执行另一个命令时，shell脚本中的参数替换

Question

I have written a script which curls and downloads a DB

DB_URL="$(curl --user $1:$2 -s https://myurl.com/ | grep -o \\http\\:[a-zA-Z.0-9\\:/-]* | grep DBNAME$ | tail -1)"

echo -e "\\033[33;31m The database is here :" echo $DB_URL

But, the $1 and $2 are not being substituted with the two params that I am passing. How do I do it? I have tried many different ways to substitute but still no success.

Works fine when executed directly by replacing $1 and $2 with correct username and password

Answer 1

To answer my own question, this worked :

DB_URL=$(curl --user "$1":"$2" -s https://myurl.com/ | grep -o \\http\\:[a-zA-Z.0-9\\:/-]* | grep DBNAME$ | tail -1)

Had to put the variables in quotes.

See glenn's comment below. The quotes prevented special characters in the password from being interpreted.