sql statement in a bash script "command not found"

Question

I've got a bash script that runs a series of sql statements:

#!/bin/bash

mysql -u root -p << QUERY_INPUT

CREATE DATABASE dba_first;
CREATE DATABASE dba_second;
CREATE DATABASE dba_third;

GRANT ALL PRIVILEGES ON `dba%`.* TO `dbuser`@`localhost`;

QUERY_INPUT

When I run the script, I get: ./quick.sh: line 20: dba%: command not found

Any suggestions please?

What is on line 20?

rghome
– rghome

2016-05-31 14:34:31 +00:00
Commented May 31, 2016 at 14:34 — rghome
– rghome, Commented May 31, 2016 at 14:34
It was the `dba%` backticks not being escaped properly.

leshank
– leshank

2016-06-01 05:48:18 +00:00
Commented Jun 1, 2016 at 5:48 — leshank
– leshank, Commented Jun 1, 2016 at 5:48

leshank · Accepted Answer · 2016-05-31 13:58:20Z

2

Answered by a nice person on IRC; have to escape the `

GRANT ALL PRIVILEGES ON \`dba%\`.* TO \`dbuser\`@\`localhost\`;

answered May 31, 2016 at 13:58

leshank

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