Command substitution: file or command not found

Question

The following script

#!/bin/bash
QUERY='select * from cdr;'
MYROWS=$("sqlite3 -list -nullvalue NULL -separator ',' /var/log/asterisk/master.db '${QUERY}'")

gives me

./bla.sh: row 35: sqlite3 -list -nullvalue NULL -separator ',' /var/log/asterisk/master.db 'select * from cdr;': file or directory not found

If I execute directly

sqlite3 -list -nullvalue NULL -separator ',' /var/log/asterisk/master.db 'select * from cdr;'

then it works. I guess there is some error with the quotes that can't be seen on the error message. Please note that I need the single quotes around

select * from cdr;

Thanks for any hint on what I am doing wrong!

Answer 1

What's inside $(…) is a shell command with the usual syntax. The double quotes around the whole snippet you intend as a command make it parsed as a single word, which is the first word of the command so it's interpreted as a command name.

Furthermore, your quoting of $QUERY isn't right: you need double quotes around it, so that the variable is expanded.

MYROWS=$(sqlite3 -list -nullvalue NULL -separator ',' /var/log/asterisk/master.db "${QUERY}")