Loop through result set of mysql in Bash

Question

I have a simple bash script. I wish to get an exact count of the number of rows in each table of the database.

#!/bin/bash

TABLES_OLD=$( mysql -u user -ppassword MySchema --batch --skip-column-names -e"SHOW TABLES FROM MySchema" )

for table in "${TABLES_OLD[@]}"
do
    QUERY="SELECT COUNT(*) FROM ${table}"
    echo "${QUERY}"
done

The script prints:

SELECT COUNT(*) FROM Table 1
Table2
Table3
Table4
etc...

Clearly this is not what I want, and I don't even understand how what is happening is possible. What am I doing wrong?

Answer 1

Try this, put the tables into an array then loop thru the results
db_host='host'
db_user='user'
db_pass='password'
db='your_db'

read -ra var_id <<< $(mysql -h $db_host -u $db_user -p$db_pass $db -sse "show tables from $db")
for i in "${var_id[@]}"; 
    do
results=$(mysql -h $db_host -u $db_user -p$db_pass $db -sse "select count(*)from $i")
echo "$i $results"

done

Answer 2

This should do it :

#/bin/bash
mysql -u user-ppassword -e "SELECT table_name, table_rows
                     FROM INFORMATION_SCHEMA.TABLES
                     WHERE TABLE_SCHEMA = 'your_data_base_name';"

Answer 3

Replace echo with eval

The code will be

#!/bin/bash

TABLES_OLD=$( mysql -u user -ppassword MySchema --batch --skip-column-names -e"SHOW TABLES FROM MySchema" )

for table in "${TABLES_OLD[@]}"
do
    QUERY="SELECT COUNT(*) FROM ${table}"
    eval "${QUERY}"
done