Creating function to perform display-time arithmetic - bash

Question

Trying to create a function to get a number for hours(0-24) and a number for minute(0-60) and add 15 minutes to it. Issue is when I try to put a number 45 and up, I get a number over 60.
Created if then statement that takes my input and if the minutes are over 60, it will subtract 60 from minutes variable and add 1 to hour variable.
Expected results are that it will convert the time if over 60 minutes into 1 hour and remainder into minutes. The actual results are that sometimes the if then statement doesn't run correctly and it does both if AND then statements or sometimes I get a negative number.
#!/bin/bash
read -p "Hours: " hr1
read -p "Minutes: " min1
min1=$((min1+15))
Time() {
echo $min1
if $min1 > 60
then
min1=$((min1-60))
if $min1 < 0
then $((min1*-1))
fi
hr1=$((hr1+1))
else
$min1 + 15
fi
}
Time
echo $min1
echo $hr1

Answer 1

You can greatly simplify your Time() function by using shell arithmetic expressions delimited by (( and )). Here stand-alone arithmetic expressions are only valid with Bash shell but not posix shells where it is undefined
(( hr1 += min1 / 60, min1 %= 60 ))
Splits into two expressions separated by a comma ,:
hr1 += min1 / 60 ⇔ hr1 = hr1 + min1 / 60
min1 / 60 is evaluated in priority to the addition.
Shell arithmetic is integer only
min1 %= 60 ⇔ min1 = min1 % 60
The modulo reminder of min1 / 60
#!/usr/bin/env bash
read -r -p 'Hours: ' hr1
read -r -p 'Minutes: ' min1
# Bash stand-alone arithmetic expression add 15 to min1
(( min1+=15 ))
Time() {
echo "${min1}"
# Two Bash stand-alone arithmetic expressions separated by ,
# Add integer division of min1 by 60 to hr1
# Truncate min1 to the integer division reminder of itself by 60
(( hr1 += min1 / 60, min1 %= 60 ))
}
Time
echo "${min1}"
echo "${hr1}"

Answer 2

There are many, many ways to add minutes to a given time and then obtain the resulting hours and minutes. A systematic way is using the date function as Oliver Gaida shows. A manual conversion is fine for learning purposes, but you will want to ensure you deal with the addition of time that causes the hours to roll into the next day.
In order to handle all aspects of the conversion, it is useful to convert the total time to time into seconds. At that point, you can perform all necessary tests to determine if the total time has rolled the hours into the next day, etc..
Since you seem to only want to capture the hours and minutes of the resulting time and are not concerned with the number of days, you can simply test the number of seconds against the seconds-per-day and if the number of seconds exceeds seconds-per-day, simply reduce the number of seconds modulo by seconds-per-day.
A short function getHM() (your Time() function) to update the values in hr1 and min1 could be similar to the following:
## function converting number of seconds to hours, minutes (discarding days)
getHM() {
test -z "$1" && { ## validate input given
printf "error: insufficient arguments getHM()\n" >&2
return 1
}
local secs="$1" ## local variables seconds
local secsperday=$((3600 * 24)) ## seconds-per-day
local days=$((secs / secsperday)) ## days
(( days > 0 )) && { ## seconds exceed seconds-per-day
printf "error: time exceeds 24 hours, days discarded.\n" >&2
secs=$((secs % secsperday)) ## reduce secs modulo by secsperday
}
hr1=$((secs / 3600)) ## update hr1 & min1 values
min1=$(((secs - hr1 * 3600) / 60))
}
As noted in the function, it simply discards any additional days in order to return the resulting hours (0-23) and minutes (0-59).
Adding that to a short example and you could do:
#!/bin/bash
declare -i hr1=-1 min1=-1 addminutes=15 ## initialize variables
## function converting number of seconds to hours, minutes (discarding days)
getHM() {
test -z "$1" && { ## validate input given
printf "error: insufficient arguments getHM()\n" >&2
return 1
}
local secs="$1" ## local variables seconds
local secsperday=$((3600 * 24)) ## seconds-per-day
local days=$((secs / secsperday)) ## days
(( days > 0 )) && { ## seconds exceed seconds-per-day
printf "error: time exceeds 24 hours, days discarded.\n" >&2
secs=$((secs % secsperday)) ## reduce secs modulo by secsperday
}
hr1=$((secs / 3600)) ## update hr1 & min1 values
min1=$(((secs - hr1 * 3600) / 60))
}
## loop until valid input received
while ((hr1 < 0)) || ((hr1 > 23)) || ((min1 < 0)) || ((min1 > 59)); do
read -p "Hours: " hr1
read -p "Minutes: " min1
done
## convert all to seconds adding desired 15 minutes
secs=$((hr1 * 3600 + (min1 + addminutes) * 60))
getHM "$secs" ## call function to update hr1 & min1
printf "\nHours : %d\nMinutes : %d\n" "$hr1" "$min1"
(note: when using the arithmetic operator for comparison, e.g. ((...)), any non-integer values are evaluated as 0, so if you want to validate the use provides only integer input, you need to do that in the while loop after the read is complete -- and reset either variable to -1 if a non-integer value is detected)
Example Use/Output
No adjustment in hour required:
$ bash gethm.sh
Hours: 23
Minutes: 44
Hours : 23
Minutes : 59
Addition causing total time to land precisely at the start of a day:
$ bash gethm.sh
Hours: 23
Minutes: 45
error: time exceeds 24 hours, days discarded.
Hours : 0
Minutes : 0
(note: the error message provided if the total time causes the hours to roll into the next day. It is informational only and you can remove it to suit your needs)
Example showing the roll to 1 minute past the new day:
$ bash gethm.sh
Hours: 23
Minutes: 46
error: time exceeds 24 hours, days discarded.
Hours : 0
Minutes : 1
Look things over and let me know if you have further questions

Answer 3

With date it is easy. Convert the date to epoche-seconds, add 900 seconds and convert it back.
date --date="#$(echo $(($(date --date="22:53" +"%s")+900)))" +"%H:%M"
23:08