How do I compare the day of the week?

Question

I'm trying write a script which shows right statement based on which day of the week it is. Two examples:

1. If today is 4th day of the week. echo Today is a working day.
2. If today is 6th day of the week. echo Today is a weekend.

I wrote this however it doesn't work

echo Hello!
echo Today's date is: date
DAY=$(date +"%u")
if [ "${DZIEN}" -ge 1 && "${DZIEN}" -le 5 ]
then 
   echo WORKING DAY;
else
   echo WEEKEND;
fi

Solution

Try this :

echo "Today's date is: $(date)"
day=$(date +"%u")

if ((day > 5)); then
   echo "WEEKEND"        
else
   echo "WORKING DAY"
fi

I use (( )) bash arithmetic

Or less readable :

echo "Today's date is: $(date)"
day=$(date +"%u")

if [[ day -gt 5 ]]; then
   echo "WEEKEND"        
else
   echo "WORKING DAY"
fi

Comments

[Admin]

you set DAY, then check DZIEN... pick one. And perhaps try -a instead &&.

[Admin]

You need [..] && [..] or [[.. && ..]] - why don't you google for shell multiple conditions in if ?? It's much shorter to type than all the above...

[Admin]

@frostschutz I made a mistake in copying variable name.

[Admin]

@don_crissti I have to do this in that way, this is a task from the teacher :/ Thanks for help, now it's working.

[Michael Hoffmann]

Why does this work? What is %u?

[Janek Bogucki]

From man date, % is the format specifier for day of week: %u day of week (1..7); 1 is Monday