need to “build” a mysql command and exec it in a shell script

Question

I'm new to writing shell scripts.

I am attempting to create a database using a shell script. Here's the script:

#!/bin/bash
#create a new db
a="mysql -uuser -ppassword -e'create database $1;'"
exec $a

The command exec mysql -uuser -ppassword -e'create database databaseName;' works in a shell, but when I sh the script, I get the mysql help open...

Answer 1

I think the problem is in the quotes, the simple quote prevent the variable expansion. You can simply do like this in your script:

    #!/bin/bash
    #create a new db

    mysql -u user -p password -e "create database $1;"

Or you can try to place all your mysql commands in a file, let's say "dbname.sql". And do this:

    #!/bin/bash
    #create a new db

    mysql -u user -p password "$1" < "$1.sql"

Answer 2

if you like use exec to run commands this can be a possible solution

    #!/bin/bash
    #create a new db
    programm="mysql"
    parameter[0]="-ppassword"
    parameter[1]="-uuser"
    parameter[2]="-ecreate database $1;"
    exec "$programm" "${parameter[@]}"

exec parameter are

exec [-a NAME] [-cl] [COMMAND] [ARG...] [REDIRECTION...]

command is $programm and the array parameter is the argument list.

Answer 3

听起来您需要使用'cat'命令和管道，而不是使用'exec'。

cat /path/to/my/file | mysql -h localhost -u root -padmin