Compare files that are in directory 1 but not directory 2?

Question

I'm having trouble with a bash script I want to make

I know ls will list files that are in a directory but I want it to list directories that are in directory1 but NOT in directory2, and then list files in directory2 that are NOT in directory1.

In a feeble attempt, i tried:

ls -al | diff directory1 directory2

Quickly I realized why it didn't work. Can anyone help out a total bash nerd out? Thanks!

Answer 1

Given bash, this might be easiest as

$ comm <(ls -a dir1) <(ls -a dir2)

The <(command) expression runs command over a pipe and substitutes a /dev/fd reference:

mress:10018 Z$ echo <(ls)
/dev/fd/11

So the command above runs ls -a on each directory and feeds their outputs as file arguments to comm, which outputs up to 3 columns, tab-indented: entries only in the first, entries in both, entries in the second. (That is, if it's in both then it is indented by a tab, if it's only in the second it's indented by 2 tabs.) You can also suppress columns by number: comm -1 foo bar displays only the lines in both and lines in the second file, the latter indented by one tab. (This is most commonly used by suppressing all but the column you want: comm -13 foo bar shows only the lines in common.)

Given that you want those in the first directory, that translates to

$ comm -23 <(ls -a dir1) <(ls -a dir2)

If you need more than just whether it is present, use diff -r, which will output diffs for the files in common and a one-line message for files found only in one or the other.

Answer 2

And here a pure script. These are the directories a and b:

find a b
a
a/a
a/b
a/c
a/f
a/f/h
a/f/i
b
b/b
b/c
b/d
b/f
b/f/g
b/f/h

Here is the command:

cd a
find ./ -exec test ! -e ../b/{} ";" -print 

output:

./a
./f/i

Swap a and b for files in a but not in b. The ! is a negation. -e tests for -existance. In prosa: "Test if not exists the file found in a in ../b".

Note: You have to dive into a first, to get names without 'a'. For the second comparison you have to cd ../b.

Answer 3

If you prefer a graphical tool, use

xxdiff dir1 dir2 

you might need to install it first. Similar programs are

gtkdiff
tkdiff

The midnight commander has a compare directories command build in, which works nice, if you don't go for subdirs.

Answer 4

You could use the neglected join command. Here some setup for two example directories, d1/ and d2/, each of which has some files with names unique to the directory, and some files with names in common with the other directory. This is only an example, so I used single-letter file names to illustrate file names unique to one or the other, and file names in common.

# set up for example
mkdir d1 d2
for name in a  b  c  d  e  f  g  h
do
    touch d1/$name
done
for name in e f g h i j k l
do
    touch d2/$name
done

ls -1 d1 > d1.out   # That's "minus one" not "minus ell"
ls -1 d2 > d2.out
join d1.out d2.out       # files common to both d1/ and d2/
join -v 1 d1.out d2.out  # files only in directory d1/
join -v 2 d1.out d2.out  # files only in directory d2/

For me, it shows files like this:

 5:51PM 100 % join d1.out d2.out
e
f
g
h
 5:51PM 101 % join -v 1 d1.out d2.out
a
b
c
d
 5:52PM 102 % join -v 2 d1.out d2.out
i
j
k
l

UPDATE: You'd want to do different things in real life to accomodate files with whitespace in them, as join uses the first field "delimited by whitespace" to decide which lines are uniqe and which lines are common.

Answer 5

You can use find and awk to solve this.

With following layout:

$ mkdir a b a/1 b/1 b/2 a/3
$ touch a/f1 b/f1 a/f2 b/f3

Part one:

$ find a b -mindepth 1 -maxdepth 1 -type d | \
    awk -F/ ' { if (!w[$1]) w[$1]=++i; if (w[$1]>1) b[$2]=1; else a[$2]=1; }
          END { for (x in a) if (!b[x]) print x }'
3

Part two:

$ find b a -mindepth 1 -maxdepth 1 -type f | \
    awk -F/ ' { if (!w[$1]) w[$1]=++i; if (w[$1]>1) b[$2]=1; else a[$2]=1; }
          END { for (x in a) if (!b[x]) print x }'
f3

This compares to a comm solution:

$ comm -23 <(ls a) <(ls b)    
3
f2
$ comm -13 <(ls a) <(ls b)
2
f3

And to a join solution:

$ join -v1 <(ls a) <(ls b)
3
f2
$ join -v2 <(ls a) <(ls b)
2
f3