Rename files matching a list

catkin · 07-15-2009, 03:15 PM

Quote:

Originally Posted by Qu3ry

1.new year.mp3
2.linux.mp3
3.hello.mp3

etc...

Sorry, I'm still not clear exactly what you want to do here. You want to rename which *.mp3 files? The "1-01 track.mp3" etc ones as "new year 01 track.mp3" etc?

Qu3ry · 07-15-2009, 03:53 PM

Quote:

Originally Posted by catkin

Sorry, I'm still not clear exactly what you want to do here. You want to rename which *.mp3 files? The "1-01 track.mp3" etc ones as "new year 01 track.mp3" etc?

1-01 track.mp3 is one of the split files of 1.new year.mp3.

first to merge, then to rename them.

catkin · 07-15-2009, 04:25 PM

Quote:

Originally Posted by Qu3ry

1-01 track.mp3 is one of the split files of 1.new year.mp3.

first to merge, then to rename them.

How do you merge them?

Qu3ry · 07-16-2009, 04:02 PM

cat 01-*.mp3 > 01.mp3
cat 02-*.mp3 > 02.mp3

etc..

Writing a for loop to automate the task. The problem is that you have to know number of tracks to be merged in each directory.

Code:

i = 01
j = 32 # number of tracks to be merged

for i <= j ; do 

   cat "$i"-*.mp3 > "$i".mp3; 
   i = $(i + 1) 

done

ntubski · 07-16-2009, 08:50 PM

What language are those for loops in? I'm pretty that is not valid bash.

This should work, assuming the song number is always separated by a "-" and is the first thing in the file name. Also it might screw up if there are other files in the directory.

Code:

for song in $(ls *-*.mp3 | cut -d- -f1 | sort -u) ; do
   cat "$song"-*.mp3 > "$song".mp3
done

Qu3ry · 07-17-2009, 01:55 PM

ntubski, your script should work. However, I found a new pattern in other directories.. Some songs aren't grouped together by their number prefixes. More precisely they are grouped by their names.. for instance..

01-1.song.mp3
01-2.song.mp3
01-3.pop.mp3
01-4.pop.mp3

There are two songs in those 4 files. How to change the script accordingly?

catkin · 07-17-2009, 03:23 PM

Hello Qu3ry

Quote:

Originally Posted by Qu3ry

cat 01-*.mp3 > 01.mp3

Code:

i = 01
j = 32 # number of tracks to be merged

for i <= j ; do 

   cat "$i"-*.mp3 > "$i".mp3; 
   i = $(i + 1) 

done

Bash requires no spaces either side of the = command in an assignment. Thus i = 01 does not assign 01 to i. Instead it tries to run the command i with operands = and 01.

Bash interprets integer constants beginning with 0 as octal. Doesn't make any difference for 01 to 07 but 09 would be invalid.

The syntax of the "for" statement is either

Code:

for name [in words ...]

or

Code:

for (( expr1 ; expr2 ; expr3 ))

It looks as if you wanted

Code:

while test-commands

The <= comparison operator is for strings unless you are doing shell arithmetic in a let <arithmetic expression>, ((<arithmetic expression>)) or $((<arithmetic expression>)). The bash test numeric comparison operator is -le.

In cat "$i"-*.mp3 > "$i".mp3; the ";" is OK but not necessary because the line end serves the same purpose.

i = $(i + 1) will not do what you want, first because of the spaces either side of the "=" and second because $(i + 1) will try to run command i with operands + and 1. i=((i + 1)) would do what you want because bash evaluates expressions within (( )) as arithmetic and substitutes their value. Assuming $i is 5 then i=((i + 1)) becomes i=6

This works:

Code:

for (( i=1; i<=32; i++ ))
do
    cat "$i"-*.mp3 > "$i".mp3
done

BTW, I'm amazed and educated to learn that you can cat *.mp3 files together and play the result. Please confirm you have tested this.

Best

Charles

ntubski · 07-17-2009, 03:47 PM

Quote:

Originally Posted by Qu3ry

01-1.song.mp3
01-2.song.mp3
01-3.pop.mp3
01-4.pop.mp3

There are two songs in those 4 files. How to change the script accordingly?

changes in red.

Code:

for song in $(ls *-*.*.mp3 | cut -d. -f2 | sort -u) ; do
   cat *."$song".mp3 > "$song".mp3
done

Quote:

Originally Posted by catkin

BTW, I'm amazed and educated to learn that you can cat *.mp3 files together and play the result.

You know, now that you mention it, it does seem surprising.

Qu3ry · 07-17-2009, 04:15 PM

I have been using the cat command to merge avi and mp3 files for awhile with no problem.

