Bash passing arguments to a command

I need to execute in Bash the following command:

beeline -u "${DATALAKE_NAMENODE_HIVE}:${DATALAKE_NAMENODE_PORT_HIVE}/pcgexp; -hiveconf ....  -f ..... "

The thing is that in Linux the ‘;’ character is indicator of end of the command and I also need to use environment variables. How can I do this? How can I achieve this?

Answer

The answer from @ilkkachu is on the money, as is chepner’s comment — quoted semicolons have no special meaning, and quoted arguments make more sense than a quoted SET of arguments.

That said, another handy way of handling arguments in bash is to use arrays.

declare -a b_args=()

b_args+=( -u "${DATALAKE_NAMENODE_HIVE}:${DATALAKE_NAMENODE_PORT_HIVE}/pcgexp;" )
b_args+=( -hiveconf "...." )
b_args+=( -f "....." )

beeline "${b_args[@]}"

This helps with manageability — you can comment out a part of the argument list, have a clear view of what’s being set without having a super long command line to figure out.

In a normal (non-associative) array, arguments are always output in the numerical order of their index, which always increments as you append to the array.