Creating directories from an input file - unix

Question

I have below mentioned directories in an input file and i need them to create in a loop.
data/app_rt_ws/Request/2017_06_27
data/app_rt_ws/Response/2017_06_19
data/app_rt_ws/RTWS
data/app_rt_ws/SDP
data/edge/response/9-20-2016
data/edge/response/9-22-2016
Problem is that i don't need the directories in the yyyy_mm_dd or dd-mm-yyyy format which get created at run time on the server. I need them to be discarded and have the rest of the static path of the directories created.
I am using below mentioned code but can't seem to figure out how to omit the above mentioned part
for i in `cat /tmp/inputfile.txt`
do
echo $i
cd /opt/app/app
awk '/date_year/{print $1}' (need to filter out the entries with date)
mkdir -p $i ( need to create rest of the directories)
done

Answer 1

You may modify your script as followed,
for i in `awk '$0 !~ (/[0-9]{1,2}-[0-9]{1,2}-20[0-9]{2}/ && /20[0-9]{2}_[0-9]{1,2}_[0-9]{1,2}/){print}' inputfile.txt`;
do
echo $i
cd /opt/app/app
mkdir -p $i
done
And the output of the awk command is like this,
$ awk '$0 !~ (/[0-9]{1,2}-[0-9]{1,2}-20[0-9]{2}/ && /20[0-9]{2}_[0-9]{1,2}_[0-9]{1,2}/){print}' inputfile.txt
data/app_rt_ws/RTWS
data/app_rt_ws/SDP

Answer 2

With bash for regexp matching:
$ cat tst.sh
while IFS= read -r path; do
if [[ $path =~ /([0-9]{4}(_[0-9]{2}){2}|([0-9]{1,2}-){2}[0-9]{4})$ ]]; then
path="${path%/*}"
fi
echo mkdir -p "$path"
done < "$1"
$ ./tst.sh file
mkdir -p data/app_rt_ws/Request
mkdir -p data/app_rt_ws/Response
mkdir -p data/app_rt_ws/RTWS
mkdir -p data/app_rt_ws/SDP
mkdir -p data/edge/response
mkdir -p data/edge/response
Remove the echo once you've tested and are happy with the result. Check the regexp - you said you wanted to match dd-mm-yyyy but then your input contained dates as d-mm-yyyy so idk what you really wanted and so I guessed you'd be happy with 1 or 2 digits, hence [0-9]{1,2}.
With other shells use a case statement or similar to match the date at the end of line as a globbing pattern. You do NOT need to call an external tool to do this check.
Using:
for i in `some command`
is always to be avoided because it breaks when the output of some command contains spaces and it's using deprecated backticks instead of modern notation $(some command), and doing:
echo $i
is always to be avoided because you should ALWAYS quote your shell variables (echo "$i") unless you have a specific reason not to so as to avoid accidental word splitting and file name generation.

Answer 3

You could use the bash string substitutions to get rid of the date.
for i in `cat /tmp/inputfile.txt`
do
echo $i
cd /opt/app/app
if [[ $i =~ [0-9] ]]; then
mkdir -p ${i%/*}
else
mkdir -p $i
fi
done
The substitution cuts off everything after the last / so the date is gone.
However if you just want all the ones without numbers then you could do:
for i in `cat /tmp/inputfile.txt`
do
echo $i
cd /opt/app/app
if [[ $i =~ [0-9] ]]; then
:
else
mkdir -p $i
fi
done
Output of the second version:
data/app_rt_ws/RTWS
data/app_rt_ws/SDP
I hope this is what you looked for :)