Selecting "ALL" in a query using a list of variables - mysql

Question

My first (simplified) script mysql_script.sh is:
#!/bin/bash
/usr/bin/mysql --password=**** --database=mydb<<EOMYSQL
SELECT SUM(usd) FROM mytable WHERE country=$VAR
INTO OUTFILE "oufile_$VAR.txt" FIELDS TERMINATED BY ';';
EOMYSQL
The first script is called from the second one:
#!/bin/bash
list=( 30 50 60 )
for L in "${list[#]}"
do
VAR=${L} ./mysql_script.sh
done
It works fine, but now I need to get a 4th query with the results for $VAR=*, or $VAR=30 or 50 or 60. As VAR is the code of every country in my DB, I need the value of SUM(usd) for "all countries".
How can I express that?
I tested '%%' and '*' without success.
Thanks,
EDITED for (I hope) better understanding

Answer 1

Try this:
#!/bin/bash
list=( 30 50 60 country)
for L in "${list[#]}"
do
VAR=${L} ./mysql_script.sh
done
this will result in where country=country which will always be true and thus return a sum of all rows.
Handling ORs would get trickier and I'd say at that point you'd be well served to move past bash into as scripting language like perl or python.