Question

$st = $this->db->prepare("SELECT * FROM invoices WHERE group_id=?");
$st->execute(array($id));

    if($st->rowCount() >= 1){
        foreach ($st as $row) {
            $counter = $row["paymentAmount"];
            $start = 1;

            for($start; $start < $st->rowCount(); $start++) {
                $counter = $counter + $row["paymentAmount"];
                }
        }

实际上会打印$row["paymentAmount"] + $row["paymentAmount"]等等，具体取决于$row["paymentAmount"]的数量$row["paymentAmount"]。但问题是onlyOneEl e x = fromJust (lookup x counts) == 1 where gs = group (sort e) counts = zip (map head gs) (map length gs)的最后一个输出是2500。

有： 10000 10000 2500

结果是：7500 我希望它是：22500

如果最后的结果是3000，它应该是23000.所以我只需要这个代码从数据库中获取每一行，而不是最新的一行。

编辑：我希望它在SQL查询之外

Answer 1

你不需要像这样的PHP逻辑。该功能内置于SQL中。

SELECT SUM(paymentAmount) FROM invoices WHERE group_id=?

Answer 2

您应该让数据库处理总和，除非您有正当理由需要在PHP中处理它。使用此类操作可以提高数据库的效率，避免PHP中的循环。

SELECT SUM(paymentAmount) AS TotalPaymentAmount FROM invoices WHERE group_id = ?

然后，您可以将PHP更改为仅返回一行：

$row = $st->fetch();
echo $row["TotalPaymentAmount"];

如果您需要在SQL之外进行此计算，只需更改循环：

if($st->rowCount() >= 1){
        //init the counter to 0 before you loop through your rows
        $counter = 0;
        //the foreach will iterate over your result set and add the paymentAmount to $counter.
        foreach ($st as $row) {
            $counter += $row["paymentAmount"];
        }
        //echo results outside of the loop
        echo $counter;
}

Answer 3

如果您需要故意编写外部SQL代码（例如，因为您需要对每一行进行进一步处理），那么我将按如下方式对其进行编码：

if ($st->rowCount() >= 1) {
    $counter = 0;
    foreach ($st as $row) {
        $counter += $row["paymentAmount"];
    }
}

来自数据库的PHP循环和

3 个答案: