подсчитать сумму столбца table2 в сравнении столбца table1

Как-то так?

$men = $women = child = c_contribution = 0;

$contributions = $pdo->query("
    SELECT c.books, c.uniform, c.food, bd.m_w_c
    FROM contribution c
    LEFT JOIN basic_detail bd ON c.bd_id = bd.id
")->fetchAll();

foreach($contributions as $contribution) {
    if ( $contribution['m_w_c'] == 'c' ) {
        $c_contribution += $contribution['books'];
        $c_contribution += $contribution['uniform'];
        $child += $contribution['food'];
    } else {
        $men += $contribution['books'];
        $women += $contribution['books'];
        $men += $contribution['uniform'];
        $women += $contribution['uniform'];
        $c_contribution += $contribution['food'];
    }
}

print("Men Total: $men");
print("Women Total: $women");
print("Child Total: $child");
print("Combined Total: $c_contribution");

Этот код помогает мне просматривать сумму вкладов мужчин, женщин, детей в каждой области, но я все еще не могу ДОБАВИТЬ вклад мужчин или женщин в питание в сумме детей.

require "connpdo.php";
        $connection = new PDO($dsn, $username, $password, $options);
$men = $women = $child = $c_contribution = 0;

$contributions = $connection->query("
    SELECT c.books, c.uniform, c.food, bd.m_w_c
    FROM contribution c
    LEFT JOIN basic_detail bd ON c.bd_id = bd.id
")->fetchAll();

foreach($contributions as $contribution) {
    if ( $contribution['m_w_c'] == 'w' ) {
        $women += $contribution['books'];
        $women += $contribution['uniform'];
    } elseif ( $contribution['m_w_c'] == 'm' ) {
        $men += $contribution['books'];
        $men += $contribution['uniform'];
    } elseif ( $contribution['m_w_c'] == 'c' ) {
        $child += $contribution['books'];
        $child += $contribution['uniform'];
        $child += $contribution['food']; 
}
}
print("Men Total: $men");
print("Women Total: $women");
print("Child Total: $child");
print("Combined Total: $c_contribution");
?> 

Пожалуйста ...