Shell Scripts - Show all directories with full information ( and no files)


 
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# 1  
Shell Scripts - Show all directories with full information ( and no files)

Hello all, i'm stumped.... I need to list all directories with all there info and exclude the files, then vice versa. I am not sure if I need to string several ls commands together or how to even do that. I believe I need to do some variation of ls -l but need to figure out how to take out the files from this list and leave the directories and vice versa? I am slowly going mad. Can I pipe options together?

I am trying to write a shell script to do this. Once I know the command I need to use would I echo this into my file then run it? This stuff is not coming easy for me at all......

I appreciate any and all help. Struggling in Seattle, citizencro
# 2  
Code:
ls -ld */

# 3  
Thank you very much.....

Now how do I get it to do the reverse, showing all files with full info and no directories? Thanks for responding, citizencro
# 4  
One way ..
Code:
ls -l | grep -v ^d

# 5  
Code:
find -type f -exec ls -l {} \;

# 6  
For "full" information:
Code:
cd /start_dir
find . -type f -exec ls -lisad {} \;
find . -type d -exec ls -lisad {} \;


Last edited by methyl; 04-11-2011 at 04:37 PM..
 

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