My last question expanded, so let's analyse things separately.
AND-operater, Intersection?
I will give an example:
$ find . | awk -F"/" '{ print $2 }'
.zcompdump
.zshrc
.zshrc_copy
.zshrc_somequy
.bashrc
.emacs
$ find ~/bin/FilesDvorak/.* -maxdepth 0 | awk -F"/" '{ print $6 }'
.bashrc
.emacs
.gdbinit
.git
.profile
When I save the outputs to separate lists, I cannot understand why the command does not take the common things:
find -f all_files -and -f right_files .
I want only:
.bashrc
.emacs
To slightly expand on Dave's answer:
You appear to want the intersection of two find commands. Find doesn't do set manipulation. Find traverses down (or up depending on how you look at it) a set of paths and applies an expression to each item it encounters. The default action it takes is to print the path of the items it finds that evaluate to true against the expressions. (I believe some old versions of find required you to explicitly add a -print expression.) It doesn't collate the results. For intersection analysis you can use tools like diff, sdiff, comm.
I assume you are trying to find the items with the same name in two separate directories, not in sub directories.
Assuming bash you can do something like
comm -12 <(find . -maxdepth 1 | sort) <(cd ~/bin/FilesDvorak/; find . -maxdepth 1 | sort)
I believe the -and in find commands is almost always superfluous. Ex.
find . -type f ! -type d
Is the same as
find . -type f -and ! -type d
Also the -f flag is an option to add to the paths to traverse. I don't believe it is an expression. Please 'man find' for clarification.
-and only works on tests. If you want to find the common elements of the two lists you should do something like
sort all_files > all_files.sorted
sort right_files > right_files.sorted
comm -12 all_files right_files > common_files
The AND -option in the man
The -and operator is the logical AND
operator. As it is
implied by the juxtaposition of two expressions it
does
not have to be specified. The expression evaluates
to
true if both expressions are true. The second
expression
is not evaluated if the first expression is false.
This part of the manual is rather cryptic for me. However, let's analyse it.
1st sentence: like AND operator in Math. Two two 1s only!
2nd sentente: This suggests me that you do not need to use the command, as you can see in Dave's answer. Great! - - It seems to juxpose the 1st expression with -AND to and Expression without -AND. Please, clarify this Greek to me.
3rd sentence: repeation: It says the same in the 1st sentence by the word "logical".
4th sentence: repeation: It says that 01 implies false
The same in English:
Let p and q be two expressions with the following truth valuas
p q pANDq
1 1 1
1 0 0
0 1 0
0 0 0
where 1 implies true, while 0 false.
Note the manuals contain often superfluous description and sometimes mistakes.
