Php 为什么标题在这段代码中不起作用？

执行查询后到达标头的php代码

<?php
if (isset($_POST['Submit1'])) {
    $con = mysqli_connect("localhost:3306", "root", "", "travels");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $fname1 = $_POST['fname'];
    $lname1 = $_POST['lname'];
    $pnum1 = $_POST['pnum'];
    $email1 = $_POST['email'];
    $fcode = $_POST['fcode'];
    $sql = "insert into customer_info(fname,lname,pnumber,email) values ('$fname1','$lname1','$pnum1','$email1')";
    $sql1 = "insert into booking_info(fname,lname,pnumber,email,f_code) values ('$fname1','$lname1','$pnum1','$email1','$fcode')";
    $sql2 = "update flight_info set seats_available=seats_available-1 where flight_code='$fcode'";
    mysqli_query($con, $sql);
    mysqli_query($con, $sql1);
    mysqli_query($con, $sql2);
    header("Location: Booking_confirm.php");
}
?>

---

确保header（）函数之前没有空格、html标记或任何输出

上传整个代码结构，以便我们可以帮助您找到解决方案。您有两个嵌套表单，第一个表单包含所有字段，而第二个嵌套表单仅包含提交按钮和一个
无
字段

替换

<form action="Booking_confirm.php" method="POST" target="_blank">
  <button type="Submit" name="Submit1" class="btn btn-primary">Book</button>
</form>

与



取决于“标题”/代码的放置位置
欢迎。我们这里说的是什么“头”？你有两个嵌套表单，为什么是两个？仅供参考，您的第二个表单（
action=“Booking\u confirm.php”
）对您在第一个表单（
action=“Booking.php”
）中输入的表单字段一无所知@kerbhloz我所说的标题在php代码中，我希望执行三个sql查询并转到Booking\u confirm.php
<form action="Booking_confirm.php" method="POST" target="_blank">
  <button type="Submit" name="Submit1" class="btn btn-primary">Book</button>
</form>

<button type="Submit" name="Submit1" class="btn btn-primary">Book</button>

<form action="Booking.php" method="POST">

<form action="Booking_confirm.php" method="POST" target="_blank">