Create a file in all sub-directories if it does not exist 1

[himalay]

Hello everyone,

I need to make a task automated using scripting.
I need to search for a file named ".archlock" in all the subdirectories of a directory up to infinite level and then create a file if it does not exist.

until now, I did the following.

1. ls -l|grep ^d | wc -l --> returns number of sub directories in that directory.
2. ls -ltr /*.arch* | wc -l --> verifies how number of sub-dir have that file
3. if the count of step 1 and 2 are not same, I used to go to the individual directory and see if the file exist.

Clearly there must a much better way using scripting, please help.

Thank you,
Himalay

 

[PHV]

One way:
find /path/to/a/directory -type d | while read d; do cat /dev/null >>"$d/.archlock"; done

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[PDreyer]

find /mydir -type d -exec ksh -c 'test -f $1/.archlock || touch $1/.archlock' create_archlock {} \;

 

[PDreyer]

And another way:

for d in `find d1 -type d`
do
touch $d/.archlock
done

 

[himalay]

Thank you for the timely reply. Can anyone of you please direct me to an appropriate tutorial to learn it myself.

I was also trying this
ls -ltr /*.arch* | wc -l --> returns number of sub-dir have file starting with ".arch"

how do I find the number of sub-dir that do not have the file. I tried ! operator but it dint work for me.

Thank you
-Himalay

 

[Annihilannic]

I was also trying this
ls -ltr /*.arch* | wc -l --> returns number of sub-dir have file starting with ".arch"

I don't think it does actually. It will list the contents of directories under root containing the string ".arch", and then count those contents.

I would use something like:

# count the number of directories under /d1
DIRS=$(find /d1 -type d | wc -l)
# count the number of lock files under /d1
LOCKFILES=$(find /d1 -type f -name .archlock | wc -l)
# subtract the two and print the result
echo "$(( $DIRS-$LOCKFILES )) of $DIRS directories do not contain lock files"

Annihilannic.