Sort absolute directory path alphabetically and numerically

Question

I am having difficulty finding solutions online to sort an absolute directory path using letters and numbers.

Example is the below in a file called test.txt

 /opt/informix/data/dcdwhdev/rootdbs.2
 /opt/informix/data/dcdwhdev/db1.1
 /opt/informix/data/dcdwhdev/db1.10
 /opt/informix/data/dcdwhdev/db1.11
 /opt/informix/data/dcdwhdev/rootdbs.1
 /opt/informix/data/dcdwhdev/db1.12
 /opt/informix/data/dcdwhdev/db1.19
 /opt/informix/data/dcdwhdev/db1.2
 /opt/informix/data/dcdwhdev/db1.21
 /opt/informix/data/dcdwhdev/db1.22
 /opt/informix/data/dcdwhdev/db1.23
 /opt/informix/data/dcdwhdev/db1.3
 /opt/informix/data/dcdwhdev/db1.31
 /opt/informix/data/dcdwhdev/db1.32
 /opt/informix/data/dcdwhdev/db1.33

Below is what I want to achieve:

/opt/informix/data/dcdwhdev/db1.1
/opt/informix/data/dcdwhdev/db1.2
/opt/informix/data/dcdwhdev/db1.3
/opt/informix/data/dcdwhdev/db1.10
/opt/informix/data/dcdwhdev/db1.11
/opt/informix/data/dcdwhdev/db1.12
/opt/informix/data/dcdwhdev/db1.19
/opt/informix/data/dcdwhdev/db1.21
/opt/informix/data/dcdwhdev/db1.22
/opt/informix/data/dcdwhdev/db1.23
/opt/informix/data/dcdwhdev/db1.31
/opt/informix/data/dcdwhdev/db1.32
/opt/informix/data/dcdwhdev/db1.33
/opt/informix/data/dcdwhdev/rootdbs.1
/opt/informix/data/dcdwhdev/rootdbs.2

Answer 1

If you have GNU sort (might be called gsort on your system, or located in a /opt/gnu/bin directory, or from the coreutils RPM at IBM.com):

sort -V < test.txt

If zsh is available (RPM package at IBM.com):

print -rC1 -- /(Nne['reply=(${(f)"$(<test.txt)"})'])

Or listing the files in that directory directly:

print -rC1 /opt/informix/data/dcdwhdev/*(Nn)

If not, you could use perl for instance as:

perl -e '
  print $_->[1] for 
    sort {$a->[0] cmp $b->[0]}
      map {[s/\d+/sprintf "%06d", $&/ger, $_]} <STDIN>' < test.txt

Where we sort the list by comparing the lines after all the sequences of decimal digits have been zero-padded to 6 digits.

If the lines in the file are always shaped like that and you only need to compare the part before the one and only . lexically, and the part after it numerically, then, you can just do:

sort -t. -k1,1 -k2,2n < test.txt

Answer 2

My process is to break the name into 2 and then sort and bring back together

for fp in `cat test.txt | awk -F\. '{print $1}' | sort | uniq`; do grep $fp test.txt | sed 's/\./\ /g' | sort -nk2 | sed 's/\ /\./g' ;done
./opt/informix/data/dcdwhdev/db1.1
./opt/informix/data/dcdwhdev/db1.2
./opt/informix/data/dcdwhdev/db1.3
./opt/informix/data/dcdwhdev/db1.10
./opt/informix/data/dcdwhdev/db1.11
./opt/informix/data/dcdwhdev/db1.12
./opt/informix/data/dcdwhdev/db1.19
./opt/informix/data/dcdwhdev/db1.21
./opt/informix/data/dcdwhdev/db1.22
./opt/informix/data/dcdwhdev/db1.23
./opt/informix/data/dcdwhdev/db1.31
./opt/informix/data/dcdwhdev/db1.32
./opt/informix/data/dcdwhdev/db1.33
./opt/informix/data/dcdwhdev/rootdbs.1
./opt/informix/data/dcdwhdev/rootdbs.2