How to list only files on HP-UX B.11.11 U 9000/800

[diwa]

Hi,

One dir is having files and dirs. I want to list only files. Can any one suggest how to do this?

Thanks

[Ivan Krastev]

use:

find . -type f

regards,
ivan

[Steven E. Protter]

Shalom,

find -type f

This will show only files, not directories.

SEP

[diwa]

Hi,

Soory but i want something like that:

Suppose i am under dir1:
under this i am having files f1.txt,f2.txt,f3.txt and subdir subdir1,subdir2.

Now if i use ls -lrt it will list all the files as well as dir.

If i use ls -lrt | grep "^d" it will list only dir.

Similarly i want some command which will display f1.txt,f2.txt.f3.txt and nothing else menas no files under subdir1 and so on.

Thanks.

[V. Nyga]

Hi,

add option '-p' to your ls command, then pipe it to grep: '| grep -v /'

This exludes directories listed with '/' at the end (option -p of ls command)

HTH
Volkmar

[V. Nyga]

ls -lrt | grep -v "^d" will work too

V.

[Sandman!]

# ls -ltr | grep ^-

[Peter Nikitka · Accepted solution]

Hi,

first I want to remember, that there is more than (plain) files and directories:
sockets, named pipes, symbolic links, block and character devices, ... (*)
Excluding directories is one point, listing only plain files another.

Second: It seems necessary for me to get just filenames without additional stat information.

I want to highlight some aspects of the two solutions of Volker and Sandman (dropped options 'rt' which are meaningless here):
1) ls -p | grep -v /
- prints no additional status information
- prints anything but directories (including (*))
This is the best solution if you do not want to get plain files only.

2) ls -l | grep -v "^d"
- prints additional status information
- like 1)

3) ls -l | grep ^-
- prints additional status information
- list just plain files, which are not symlinks

Though you could use the option 'ls -F' and filter its additional markers out, this won't seperate the block+character devices.
My favourite (the 'sub' has six spaces in it):

ls -go | awk '/^-/ {for(i=1;i<7;i++) $i="";sub("^ ","");print}'

This will lead to correct results even when files containing whitespace are found.

mfG Peter

[diwa]

Hi,

Thanks a lot for your sloutions.

Special thanks to Peter.

[diwa]

Thanks a lot,

I got satisfactory answers for my query.