List all cronjobs if there are any without superfluous info - bash

Question

Trying to output a list of cronjobs, not a list of users. The raw output of crontab -l is way too dirty and I can't seem to clean it up. I run this with sudo script.sh or su and then run it. I've tried invoking it sudo script.sh | grep -v no also. I'm mystified why this doesn't work:
#!/bin/bash
#Trying to show all cronjobs but no extraneous info
#
# This shows "no crontab for USER" for every USER without
# a crontab - I only want to see actual cronjobs, not a long
# list of users without crontabs
echo "Here is the basic output that needs manipulation:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER
done
#
# grep -v fails me
# (grep'ing the output of the script as a whole fails also)
echo "
trying with grep -v no on each line:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER | grep -v no
done
echo "
maybe with quotes around the no:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
crontab -l -u $USER | grep -v "no"
done
# string manipulation - I can't even get started
echo "
And here I try to put the commmand output into a string so I can manipulate it further, and use an if/then/fi on the product:
"
for USER in `cat /etc/passwd | cut -d":" -f1`; do
STRING="$(crontab -l -u $USER | grep -v no)"
echo "STRING: $STRING"
done
BTW, is there an easier way to get code to format correctly here than pasting in 4 spaces at the beginning of each line? I must have experimented for 40 minutes. Not complaining, just asking.

Answer 1

crontab is wiring the "no crontab for user" to standard error. To get rid of those messages, you can run
crontab -l -u $USER 2>/dev/null
within your loop.
I would also suggest renaming USER into something else. The USER variable name is reserved, and should be set to your user (login) name. Generally, you should use lower case variable names, to avoid this sort of name clash.