unary operator expected

[Erik-joix]

I have this in my shell script.

read -p "Number ? " choice ; tput cuu1

    if [ $choice = 1 ] && [ $choice1 = 0 ] ; then
	while [ "$target" != 2 ] ; do

When i just hit "Enter" i get following message.

./menu.sh: line 623: [: =: unary operator expected

This doesn't prevent the script from running but i would like the script to handle this kind of input better. How could i achieve that?

[wolveso]

Placing quotes around things removes the "unary operator expected" message when just pressing enter..

read -p "Number ? " choice ; tput cuu1

    if [ "$choice" = "1" ] && [ "$choice1" = "0" ] ; then
	while [ "$target" != "2" ] ; do

Hope this helps?!

[Erik-joix]

Yep, that's the fix.
Thanks wolveso!

BTW: What's the "policy" on these quotation marks? In the books i use, people sometimes use them and sometimes don't. They're obviously important but i can't seem to find any words about their importance and when to use them and when not.

[crackers]

If you read "man test" you'll note that the "=" comparison operator is for strings, while "-eq" is used for numerical comparisons.

If you want to get really fancy, you can also write your code as (not tested, so try at your own risk):

[ $choice -eq 1 -a $choice1 -eq 0 ] && while [ $target -ne 2 ] ; do

But it's not as easily readable, as you can see...

[Erik-joix]

Well i tried this.

read -p "Number ? " choice ; tput cuu1

    if [ $choice -eq 1 ] && [ $choice1 -eq 0 ] ; then
	while [ $target -ne 2 ] ; do

The script also worked fine but i also got that error message again. I then put quotation marks around it but the error message just changed to something like "integer expected". Well i have an idea what causes this but don't really want to hunt that problem down since the script works fine with wolveso's suggestion. Thanks crackers! Man test helped anyway.

[theurge]

Another solution is to enclose the test condition inside of double paranthesis, such as:

read -p "Number ? " choice ; tput cuu1

if (( choice = 1 && choice1 = 0 )) ; then
	while (( target != 2 )) ; do

This also saves you from having to precede your variable names with the $, if that bothers you.

[bradthemad]

The "policy" on quotation marks is to put them around any variable that might be empty. That way, after variables have been expanded, there will at least be a set of empty quotes to denote a null value, rather than absolutely nothing, generating a syntax error.

# Script without quotes:
if [ $choice = 1 ] ;
# after variable expansion, when $choice is null:
if [ = 1 ] ;
# error!

# Script with quotes:
if [ "$choice" = 1 ] ;
# after variable expansion, $choice is null:
if [ "" = 1 ] ;
# syntax is okay, test fails since "" does not equal 1

Your integer comparison fails when $choice is empty because it is expecting an integer to compare to an integer, and can't compare a blank space to an integer. Considering you're looking at user input, you should probably stick to string (unary) comparison, since a user could type in anything and cause the script to error out.

It helps to put a "set -x" before a section of script that's giving you problems.