UNIX for Dummies Questions & Answers

HI,

I require only the file name and the user information from the ls output into a file. The layout of the file is already defined as:
User id is in the first 7 bytes
FName is in the next 50 bytes

Assuming that user id is less than or = 7 bytes and
File name is less than or = 50 bytes

Can the above be achieved?

Thanks
Meghna

Code:

ls -al | awk {'print $3,$9'} | sed '1,3d' > file

Sed was included to remove the first two lines ("." and ".."). If you want "." and ".." to be included, remove "sed '1,3d'".

Thanks sun2ecliptic.

But:
ls -al | awk {'print $3,$9'} | sed '1,3d' > file
will output the user & file name information into the file. How can I format this information such that the user id is in the first 7 bytes and file name is in the next 50 bytes. Can I do this simultaneously in the awk command or do I have to manually do this and create a file 2.

Also, I am sorry but I am not clear what you mean by:
first two lines ("." and "..").


Thanks

embedded pipes to highlight field terminations:

Code:

 ls -l |while read a b c d e f g h i ;do printf "%7.7s|%50.50s|\n" $c $i ;done

you could also force left justified:

Code:

ls -l |while read a b c d e f g h i ;do printf "%-7.7s|%-50.50s|\n" $c $i ;done

Code:

ls -al | awk '{print substr($3,0,7) " " substr($9,0,50)}' | sed '1,3d'

In this example, awk uses the substr function to print the first 7 characters of the username field, include whitespace as a field separator, and the first 50 characters of the directory/filename field.

"." and ".." represent the current working directory and the parent directory. They do not represent any files within the directory you are looking for, just the current and parent directories themselves (of the listing).

Thanks a lot sun2ecliptic and curleb