Question

我知道这种问题已经回答，但仍然不能解决我的问题。我有2张桌子要插入。第一个表插入工作正常。但是第二个没有。 PHP不会引发任何错误。可能是我的MySQL查询未正确编写。

我需要尽快修复此问题。感谢您的任何帮助。

$numb = 0; $af_f = 0; $fn = 0; $an = 0; $nt = 0; $intn = 0; $med_name = "";

$q2 = "INSERT INTO prescriptions (mr_id, mb_id, number, af_food, fn, an, nt) ";
$q2 .= "SELECT ?, MB.id, ?, ?, ?, ?, ? FROM med_batch MB INNER JOIN medicines M ON M.id = MB.med_id ";
$q2 .= "WHERE M.name = ? AND DATE(CURDATE()) < DATE_SUB(DATE(MB.exp_date), INTERVAL ? DAY) ";
$q2 .= "AND MB.stock_num > ? ORDER BY MB.arr_date LIMIT 1";
$q2 = $conn->prepare($q2) or die($conn->error);
$q2->bind_param("diiiiisii", $id, $numb, $af_f, $fn, $an, $nt, $med_name, $intn, $numb) or die($q->error);


$n = 1;
while(isset($_POST["mr-med" . $n]))
{
    $intn = 0;
    $med_name = $_POST["mr-med" . $n];
    $numb = $_POST["mr-med-num" . $n];
    $af_f = (isset($_POST["mr-af" . $n]))? 1 : 0;
    $fn = (isset($_POST["mr-fn" . $n]))? 1 : 0;
    $an = (isset($_POST["mr-an" . $n]))? 1 : 0;
    $nt = (isset($_POST["mr-nt" . $n]))? 1 : 0;
    $intn = $numb / ($fn + $an + $nt);
    $intn = ceil($intn);

    $q2->execute() or die($q2->error);
    $n = $n + 1;
}
$r = $q2->store_result();

if($q2->affected_rows > 0)
{
    echo "Successfully Added";
}
else
{
    echo "Error";
}

在PHP mysqli预准备语句中的INSERT INTO SELECT中合并变量

0 个答案: