Parsing part of a file name to use as a variable in bash - linux

Question

I need to get the DB name and use it as a variable from part of a file name. Currently I have it pulling in the name of the file from a ls command. This is what it returns:
Test-20150311-1200.sql
I need the first part of the filename up to the "-" so for this example it would be Test. How can I get just part of the file name for my variable? Below is the full script I am working on. Any help would be great. Thanks.
#!/bin/bash
DB_USER=x
DB_PASS=x
DB=
for DB_GZFILE in $( ls *.sql.gz ); do
gunzip $DB_GZFILE
echo item: $DB_FILE unzipped
done
for DB_FILE in $( ls *.sql ); do
#Use this statement to insert dump into a new server
mysql $DB <$DB_FILE
done
#Use this command to insert into a server already in use
# mysql -u$DB_USER -p$DB_PASS <$DB_FILE
echo $DB_FILE inserted into database
done
#Remove sql files used to insert into this server
# rm $DB_FILE
# echo $DB_FILE removed
done
echo restarting the mysql process .....
# /etc/init.d/mysql restart
echo mysql restarted
done

Answer 1

Don't parse ls
for DB_GZFILE in *.sql.gz; do
for DB_FILE in *.sql; do
To get just the first part, use parameter substitution to remove the first - and all following characters:
first_part=${DB_FILE%%-*}
You should quote all your "$vars", especially any whose value you get from the user or from the filesystem: you never know when you'll get a filename with a space in it. Example:
gunzip "$DB_GZFILE"
I'd recommend you do not use ALL_CAPS_VARNAMES: one day you'll accidentally use PATH=... and then wonder why your script is broken. Leave ALL_CAPS for system environment variables and shell special vars.