How to store results of multiple sql queries in shell variables in Bash?

Question

I need to store in array the results retrieved from the database. So far this is basically what I'm trying with no success.

 Myarray=$(echo "SELECT A, B, C FROM table_a" | mysql database -u $user -p$password)

My understanding of bash commands is not very good as you can see. thank you.

Answer 1

This works:

#!/bin/bash
results=($(mysql --user root -pwelcome ts -Bse "SELECT type, network_id, subnet_msk FROM remote_subnet;"))

cnt=${#results[@]}
for (( i=0 ; i<cnt ; i++ ))
do
    echo "Record No. $i: ${results[$i]}"
    fieldA=${results[0]};
    fieldB=${results[1]};
    fieldC=${results[2]};
done

Answer 2

The following worked for me:

results=($(executeQuery "SELECT fieldA, fieldB FROM tableName;")) count=${#results[@]} echo $count for (( i=2 ; i<count ; i++ )) do fieldA=${results[$i]}; echo $id i=$((i+1)) fieldB=${results[$i]}; echo $fieldB done

Answer 3

#! bin/bash
query="select * from test;"

myarray=$(mysql -h hostname -u user -ppass database << eof
$query
eof)

echo $myarray