For further reference: an article on looping filenames with spaces:

http://www.cyberciti.biz/tips/handli...s-in-bash.html

catkin · 07-17-2009, 04:51 PM

Hello Qu3ry

Quote:

Originally Posted by Qu3ry

I have been using the cat command to merge avi and mp3 files for awhile.

For further reference: an article on looping filenames with spaces:

http://www.cyberciti.biz/tips/handli...s-in-bash.html

Thanks for the info.

Slightly scary page on the link though ...

"By default $IFS is set to the space character" is not correct. By default it is set to space, tab and newline.

There's no need to adjust IFS for this code to work when files in the current directory contain spaces (and tabs and newlines!)

Code:

for f in *
do
  echo "$f"
done

It works just fine with the default IFS!

I'm surprised that IFS=$SAVEIFS works. What if there are current IFS characters in $SAVEIFS? Wouldn't the shell evaluate $SAVEIFS into more than one word? IFS="$SAVEIFS" is safer. Mmm ...

In IFS=$(echo -en "\n\b"), why put the backspace character in IFS? Mmm ...

while IFS=: read userName passWord userID groupID geCos homeDir userShell is perfectly correct but neither transparent nor efficient; its only merit is to save a line. Functionally this is identical

Code:

IFS=:
while read userName passWord userID groupID geCos homeDir userShell

Best

Charles

Qu3ry · 07-18-2009, 07:54 AM

Quote:

Originally Posted by ntubski

changes in red.

Code:

for song in $(ls *-*.*.mp3 | cut -d. -f2 | sort -u) ; do
   cat *."$song".mp3 > "$song".mp3
done

You know, now that you mention it, it does seem surprising.

ntubski, the new script didn't work. There is no mp3 matches the pattern to begin with. I have changed the code to (ls *-*.mp3) to match the pattern and ran the code with the echo command instead of the cat command. It didn't work neither. Here is the result:

*.mp3.mp3 to mp3.mp3

where "to" is the replacement of ">"

Qu3ry · 07-18-2009, 08:24 AM

catkin, I find using files=($(ls *.mp3)) as an array assignment not able to deal with filenames with spaces. Changing it to files=( *.mp3 ) can solve the problem.

Code:

files=( *.mp3 )
i=0
cat test.txt | while read j
do
    mv "${files[ $i ]}" "$j".mp3 
    let i++
done

BTW, for those of you who are lucky to have id3 tag attached to their mp3es, there is a program to rename mp3es according to their id3 tag and vice versa.

http://pwp.netcabo.pt/paol/tagtool/

catkin · 07-18-2009, 09:22 AM

Hello Qu3ry

Quote:

Originally Posted by Qu3ry

catkin, I find using files=($(ls *.mp3)) as an array assignment not able to deal with filenames with spaces. Changing it to files=( *.mp3 ) can solve the problem.

Thanks for the correction and sorry for the mistake.

Best

Charles

ntubski · 07-18-2009, 12:15 PM

Quote:

Originally Posted by catkin

while IFS=: read userName passWord userID groupID geCos homeDir userShell is perfectly correct but neither transparent nor efficient; its only merit is to save a line. Functionally this is identical

Code:

IFS=:
while read userName passWord userID groupID geCos homeDir userShell

I agree with all your other criticisms, but using while IFS=: read userName passWord userID groupID geCos homeDir userShell does have the advantage that IFS gets reset back to previous value after read command finishes.

Quote:

Originally Posted by Qu3ry

ntubski, the new script didn't work. There is no mp3 matches the pattern to begin with. I have changed the code to (ls *-*.mp3) to match the pattern and ran the code with the echo command instead of the cat command. It didn't work neither. Here is the result:

*.mp3.mp3 to mp3.mp3

where "to" is the replacement of ">"

It works for me:

Code:

~/tmp/songs$ ls
01-1.song.mp3  01-2.song.mp3  01-3.pop.mp3  01-4.pop.mp3
~/tmp/songs$ for song in $(ls *-*.*.mp3 | cut -d. -f2 | sort -u) ; do
> echo *."$song".mp3 to "$song".mp3
> done
01-3.pop.mp3 01-4.pop.mp3 to pop.mp3
01-1.song.mp3 01-2.song.mp3 to song.mp3

catkin · 07-18-2009, 12:22 PM

Quote:

Originally Posted by ntubski

I agree with all your other criticisms, but using while IFS=: read userName passWord userID groupID geCos homeDir userShell does have the advantage that IFS gets reset back to previous value after read command finishes.

Thanks

Interesting! How does that work? AIUI, it would only happen if "while" set up a subshell -- but, if that's the case, how are any variables set in the loop available in the parent shell?