unknown option '--print-defaults' from bash script

Question

I'm trying to run a query from a bash script:

#!/bin/bash

query="\"show databases\""
command="mysql --defaults-file=/user/.my.cnf -e "
outputfile=" > query_result.txt"
command=$command$query$outputfile
$($command)

the result is this:

# ./query_test
mysql: unknown option '--print-defaults'

what I'm doing wrong?

The command:

 mysql --defaults-file=/user/.my.cnf -e "show databases"

works without any issue from the shell

Where does `--print-defaults` come from? It's nowhere in your script. — Benjamin W., Nov 28 '18 at 17:20

score 1 · Accepted Answer · edited Nov 28 '18 at 18:40

1

thanks to the comment of @benjamin-w I solved with this:

#!/bin/bash
args=(--defaults-file=/users/.my.cnf)
args+=(-e "show databases")
outputfile="query_result.txt"
mysql "${args[@]}" > "$outputfile"

other examples in this link: https://mywiki.wooledge.org/BashFAQ/050

edited Nov 28 '18 at 18:40

chepner

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answered Nov 28 '18 at 17:56

Kintaro

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unknown option '--print-defaults' from bash script

1 Answers1