ for n in $(seq 10); do echo $n; done
1
2
3
4
5
6
7
8
9
10

If you set a format you can keep all number with the same amount of digits and proper padding zeros:

$ for n in $(seq -f "%02.0f" 10); do echo $n; done
01
02
03
04
05
06
07
08
09
10

In your case you'd need something like "%05.0f"

Hope it helps. Let us know how it goes.
Regards.

[steeldriver]

Just for completeness, the built-in bash sequence generator will pad field widths as well i.e.

$ for i in {0006..0012}; do echo $i; done
0006
0007
0008
0009
0010
0011
0012

[alfirdaous]

If you are set on using your current loop solution, you could use printf to create the new name with a zero-padded number i.e. (simple illustration)

Thanks in advance, I did not get the printf way

There is a regex rename tool already ... called rename.

In this case I need to use 2 commands line

EDIT: this solves your sequence output, but not your renaming problem.

I liked the code, quite simple, but still don't understand the printf

Another thing, If I would like to increment the names, for exemple:

0001 ==> 0005
0002 ==> 0006
0003 ==> 0007

and so on

[steeldriver]

Well you basically have 3 approaches available to you

1. (as per your original post) loop over the *.mp4 files using a shell glob, and separately increment a counter that will form the basis of the new file name. In that case, if you want the number to have a particular format, you can do that using 'printf'. The bash printf builtin is usually used to write formatted output to the terminal, but the optional form 'printf -v var ... ' allows you to write it to shell variable var, which you can use in your script. Type 'help printf' at the bash prompt for more info.
2. you can create a suitable numeric sequence in the desired (zero padded) format using either 'seq -f' or like 'for n in {00000...00199}', and then look for a file with that number in its name, and rename it
3. you can use rename (or sed) to create the new name from the old name by substitution

[Vaphell]

printf is a classic command/function used in many languages. Format string has % placeholders in which values of provided parameters are put. %05d is a placeholder that means 0-padded 5 digits wide decimal number, so eg

i=5; printf "filename-%05d" $i

will produce "filename-00005" string. optionally -v variable stores the result in variable

and if you want to shift numbers, rename is out of question - regular expressions rock but they don't know concept of math (it's all characters)
Assuming transition from 99999-garbage.mp4 to title_99999.mp4, you'd need something like

for f in *.mp4; do n=${f%%-*}; n=$(( 10#$n+4 )); mv "$f" "$( printf "title_%05d.mp4" "$n" )"; done

n=${f%%-*} = strip rightside '-'+anything from filename, leaving the digits
n=$(( 10#$n+4 )) = by default leading 0s suggest octal number, so 10# makes sure it is considered decimal, add 4.

but be warned - if you want to change numbers by adding N but preserve the filename format you need to be careful. Going in ascending order means overwriting files with higher indices ( file-0001 becomes file-0005 overwriting old file-0005 before it gets processed), you need to go in descending order or use temporary format to avoid naming collisions

[GrouchyGaijin]

This is pretty cool, although I don't understand how it works.
For a long time I've been wanting to be able to batch rename photos from the command line.

If I have a bunch of photos with names of various lengths and ending in .jpg or .JPG,
how could I rename the entire batch to something like September-19-2013-1.jpg September-19-2013-2.jpg etc?

[Vaphell]

does order or any part of original filenames matter?
if not you simply do something like

i=1; for f in *.[jJ][pP][gG]; do mv -- "$f" "$( printf "September-19-2013-%04d.jpg" $i )"; ((i++)); done

printf is only for 0-padding so if you dont need it you can simply use plain "September-19-2013-$i.jpg" instead
how it works: take all files matching given glob and rename them in loop to format with $counter plugged in, increment $counter after each file

$ ls
garbage.jpg  something.JPG  trolololo.JPG
$ i=1; for f in *.[jJ][pP][gG]; do mv -- "$f" "$( printf "September-19-2013-%04d.jpg" $i )"; ((i++)); done
$ ls
September-19-2013-0001.jpg  September-19-2013-0002.jpg  September-19-2013-0003.jpg