getops still performs default actions when arguments are provided - bash

Question

I've recently started working with the getopts command in bash. I am confused as to why my script runs the dafult action "cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl" when arguments have been provided. I only want that to run if no arguments were passed to the shell script. I've used getopts:Std in perl where I was able to code somthing like:
unless ($opts{d}) {
do something...}
How would I code something like that in a shell script? Also, how would I code logic such as this:
if ($opts{c}) {
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -c
}
elsif ($opts{d} {
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -d
My code:
#!/bin/sh
while getopts cd name
do
case $name in
c)copt=1;;
d)dopt=1;;
*)echo "Invalid arg";;
esac
done
if [[ ! -z $copt ]] #Specifies what happens if the -c argument was provided
then
echo "CSV file created!"
cat "~/bin/Temp/log.txt" | ~/bin/Scripts/vpnreport/report.pl -c
fi
if [[ ! -z $dopt ]] #Specifies what happens if the -d argument was provided
then
echo "Debug report and files created"
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -d
fi
if [[ ! -z $name ]] #Specifies what happens if no argument was provided
then
echo "Running standard VPN report"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl
fi
shift $(($OPTIND -1))
My Output:
[~/bin/Scripts/report]$ sh getoptstest.sh
Running standard report
[~/bin/Scripts/report]$ sh getoptstest.sh -d
Debug report and files created
Running standard report
[~/bin/Scripts/report]$
The two getopts commands are vasty different from bash to perl and I just can't seem to get the hang of the bash varient even after reading several tutorials. Any help would be greatly appreciated!

Answer 1

On the final run of getopts, your variable (name) will be set to "?".
#!/bin/bash
while getopts abc foo; do :; done
echo "<$foo>"
Output of the above:
$ ./mytest.sh
<?>
$ ./mytest.sh -a
<?>
Insead, use elif, which is like Perl's elsif:
if [[ ! -z $copt ]]
then
# ...
elif [[ ! -z $dopt ]]
then
# ...
else
# ...
fi
Or test if [[ -z $copt && -z $dopt ]], or so forth. Other notes:
See the official if and case documentation in the Bash manual under "Conditional Constructs".
[[ ! -z $name ]] means the same as the more-direct [[ -n $name ]].
Use #!/bin/bash instead of #!/bin/sh, or switch off of [[ in favor of [. The double square bracket (and your use thereof) is specific to bash, and rarely works with sh.

Answer 2

I took Jeff's answer and rewrote my script so it works now:
#!/bin/bash
while getopts cd name
do
case $name in
c)carg=1;;
d)darg=1;;
*)echo "Invalid arg";;
esac
done
#Specifies what happens if the -c argument was provided:
if [[ ! -z $carg ]]
then
if [[ -z $darg ]]
then
echo "CSV created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -c
else
echo "Debug CSV created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -cd
fi
fi
#Specifies what happens if the -d argurment was provided:
if [[ ! -z $darg ]]
then
echo "Debug report created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -d
#Specifies what happens if no argument was provided:
else
echo "Standard report created"
cat ~bin/Temp/logs.txt | ~bin/Scripts/report.pl
fi
Thank you again for your assistance!