Question

我正在NGO数据库中工作，那里有很多表和许多列，现在我停留在每种性别类型（m =男性，w =女性，c =儿童）的计算/总缴费额上

以下是我的表及其列

Table1 "basic_detail"
+---+------------------+------+
|id | name             | m_w_c 
----+------------------+------+
|1  |jhon              |m
|2  |sara              |w
|3  |mike              |c
|4  |simon             |m
|5  |frank             |c
|6  |suzi              |w
"and so on more than 1000 entries"

Table2 "contribution" (c_id= contribution table_id auto incremental) (bd_id= basic_detail table_id)
+------+------+--------+-----------+-----------+
|c_id  |bd_id | books  | uniform   | food      |
+------+------+--------+-----------+-----------+
|1     | 1    | 50     | 40        | 0         |
|2     | 3    | 0      |  0        | 5         |
|3     | 2    | 50     | 45        | 0         |
|4     | 5    | 2      | 4         | 3         |
|5     | 4    | 60     | 75        | 0         |
|6     | 6    | 35     | 50        | 10        |
and so on more than 1000 entries equals to basic_detail table

我想分别计算男人，女人和孩子的捐款，

如果孩子只能贡献食物，如果孩子贡献书籍和校服，则此贡献应计入c_contribution，

其中m = men，w = women仅能贡献书籍和校服食品中，以下金额应计入c_contribution，

如果您不满意，请大家在我的问题上帮助我，请不要标记“-”，谢谢。

Answer 1

像这样吗？

$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");

Answer 2

此代码可帮助我查看各个领域中男性，女性，儿童贡献的总和，但是我仍然无法添加男性或女性在食物中贡献的总和。

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");
?>

请...

在table1列的比较中计算table2列的总数

2 个答案: