Closed maheshjag closed 10 years ago
Thanks—this was fixed with 1ec91ed2872977e55963c52c9a9287b313d110c4; the if
statement checking $l
didn't quote the expanded $()
expression, only the internal "$l"
, which wasn't obvious. You were completely right, the space was the issue, but you should be able to use spaces now.
Got this error when trying to invoke the script like this:
./bashweather.sh -l "London, UK"
The space character between the comma and UK seems to be the culprit here.