Ceil only floating point numbers in bash linux

Question

In linux bash, I want to get the next integer for a given number only if it is a floating point number. I tried with

count=$((${count%.*} + 1));

But with the above code, all the time (even if the number is not floating point), it is giving next integer.

Expected result :

345.56 => 346
345.12 => 346
345 => 345

Can anyone help me to find the solution?

Thanks in advance.

Answer 1

you can use perl

NUMBER=365
perl -w -e "use POSIX; print ceil($NUMBER/1.0), qq{\n}"

for assigning to a variable

MYVAR=$(perl -w -e "use POSIX; print ceil($NUMBER/1.0), qq{\n}")

Answer 2

In awk. Some test records:

$ cat file    # dont worry about stuff after =>
345.56 => 346
345.12 => 346
345 => 345
345.0
0 
-345.56 => 346
-345.12 => 346
-345 => 345
-345.0

The awk:

$ awk '{print $1 " => " int($1)+($1!=int($1)&&$1>0)}' file
345.56 => 346
345.12 => 346
345 => 345
345.0 => 345
0 => 0
-345.56 => -345
-345.12 => -345
-345 => -345
-345.0 => -345

Answer 3

You'll have to test for the presence of a dot:

case $count in
    *.*) count=$(( ${count%.*} + 1 )) ;;
      *) : nothing to do
         ;;
esac

or

[[ $count == *.* ]] && count=$(( ${count%.*} + 1 ))