mjhjmtu

I use the following command to convert a decimal value into a time value:

echo "1.5" | awk -F'.' 'printf $1 ":" "%.0f", $2 / 100 * 60'

outputs: 1:3

how can I make awk to add a trailing zero to the output so I would get: 1:30?

I don't think that adding a trailing 0 is a good approach.
It may produce the intended output for the 1.5 as input,
but if you need a general solution for other inputs,
this approach will probably not work well.

A better approach is to not split the integer part and the decimal parts,
but to work with minutes, using the / and % operators to compute the correct hours and minutes, for example:

awk 'printf "%d:%02d", ($1 * 60 / 60), ($1 * 60 % 60)' <<< 1.5
# prints 1:30

awk 'printf "%d:%02d", ($1 * 60 / 60), ($1 * 60 % 60)' <<< 1.50
# prints 1:30

awk 'printf "%d:%02d", ($1 * 60 / 60), ($1 * 60 % 60)' <<< 1.7
# prints 1:42

awk 'printf "%d:%02d", ($1 * 60 / 60), ($1 * 60 % 60)' <<< 1.05
# prints 1:03

To handle negative values,
you can introduce an abs function:

awk 'function abs(v) return v < 0 ? -v : v 
 printf "%d:%02d", ($1 * 60 / 60), ($1 * 60 % 60)' <<< -1.7
# prints -1:42

Since you don't know how many fractional digits you have, you don't know that you need to divide by 100. @janos has the correct answer, but if you insist on splitting on the decimal point, you need to force convert the fractional part back into a fractional number:

echo 1.5 | awk -F'.' 'printf "%d:%02dn", $1, ("0."$2) * 60'

"0."$2 is string concatenation resulting in the string "0.5".

Then, multiply by the number 60 to get the number of minutes.

Or, you can figure out what divisor to use, based on the number of digits

awk -F'.' 'printf "%st%d:%02dn", $0, $1, $2 / 10**length($2) * 60' << END
1
1.5
1.75
1.6666667
END

1 1:00
1.5 1:30
1.75 1:45
1.6666667 1:40

But, don't do either of these things.