Sourcing date hour in Bash - bash

Question

I need to write a Bash script that source another script (config script) for hours. If the hour mentioned in config script matches the Linux past hour it needs to print the hour.
$ cat ConfHours.sh
#!/bin/bash --
Hours=(0 1 2 22 23)
$ cat Foo.sh
#!/bin/bash --
source /home/Geo/ConfHours.sh
arrayHours=( ${HOURS} )
for v in "${arrayHours[#]}"
do
HOUR=$(( $(date +%H) -1))
if [ "${HOUR}" == "v" ] ; then
HOUR = ${HOUR}
echo $HOUR
fi
done
When I run Foo.sh, I do not get anything. Could you please correct me where I am wrong?

Answer 1

Some errors:
source /home/Geo/ConfHours.sh
arrayHours=( ${HOURS} )
ConfHours defines a variable named Hours -- different variable
for v in "${arrayHours[#]}"
do
HOUR=$(date -d "1 hour ago" +%H)
You don't need to define this every time through the loop: put it before the for statement
if [ "${HOUR}" == "v" ] ; then
missing $ for the v variable
$HOUR will contain a leading 0 (due to %H)
a better test: if (( 10#$HOUR == 10#$v ))
HOUR = ${HOUR}
No spaces around the = allowed for variable assignment. Why are you trying to redefine the variable to itself?
echo $HOUR
fi
done
A more concise way to test an array contains a value is to take advantage of array string concatenation and pattern matching:
source ConfHours.sh
hour=$(date +%k) # leading space, not leading zero
if [[ " ${Hours[*]} " == *" ${hour# } "* ]]; then
echo "$hour"
fi
All spaces and quotes are required.
Don't use UPPER_CASE_VARS: here's why