how to format a list of files from ls?

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10-20-2009

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how to format a list of files from ls?

Hi,

I have am looking for advice on how to format a directory listing... Basically, I have a list of speech voices at /System/Library/speech/voices

This directory looks like:
Agnes.SpeechVoice
Albert.SpeechVoice
Alex.SpeechVoice
.. etc

I would like to print these out by splitting by "." and just displaying the first part and order them by number like:

1. Agnes
2. Albert
3. Alex
.. etc.

How is the best way to do this?

-patrick

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10-20-2009

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bash

Code:

# i=0;for file in *SpeechVoice; do i=$(( i+1 ));echo "$i ${file%%.*}"; done
1 Agnes
2 Albert
3 Alex

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10-20-2009

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Code:

ls | sort | cut -d. -f1

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10-20-2009

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you can do that with a simple command..

Code:

for dir_name in `ls *.SpeechVoice` ; do
basename  $dir_name .SpeechVoice
done

Last edited by vidyadhar85; 10-21-2009 at 12:55 AM..

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10-21-2009

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that's not quite there...yet at least. There's no option for wildcard expansion for basename the last time i checked. (if there is, then pardon me for my ignorance)

Code:

# basename  Agnes.SpeechVoice  .SpeechVoice
Agnes
# basename  *SpeechVoice  .SpeechVoice
basename: extra operand `.SpeechVoice'
Try `basename --help' for more information.
# basename  *.SpeechVoice  .SpeechVoice
basename: extra operand `.SpeechVoice'
Try `basename --help' for more information.
# basename  *.SpeechVoice
Agnes.SpeechVoice

---------- Post updated at 11:34 PM ---------- Previous update was at 10:43 PM ----------

Quote:

Originally Posted by vidyadhar85

you can do that with a simple command..

Code:

for dir_name in `ls *.SpeechVoice` ; do
basename  $dir_name .SpeechVoice
done

another way without the ls

Code:

for name in *.SpeechVoice
do
 .....
done

Last edited by ghostdog74; 10-21-2009 at 01:33 AM..

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10-21-2009

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Quote:

Originally Posted by vidyadhar85

you can do that with a simple command..

Code:

for dir_name in `ls *.SpeechVoice` ; do

Not only is ls unnecessary, but it will break the script if any filenames contain spaces. Use wildcard expansion directly.

Quote:

Code:

basename  $dir_name .SpeechVoice
done

And there's no need for the external command, basename.
Code:
for dir_name in *.SpeechVoice
do
  printf "%s\n" "${dir_name%.SpeechVoice}"
done
In bash or ksh93, it can be done much faster using an array:
Code:
x=( *.SpeechVoice )
printf "%s\n"  "${x[@]%.*}"

Last edited by cfajohnson; 10-23-2009 at 12:50 AM..

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10-23-2009

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Code:

awk -F"." '{printf("%s %s\n",NR,$1)}'

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