PHP将语句插入XAMPP MySQL数据库不会导致表更新[重复]

Question

这个问题在这里已有答案：

mysqli_fetch_assoc() expects parameter 1 to be mysqli_result or how to get MySQLi error information 1回答

我正在尝试编写一个程序，它将从html表单中获取信息，并将其上传到我本地主机上存在的mysql数据库（使用XAMPP）。为此，我使用HTML创建了一个表单，该表单将接收要插入数据库的数据，如下所示：

<!DOCTYPE html>
<html lang="en">
    <head>
        <link rel="stylesheet" href="styles/main.css">
        <meta charset="utf-8">
        <title>Presentation Registration</title>
        <script type="text/javascript">
        function Submit()
        {
            // I put input validation stuff here, but left it out to simplify code for reading
            document.getElementById("myForm").submit();
        }
    </script>
</head> 

<body>
    <form id="myForm" action="actionPage.php" method="post">
    <TABLE>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot1" value="Slot1">12/9/15, 6:00 PM – 7:00 PM</TD>
            <TD>
            <?php 
            $con = mysqli_connect('localhost','username','password', 'myDB');

            $query = "SELECT * FROM reserve";

            $result = mysqli_query($con, $query);

            $row = mysqli_fetch_array($result);
            echo $row["Open Seats"]." ";
            ?>
            Seats Remaining</TD>
        </TR>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot2" value="Slot2">12/9/15, 7:00 PM – 8:00 PM</TD>
            <TD><?php
                $row = mysqli_fetch_array($result);
                echo $row["Open Seats"]." ";
                ?>
                Seats Remaining</TD>
        </TR>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot3" value="Slot3">12/9/15, 8:00 PM – 9:00 PM</TD>
            <TD><?php
                $row = mysqli_fetch_array($result);
                echo $row["Open Seats"]." ";
                ?>
                Seats Remaining</TD>
        </TR>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot4" value="Slot4">12/10/15, 6:00 PM – 7:00 PM</TD>
            <TD><?php
                $row = mysqli_fetch_array($result);
                echo $row["Open Seats"]." ";
                ?>
                Seats Remaining</TD>
        </TR>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot5" value="Slot5">12/10/15, 7:00 PM – 8:00 PM</TD>
            <TD><?php
                $row = mysqli_fetch_array($result);
                echo $row["Open Seats"]." ";
                ?>
                Seats Remaining</TD>
        </TR>
        <TR>
            <TD><input type="radio" name="TimeSlot" class="TimeSlot" id="Slot6" value="Slot6">12/10/15, 8:00 PM – 9:00 PM</TD>
            <TD><?php
                $row = mysqli_fetch_array($result);
                echo $row["Open Seats"]." ";
                ?>
                Seats Remaining</TD>
        </TR>
    </TABLE>
    <button type="button">Student List</button>
    <TABLE>
        <TR>
            <TD>UMID&nbsp</TD>
            <TD><input type="text" class="inputText" id="UMID" name="UMID"></TD>
        </TR>
        <TR>
            <TD>First Name&nbsp</TD> 
            <TD><input type="text" class="inputText" id="fName" name="fName"></TD>
        </TR>
        <TR>
            <TD>Last Name&nbsp</TD>
            <TD><input type="text" class="inputText" id="lName" name="lName"></TD>
        </TR>
        <TR>
            <TD>Project Title&nbsp</TD>
            <TD><input type="text" class="inputText" id="projectTitle" name="projectTitle"></TD>
        </TR>
        <TR>
            <TD>E-Mail&nbsp</TD>
            <TD><input type="text" class="inputText" id="email" name="email"></TD>  
        </TR>
        <TR>
            <TD>Phone Number&nbsp</TD>
            <TD><input type="text" class="inputText" id="phone" name="phone"></TD>
        </TR>
    </TABLE>
    <button type="button" OnClick=Submit()>Submit</button>
    </form>

</body>
</html>

如代码所示，我在提交函数上使用动作标记来触发“actionPage.php”运行。该程序内部的代码如下所示：

<?php
    $con = mysqli_connect('localhost','username','password', 'myDB');
    $umid = $_POST['UMID'];
    $fName = $_POST['fName'];
    $lName = $_POST['lName'];
    $projTitle = $_POST['projectTitle'];
    $eMail = $_POST['email'];
    $phone = $_POST['phone'];
    $slotID = "1";


    $query = "INSERT INTO student (UMID, fName, lName, projTitle, e-Mail, phone#, SlotID) VALUES ($umid, $fName, $lName, $projTitle, $eMail, $phone, $slotID)";

    $result = mysqli_query($con, $query);

    echo "Database Insertion Complete";
?>

所有这些都连接到一个包含两个表（student和reserve）的数据库。当我运行代码并输入信息时，actionPage.php会触发并运行所有代码，包括最后的echo语句。但是，当我查看XAMPP时，学生表不会更新。我的代码有问题，还是我应该寻找其他问题？

Answer 1

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