List file NOT present in other directory

Shell Programming and Scripting
1

Dear community,
I have one LOG directory with some files. What I need to do is list ONLY the files that are not present in other directory.
Using Bash programming!

LOG DIR      |     SYNC DIR
FILE1        |      FILE1
FILE2        |      FILE3
FILE3        |      OTHER FILENAME
FILE4        |
FILE5        |

In this case there are 5 files under LOG directory and 3 files under SYNC directory, so the output list I'm expecting is

FILE2
FILE4
FILE5

Could someone help me? Thanks.

2 
ls dir1 | while read FILE
do
        [ -e "dir2/$FILE" ] || echo "$FILE"
done
3

Thanks for reply Corona, but it doesn't work! :frowning:

#!/bin/bash
ls /log/*.DAT | while read FILE
do
        [ -e "/log/sync/$FILE" ] || echo "$FILE"
done

This is the LOG dir:

[root@RedHat5 ~]# ls /log
FILE1.DAT
FILE2.DAT
FILE3.DAT
FILE4.DAT
sync
[root@RedHat5 ~]# ls /log/sync
FILE2.DAT

Execution:

[root@RedHat5 ~]# ./test.sh 
/log/FILE1.DAT
/log/FILE2.DAT
/log/FILE3.DAT
/log/FILE4.DAT

FILE2.DAT should not be in the list!

4

Please read what Corona688 has posted carefully. He didn't suggest globbing.
And if you need to use globbing to perform that operation on a select list of files from the directory, try a slight modification:

printf "%s\n" /log/*.DAT | while read FILE
do
   [[ -e /log/sync/${FILE##*/} ]] || echo "$FILE"
done
1 Like
5

Sorry Elixir, what do you mean with "globbing"? :confused:
I applied exactly what Corona suggest me.

6

That's the expansion of asterisk (in this case) by the shell to filenames.
Have you tried what I've suggested in my earlier post?

7

Ah ok! You had modified the previous post and I don't see the code.
Now it working fine. BTW, I'm used asterisk because under LOG directory there is the SYNC directory, so I've to use * to take only .DAT files! :wink:

Just another question, if the file doesn't exists I need to show it and then copy to SYNC dir (so next time will not be shown). How can I add the copy command to the following line?

[[ -e /log/sync/${FILE##*/} ]] || echo "$FILE" > /tmp/DAT.tmp
8

Change

[[ -e /log/sync/${FILE##*/} ]] || echo "$FILE" > /tmp/DAT.tmp

to

[[ -e /log/sync/${FILE##*/} ]] || { echo "$FILE"; cp "$FILE" /log/sync/; }
9

Ooook Thank you!
But I also need to redirect the "filtered" output to a file, that's why I add the redirect ">"

But this doesn't work:

[[ -e /log/sync/${FILE##*/} ]] || { echo "$FILE" > /tmp/DAT.tmp; cp "$FILE" /log/sync/; }
10

Using process substitution:

grep '\.DAT$' <(comm -23 <(ls log) <(ls log/sync))

A more traditional, more readable approach:

ls log > a
ls log/sync > b
comm -23 a b | grep '\.DAT$'

Regards,
Alister

1 Like