Question

我一直在努力工作几个小时，包括重建我的整个代码，我无法弄清楚出了什么问题。 HTML表单应该填充患者SQL数据库，但它不起作用。我有一个类似的形式，我在网站的另一部分使用完美无瑕地工作，但这似乎似乎并没有找出原因。

感谢您的帮助

以下是数据库字段的图片： Database fields

下面是代码：

<?php

//session_start();

//if( isset($_SESSION['user_id']) ){
//  header("Location: /");
//}

require 'connection.php';

$message = '';

if(!empty($_POST['name']) && !empty($_POST['dob']) && !empty($_POST['nok']) && !empty($_POST['nok_add']) && !empty($_POST['nok_phone'])&& !empty($_POST['doa'])&& !empty($_POST['allergies'])):
  
  // Enter the new user in the database
  $sql = "INSERT INTO patients (name, dob, nok, nok_add, nok_phone, doa, allergies) VALUES (:name, :dob, :nok, :nok_add, :nok_phone, :doa, :allergies)";
  $stmt = $conn->prepare($sql);

  $stmt->bindParam(':name', $_POST['name']);
  $stmt->bindParam(':dob', $_POST['dob']);
  $stmt->bindParam(':nok', $_POST['nok']);
  $stmt->bindParam(':nok_add', $_POST['nok_add']);
  $stmt->bindParam(':nok_phone', $_POST['nok_phone']);
  $stmt->bindParam(':doa', $_POST['doa']);
  $stmt->bindParam(':allergies', $_POST['allergies']);

  if( $stmt->execute() ):
    $message = 'Successfully created new user';
  else:
    $message = 'Sorry there must have been an issue creating your account';
  endif;

endif;



?>


<html>
  <head>
    <title>Patient Add</title>
    <meta name="viewport" content="width=device-width, initial-scale=1">
      <link rel="stylesheet" type="text/css" href="assets/css/style.css">
      <link href="//fonts.googleapis.com/css?fami=Roboto:400,300,200,100&subset=latin,cyrillic" rel="stylesheet">
<style type="text/css">      
        #headline {
        padding: 0.8em;
        color: white;
        font-family: Roboto, helvetica, arial, sans-serif;
        background-color: black;
        background-image: url(backgroundimage.jpg);
        background-size: cover;
      } 

</style>

  </head>
  <body>

    <div id="headline">
      <div class="container">
        <header>
          <h1>Patient Add</h1>
          <p></p>
        </header>



    <button onClick="window.location='manager_select.php';" class="_button">Back</button>
    <form method="POST" id="add">
          
      <h2>Enter the Patient's Details</h2>

      <label for="name">Name:</label>
      <input type="text" name="name">

      <label for="dob">Date of Birth:</label>
      <input type="text" name="dob">

      <label for="nok">Next of Kin:</label>
      <input type="text" name="nok">

      <label for="nok_add">Address:</label>
      <input type="text" name="nok_add">

      <label for="nok_phone">Phone:</label>
      <input type="text" name="nok_phone">

      <label for="doa">Date of Admission:</label>
      <input type="text" name="dob">

      <label for="allergies">Allergies:</label>
      <input type="text" name="allergies">

      <?php if(!empty($message)): ?>
        <p><?= $message ?></p>
      <?php endif; ?>
      <input type="submit">

    </form>
        <br>
      </div>
    </div>

  <!-- // [START footer] -->
      <footer>
        <div class="container">
          <p>Copyright Ghyll Court Residental Home 2017</p>
        </div>
      </footer>
      <!-- // [END footer] -->

  </script>
      <script type="text/javascript">
        function init() {
          window.matchMedia("(max-width: 600px)").addListener(hitMQ);
        }

        function hitMQ(evt) {
          sampleCompleted("GettingStarted-ContentWithStyles");
        }

        init();

      </script>

  </body>
</html>

Answer 1

正如您在duplicate post上所回答的那样（此处也是为了完整性而回答），您正在检查if语句中的“doa”字段，该字段不存在。您的表单定义了“dob”两次：一次是出生日期，一次是入学日期：

<label for="dob">Date of Birth:</label>
<input type="text" name="dob">

...

<label for="doa">Date of Admission:</label>
<input type="text" name="dob">

Answer 2

我不完全确定这是最安全的方式，但是为什么不尝试简单的mysqli_query($conn,$sql);

而不是使用MSQLI PDO

Html / Php表单不添加到SQL数据库

2 个答案: