test that argument looks like a date - regex

Question

I'm modifying a script used to test a report writer. The report writer takes optional --from and --to flags to specify the start and end dates. I'd like to modify the script function that starts up the report writer so that its date arguments are also optional.
Sadly, there are already optional arguments to the function, so I'm trying to test whether an argument is in the right format for a date (we use nn/nn/nnnn).
So, I'm echoing the candidate string and checking with grep whether it is in the correct format. Except it doesn't work.
Here is an extract from the function
# If the next argument looks like a date, consume it and use it to define
# the report start date
looksLikeDate=$(echo $1 | grep -e '[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]')
echo from -
echo \$1: \"$1\"
echo looksLikeDate: \"$looksLikeDate\"
if [ -n $looksLikeDate ]
then
echo "-n: true"
FROMFLAG="--from $1"
shift 1
else
echo "-n : false"
FROMFLAG=""
fi
# If the next argument looks like a date, consume it and use it to define
# the report end date
looksLikeDate=$(echo $1 | grep -e '[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]')
echo to -
echo \$1: \"$1\"
echo looksLikeDate: \"$looksLikeDate\"
if [ -n $looksLikeDate ]
then
echo "-n: true"
TOFLAG="--to $1"
shift 1
else
echo "-n: false"
TOFLAG=""
fi
...and here is the output with dates...
from -
$1: "09/02/2018"
looksLikeDate: "09/02/2018"
-n: true
to -
$1: "09/02/2018"
looksLikeDate: "09/02/2018"
-n: true
...and without...
from -
$1: ""
looksLikeDate: ""
-n: true
to -
$1: ""
looksLikeDate: ""
-n: true
...what have I missed? I'd expect that since looksLikeDate is demonstrable empty [ -n $looksLikeDate ] would return false and the code would go down the else path of the if statement.
Update:
Since posting, it occurs to me that the easiest thing is to not to look at the arguments in the function and get callers to pass the --from and --to with the arguments so that I can simply pass $* to the report writer as is done for the existing optional arguments.
Thank you very much for reading; I'm still curious as to why the posted code doesn't work.

Answer 1

That's because you're not quoting your variables! Use more quotes!
So here's what's happening: when Bash sees
[ -n $looksLikeDate ]
it performs parameter expansion, glob expansion, quote removal, etc., and finally sees this (I put one token on each line):
[
-n
]
and you see that the $looksLikeDate part is missing because the parameter $looksLikeDate expands to the empty string before the quote removal step. Then Bash executes the builtin [, and with the closing ], this is equivalent to the following command:
test -n
Now looking at the reference manual for the test builtin, you'll read:
1 argument
The expression is true if, and only if, the argument is not null.
And here, the argument is -n, hence not nil, hence the expression is true.
So remember:
Use more quotes! quote all your variable expansions!
This specific line should look like:
[ -n "$looksLikeDate" ]
Another possibility is to use the [[ keyword:
[[ -n $looksLikeDate ]]
But anyway, quote all your expansion!
Also, you don't need the external tool grep, you can use Bash's internal regex engine or, better yet:
if [[ $1 = [[:digit:]][[:digit:]]/[[:digit:]][[:digit:]]/[[:digit:]][[:digit:]][[:digit:]][[:digit:]] ]]; then
which is a bit long, so use a variable:
date_pattern="[[:digit:]][[:digit:]]/[[:digit:]][[:digit:]]/[[:digit:]][[:digit:]][[:digit:]][[:digit:]]"
if [[ $1 = $date_pattern ]]; then
(and here you mustn't quote the right hand side $date_pattern).