Check if the current time falls within defined time range on UNIX - unix

Question

Consider the below PSUEDO-CODE:
#!/bin/ksh
rangeStartTime_hr=13
rangeStartTime_min=56
rangeEndTime_hr=15
rangeEndTime_min=05
getCurrentMinute() {
return `date +%M | sed -e 's/0*//'`;
# Used sed to remove the padded 0 on the left. On successfully find&replacing
# the first match it returns the resultant string.
# date command does not provide minutes in long integer format, on Solaris.
}
getCurrentHour() {
return `date +%l`; # %l hour ( 1..12)
}
checkIfWithinRange() {
if [[ getCurrentHour -ge $rangeStartTime_hr &&
getCurrentMinute -ge $rangeStartTime_min ]]; then
# Ahead of start time.
if [[ getCurrentHour -le $rangeEndTime_hr &&
getCurrentMinute -le $rangeEndTime_min]]; then
# Within the time range.
return 0;
else
return 1;
fi
else
return 1;
fi
}
Is there a better way of implementing checkIfWithinRange()? Are there any inbuilt functions in UNIX that make it easier to do the above? I am new to korn scripting and would appreciate your inputs.

Answer 1

The return command is used to return an exit status, not an arbitrary string. This is unlike many other languages. You use stdout to pass data:
getCurrentMinute() {
date +%M | sed -e 's/^0//'
# make sure sed only removes zero from the beginning of the line
# in the case of "00" don't be too greedy so only remove one 0
}
Also, you need more syntax to invoke the function. Currently you are comparing the literal string "getCurrentMinute" in the if condition
if [[ $(getCurrentMinute) -ge $rangeStartTime_min && ...
I would do if a bit differently
start=13:56
end=15:05
checkIfWithinRange() {
current=$(date +%H:%M) # Get's the current time in the format 05:18
[[ ($start = $current || $start < $current) && ($current = $end || $current < $end) ]]
}
if checkIfWithinRange; then
do something
fi