[SOLVED] Add leading zero's

GPGAgent · 06-21-2020, 01:00 PM

I have a simple bash script

Code:

#!/bin/bash
COUNTER=0;
for FF in *.MP2  
  do ((COUNTER++)); 
  FFF=Ep$COUNTER.MP2 
  echo $FFF
done;

That produces this

Code:

charlie@charlie-machine:~/MP2-MP3/Test/LATEST/ThePrisoner$ ls
05241758.MP2  05311758.MP2  06071758.MP2  06141759.MP2 .....12 FILES
charlie@charlie-machine:~/MP2-MP3/Test/LATEST/ThePrisoner$ ./ProcessEpisodes.sh 
18904 (process ID) old priority 0, new priority 19
Ep1.MP2
Ep2.MP2
Ep3.MP2
Ep4.MP2
Ep5.MP2
Ep6.MP2
Ep7.MP2
Ep8.MP2
Ep9.MP2
Ep10.MP2
Ep11.MP2
Ep12.MP2
charlie@charlie-machine:~/MP2-MP3/Test/LATEST/ThePrisoner$

I want to add a leading zero to the output

Code:

Ep01.MP2
Ep02.MP2
Ep03.MP2
Ep04.MP2
Ep05.MP2
Ep06.MP2
Ep07.MP2
Ep08.MP2
Ep09.MP2
Ep10.MP2
Ep11.MP2
Ep12.MP2

It looks like it should be easy, but ........

pan64 · 06-21-2020, 01:09 PM

yes, printf is your friend. And you can find a lot of good example on the net. Like this: https://stackoverflow.com/questions/...-loop-in-shell

shruggy · 06-21-2020, 03:19 PM

If the point of this exercise is to rename files then why not do it directly?

Code:

rename -n 'our $c++;$_=sprintf "Ep%02d.MP2",$c' *.MP2

Omit -n when you're sure it does the right thing.

GPGAgent · 06-21-2020, 03:35 PM

Quote:

Originally Posted by shruggy

If the point of this exercise is to rename files then why not do it directly?

Code:

rename -n 'our $c++;$_=sprintf "Ep%02d.MP2",$c' *.MP2

Omit -n when you're sure it does the right thing.

Good point, I might just do that, and it will work.

What I'm actually going to do is to do re-encode the MP2 files to mp3 and reduce their size by using a lower bitrate on the audio.

So I would like to do this for the output something like this

Code:

ffmpeg -i 089765.MP2 Ep01.mp3

So if Ep01.mp3 could be constructed on the fly that would be great. Maybe function I guess using printf

shruggy · 06-21-2020, 03:58 PM

Code:

#!/bin/bash
for i in *.MP2
do
  printf -vo "Ep%02u.mp3" $((++c))
  ffmpeg -i "$i" "$o"
done

teckk · 06-21-2020, 04:56 PM

To the OP. You can loop on that directory of files however you wish.

Basic examples:

Code:

files=(*.mp2)

numFiles="${#files[@]}"

#Using evil eval
for i in $(eval echo {000..$numFiles}); do
    echo "Ep"$i".mp3"
    sleep .5
done
    
#Using seq
for i in $(seq -w 000 $numFiles); do
    echo "Ep"$i".mp3"
    sleep .5
done

#printf
for i in "${!files[@]}"; do
    printf "%03d\\n" "$i"
    sleep .5
done

#The file names
for i in "${files[@]}"; do
    echo "$i"
    sleep .5
done

#The array index number
for i in "${!files[@]}"; do
    echo "$i"
    sleep .5
done

etc.

GPGAgent · 06-21-2020, 05:24 PM

Quote:

Originally Posted by shruggy

Code:

#!/bin/bash
for i in *.MP2
do
  printf -vo "Ep%02u.mp3" $((++c))
  ffmpeg -i "$i" "$o"
done

Brilliant, thanks very much, but can I ask what do the -vo switches do, and I cannot understand how $o is assigned, I'm guessing -v means variable and o of -vo menas assign it to a variable named o - am I correct.

man printf didn't help.

But once again many thanks

shruggy · 06-21-2020, 05:35 PM

Quote:

Originally Posted by GPGAgent

I'm guessing -v means variable and o of -vo menas assign it to a variable named o - am I correct.

Exactly.

Quote:

Originally Posted by GPGAgent

man printf didn't help.

Don't confuse the /usr/bin/printf command from GNU coreutils with bash builtin printf. The -v option is bash-specific. In POSIX shell that would be

Code:

#!/bin/sh
for i in *.MP2
do
  : $((c+=1))
  ffmpeg -i "$i" $(printf "Ep%02u.mp3" $c)
done

GPGAgent · 06-22-2020, 05:29 AM

shruggy and teckk

Thanks guys, your tips and and solutions really help - much appreciated!!

Problem solved