how to get n-1 number

Question

I have a text file where I am keeping versions:

1.0
2.0
3.0
4.0
5.0

and so on.

I need a way to give, for example, 4.0 as an input and get3.0 as the output. Note that there would be thousands of versions.

I mean I want to give a version number and get the number of the previous version. So if I give 9.0, I would get 8.0.

I tried

  sed '1!G;h;$!d' test.txt 

But this is printing all the numbers.

Answer 1

You could do this with bc (see the GNU bc manual). You can pass bc an arithmetical expression and bc will evaluate it and return the result, e.g.

echo '9.0 - 1' | bc -l

This command returns 8.0 as its result.

Now suppose your version number is contained in the file version.txt, e.g.:

echo '9.0' > version.txt

Then you could run the following command:

echo "$(cat version.txt) - 1" | bc -l

This would produce 8.0 as its output. Alternatively, suppose that you have a list of version numbers in a file called versions.txt e.g.:

1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0

We could create such a file like so:

for i in {1..10}; do echo "${i}.0" >> versions.txt; done

In this case we could use bc inside of a while-loop:

while read line; do echo "${line}-1" | bc -l; done < versions.txt

This produces the following output:

0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0

Alternatively, you could extract the integer part from the version number and use bash to perform the arithmetic. You could use any number of text-processing tools to do this (e.g. grep, sed, awk, etc.). Here is an example using the cut command:

echo "$(( $(echo '9.0' | cut -d. -f1) - 1)).0"

This produces 8.0 as its output. Putting this into a while-loop gives us the following:

while read line; do \
    echo "$(( $(echo ${line} | cut -d. -f1) - 1)).0";
done < versions.txt