Question

我正在为我的雇主创建一个新的预订系统，其中填写一份表格，数据进入预建的MySQL数据库。

我真的不确定我做错了什么。最初数据不会发布到数据库中，但表单似乎已提交。现在，表单只是提交到白页。我将提交下面的完整页面代码，因为那里没有令人信服的数据，希望有人能够提供帮助。

<head>
<title>&nbsp;&nbsp;Moat Laptop Bookinge&nbsp;&nbsp;</title>
<?php
if (isset($_POST['submitted'])) {

    include('booking_db.php');

    $name = $_POST['name'];
    $out = $_POST['out'];
    $in = $_POST['in'];
    $sqlinsert = "INSERT INTO Future (name, out, 'in') VALUES ('$name', '$out', '$in')";

    if (!mysqli_query ($dbcon, $sqlinsert)) {
        die('error inserting new record');
            }
            $newrecord = "Laptop has been successfully Booked!";
}
?>

<link rel="stylesheet" href="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/themes/smoothness/jquery-ui.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>

 <script>
  $(document).ready(function() {
    $("#datepicker").datepicker();
  });
  </script>

   <script>
  $(document).ready(function() {
    $("#datepicker2").datepicker();
  });
  </script>
</head>

<body style="background-height: 100%;background-width: 100%;background: #141E30;background: -webkit-linear-gradient(to left, #141E30 , #243B55);background: linear-gradient(to left, #141E30 , #243B55);">
<div id="logo" style="font-family: Tw Cen MT; font-weight: Bold; position: fixed; color: white; left: 650px;top: 35px; font-size: 80px;text-shadow: 3px 3px #c7c7c7;">
Book a Laptop
</div>
<div id="content_box" style="background-color: white;position: fixed; left: 450px;top: 135px; width:60%; height: 70%; border-radius: 3px;">
<center>
<form method="post" action="book.php" style="font-family: Bodoni MT;">
<input type="hidden" name="submitted" value="true" />
         <br />
         <br />
         <b><legend>First Name and First Letter of Surname</legend></b>
         <input type="text" name="name" value="Ex. James T" />
         <br/>
         <br />

         <b><legend>When will you need to collect the device?</legend></b>
         <input id="datepicker2" name="out" />
         <br/>
         <br />

         <b><legend>When will you return the device?</legend></b>
         <input id="datepicker" name="in" />

         <br />
         <input type="submit" value="Confirm Booking" />



</center>
<?php
echo $newrecord
?>

</div>

</body>

如果您需要更多信息，请随时提出。

修改这个问题已经解决，我无法标记答案，因为这是我的回答，我必须等待2天。感谢你们所有答案。

Answer 1

行include('booking_db.php');可能存在问题。您应该在页面顶部提到error_reporting(E_ALL);，然后尝试调试：

error_reporting(E_ALL);

if (isset($_POST['submitted'])) {

    var_dump(file_exists('booking_db.php')); //check if you get true or false

    require 'booking_db.php'; // Change include to required
    echo "Test"; 
    exit;
    $name = $_POST['name'];
    echo $name; // Check
    $out = $_POST['out'];
    $in = $_POST['in'];
    $sqlinsert = "INSERT INTO Future (name, out, 'in') VALUES ('$name', '$out', '$in')";

    var_dump($dbcon); // check

    if (!mysqli_query ($dbcon, $sqlinsert)) {
        die('error inserting new record');
    }
    $newrecord = "Laptop has been successfully Booked!";
}

请在表单提交后检查您的内容。

Answer 2

将插入查询替换为：

    INSERT INTO Future 
(`name`,`out`,`in`) VALUES ('".$name."', '".$out."', '".$in."')

和 if (isset($_POST['submitted']))与if (isset($_POST['Confirm Booking']))

。{

因为您必须在POST中提交提交按钮的值

Answer 3

事实证明，我的问题是

if (!mysqli_query ($dbcon, $sqlinsert)) {
        die('error inserting new record');
            }
            $newrecord = "Laptop has been successfully Booked!";
}

我将此更改为

if (mysqli_query ($link, $sqlinsert)) {
        echo "";

} else{

echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);

}

并更改了一些变量以与之匹配，它开始工作并发布到我的数据库。感谢所有回答的人。

PHP表单未发布到MySQL数据库

3 个答案: