bash - SQL Query Outputs to variable

Taken from bash script - select from database into variable, you can read the query result into a variable.

Example

mysql> SELECT * FROM domains;+-------+---------+| user  | domain  |+-------+---------+| user1 | domain1 || user2 | domain2 || user3 | domain3 |+-------+---------+

Usage

$ myvar=$(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")$ echo $myvardomain1 domain2 domain3

echo is the bash command for output. You can then split $myvar into separate variables:

$ read var1 var2 var3 <<< $myvar$ echo $var1domain1$ echo $var2domain2

You can combine these two commands into a single one:

read var1 var2 var3 <<< $(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")

It is possible to store the results into arrays (useful if you don't know how many records there):

$ read -ra vars <<< $(mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains")$ for i in "${vars[@]}"; do$     echo $i$ donedomain1domain2domain3

Another way of doing is:

dbquery=`mysql -D$MYDB -u$MYUSER -p$MYPASS -se "SELECT domain FROM domains"`      dbquery_array=( $( for i in $dbquery ; do echo $i ; done ) )

The first line stores all the output from the query in a varriable dbquery in a array-like-way. The second line converts the dbquery into an array dbquery_array with a simple for loop.

I did this

variable=mysql -u root -ppassworrd database << EOFselect MAX(variable) AS a from table where variable2 = 'SOMETEXT' AND day(datevalue) >= 22;EOF

I hope it helps