search for directories containing one file and missing another

Question

I am currently aware of how to search directories and list those which do NOT contain a specific file like so:

find parent_directory -mindepth 1 -maxdepth 1 -type d '!' -exec sh -c 'ls -1 "{}"|egrep -i -q "^file_name$"' \; -print

but now I want to use the resulting directories to check whether or not they DO contain another file but I can't manage to do that in an extended one-line expression. Is this possible somehow?

Answer 1

You are making this far more complicated than needed. You don't seem to want to recurse into subdirectories, so all you need to find those directories that don't have a specific file is:

for dir in */; do [ ! -e "$dir"/"$filename" ] || printf '%s\n' "$dir"; done

And, to see which of those do have another file:

for dir in */; do 
    [ ! -e "$dir"/"$filename1" ] && 
    [ -e "$dir"/"$filename2" ] && 
    printf '%s\n' "$dir"; 
done

Or, in a slightly clearer syntax:

for dir in */; do 
    if [ ! -e "$dir"/"$filename1" ]; then 
        if [ -e "$dir"/"$filename2" ]; then 
            printf '%s\n' "$dir"; 
        fi 
    fi
 done

This is all done using the built-in tools of the shell. Specifically:

• [ : this, and the accompanying ] are synonyms for the test builtin (see help [ or help test if using an sh-style shell). They are ways of writing test operations in the shell.
• -e : this tests whether a file/directory etc exists. See help test. The simple format is: [ -e file ] which will return true if file exists.
• [ ! -e filename ] : the ! simply inverses the test. So [ ! -e file ] will be true if file does not exist.

Taken together, this means the command above does:

## Iterate over every directory (dirs only because of the '/' in '*/')
## saving each of them in the variable $dir.
for dir in */; do 
    ## If this $dir does not contain $filename1
    if [ ! -e "$dir"/"$filename1" ]; then 
        ## If this $dir does contain $filename2
        if [ -e "$dir"/"$filename2" ]; then 
            ## Print the directory name
            printf '%s\n' "$dir"; 
        fi 
    fi
done

To run this, of course, you need to first set $filename1 and $filename2 accordingly. For example:

filename1="unwantedFile"
filename2="wantedFile"

Answer 2

You can determine your required directories the following manner also:

find . -maxdepth 2 -path '*/*/wanted.txt' -type f \
  -execdir test ! -f unwanted.txt \; -execdir pwd \;

How this works is as follows:

• We look for entries till the depth of 2.
• The -path option will further constrain them to be exactly a depth of 2 due to the presence of 2 slashes as the maxdepth precludes from going any further and 2 explict slashes prevent from going under depth of 2.
• The wanted.txt entry found at the depth level of 2 better be a regular file ensured by -type f
• The -execdir option will elevate the operation to the directory in which the wanted.txt resides and hence the test command will be looking for the unwanted file there.
• Next it is a simple matter of printing the directory (elevated due to -execdir) where the unwanted file was not found.
• Without meaning to belabor this point, a directory can contain a particular file just once, hence the -execdir operations are run once / directory and only on those directories that are at least sure to contain wanted.txt regular files.

Answer 3

With zsh, to list the directories in the current directory that contain a musthave file and not a mustnothave file:

contain() [[ -e $REPLY/$1 || -L $REPLY/$1 ]]

printf '%s\n' *(D/e(contain musthave)^e(contain mustnothave))

Note that doing:

find ... -exec sh -c 'ls {}' \;

is not only not portable but is also a command injection vulnerability. For instance, if there's a directory called $(reboot) or ;reboot, that would run the ls $(reboot) or ls ;reboot command lines causing a reboot. {} should never be embedded in a code argument (of sh or any other language) where it may be misinterpreted. Use:

find ... -exec sh -c 'ls "$1"' sh {} \;

instead.