javascript - <select> 标签未发布到 MySql 数据库

Question

我目前正在开展一个副业项目来进行门票销售。我创建了一个简单的网页，其中包含我需要收集的信息的基础知识

<div id="session" class="tabcontent">
    <form action="test.php"  method="post" oninput="    totalamount.value = Math.round(adult.value * 10) + Math.round(student.value * 7); changedue.value = Math.round(moneygiven.value - totalamount.value);" id="ticketform"> 
        <fieldset>
            <select name=“ticketform” id="ticketform" form="ticketform">
                <option value="" disabled="disabled" selected="selected">Performance</option>
                <option value=“1“ type="number" name="showtime1">Show 1 - May 9th 2017</option>
                <option value=“2“ type="number" name="showtime2">Show 2 - May 10th 2017</option>
                <option value=“3” type="number" name="showtime3">Show 3 - May 11th 2017</option>
             </select>
            <h4>Ammount of Adults</h4>
            <input name="adult" id="adult" type="number">
            <br />
            <h4>Ammount of Students</h4>
            <input name="student" id="student" type="number">
            <br />
            <h4>Money Owed</h4>
            <output name="totalamount" id="totalamount" for="adult student"></output>
            <br />
            <h4>Money Given</h4>
            <input name="moneygiven" id="moneygiven" type="number">
            <h4>Change due</h4>
            <output name="changedue" id="changedue" for="moneyowed moneygiven"></output>
            <input type="submit" />
        </fieldset>
    </form>
    <footer>

    </footer>

但是我遇到了一些问题，作为输出的字段和标签选项不会发送到我的数据库。

$servername = "localhost";
$username = "root"; 
$password = "root";
$dbname = "TicketSales"; 


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
$showtime=$_POST['showtime'];
$adult=$_POST['adult'];
$student=$_POST['student'];
$totalamount=$_POST['totalamount'];
$moneygiven=$_POST['moneygiven'];
$changedue=$_POST['changedue'];
// Insert data into database

$sql="INSERT INTO tickets (showtime, adult, student, totalamount, moneygiven, changedue)
VALUES
('$showtime', '$adult', '$student', '$totalamount', '$moneygiven', '$changedue')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

这是我的发送表单。一切都是链接的，任何带有输入标签的东西都可以正常工作。我怎样才能让它也发送选择值和输出？

Answer 1

您正在使用 POST 方法，因此每当表单提交时 $_POST 数组都会设置，该数组使用元素名称作为键，使用元素内的值作为该键的值。

因此，当您使用 name=“ticketform” 时，$_POST 将设置为 $_POST['ticketform'] = selected_value

<select name=“ticketform” id="ticketform" >
        <option value="" disabled="disabled" selected="selected">Performance</option>
        <option value=“1“>Show 1 - May 9th 2017</option>
        <option value=“2“ >Show 2 - May 10th 2017</option>
        <option value=“3”>Show 3 - May 11th 2017</option>
 </select>

$showtime=$_POST['ticketform'];