CODEWITHSUNDEEP
mysqlshell
showkey
showkey

Reputation: 348

Why mysql command can't be replace by string in shell?

mysql -u root -ppasswd -e"use test_db;"

The command can run in shell console without entering into mysql console.

Why the following two lines can't run in the shell console?

str="use test_db;"
mysql -u root -ppasswd -e$str;


ERROR 1049 (42000): Unknown database 'test_db;'

Upvotes: 0

Views: 28

Answers (2)

Sean Zhang
Sean Zhang

Reputation: 108

Because shell split your string str. You should write like this:

mysql -u root -ppasswd -e"$str"

Or you will get this:

mysql -u root -ppasswd -euse test_db

Upvotes: 1

Tom Fenech
Tom Fenech

Reputation: 74595

The issue here is that the shell is splitting your string, so your command ends up being this:

mysql -u root -ppasswd -euse test_db;

test_db; is interpreted as a separate argument, the name of a database.

To solve this problem, use quotes:

str="use test_db;"
mysql -u root -ppasswd -e"$str"

I removed the semicolon as it seems like you're specifying it twice. For a single command, I don't think you need it at all but I guess your script will be more complicated than this in reality.

As an aside, you can skip the use db_name command by specifying a database (as you were doing by accident):

str="select col1, col2 from table"
mysql -u root -ppasswd -e"$str" test_db

Upvotes: 2

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