Q:

How to output 1-30 while searching file for those numbers and how many times they appear

I have a text file https://1drv.ms/t/s!AjlMpzDMn2h7hWKyBGBxdhHXwjm8
The last column is full of numbers from 1-30.
I need a code that will list each number from 1-30 but will also show how many times it appears in that column.
So it would look like
1=13
2=10
3=12 
.... up until 30.

currently i have a piece of code that I believe will work if i modify it correctly.
awk -F':' 'BEGIN{ split("sparkling fine fortified sweet white red", words, " ") }
       $3 in words{ c[$3]++ }
       END{ for(i in words) print words[i]"="c[i] }' file

A:

The simple approach (with the first lines from the file):
:> awk -F: '{ print $5; }' input | sort -n | uniq -c
 2 1
 5 2
 4 3
 5 4
 2 5

:> awk -F: '{ print $5; }' input | sort -n | uniq -c |
    while read count number; do echo "${number}=${count}"; done
1=2
2=5
3=4
4=5
5=2

with awk only
:> awk -F: '{a[$5]++}; END { for(i=1;i<31;i++) printf "%2d=%d\n",i,a[i]; }' input
 1=2
 2=5
 3=4
 4=5
 5=2

