Store multiple values from sql select in bash

Question

I am trying to store each value from the following sql select statement and store them in separate variables using bash.

 #!/bin/bash
 mysqlhost="thehost"
 mysqldb="thedb"
 mysqlun="theusername"
 mysqlpw="thepassword"
 mysqlconnection="--disable-column-names --host=$mysqlhost --user $mysqlun --password=$mysqlpw --database=$mysqldb"

declare -a pinIDs=$(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";)

I get the following result when I use code

echo $pinIDs

8 11 23 26

I need to store each of those values into their own variable.

Cyrus · Accepted Answer · 2016-01-02 10:26:48Z

1

Add brackets to put output in array pinIDs. Replace

declare -a pinIDs=$(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";)

by

declare -a pinIDs=( $(mysql $mysqlconnection -e "SELECT pinID FROM somewhere WHERE something = something";) )

Then see output of: declare -p pinIDs

edited Jan 2, 2016 at 10:26

answered Jan 2, 2016 at 10:23

Cyrus

90.2k15 gold badges112 silver badges173 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Frank Hayward · Accepted Answer · 2016-01-02 10:36:21Z

0

For others that are using this to learn. What I did after fixing the brackets in order to put the result into an array such as (8 11 23 26) instead of 8 11 23 26, was this:

cnt=${#pinIDs[@]}
for (( i=0 ; i<cnt ; i++ ))
do
echo "pinId: ""${pinIDs[$i]}"
done

answered Jan 2, 2016 at 10:36

Frank Hayward

1,1377 silver badges8 bronze badges

Collectives™ on Stack Overflow

Store multiple values from sql select in bash

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related