Question

我有一个带输入框的html页面。我试图基于该输入从数据库中检索数据，并使用php将其发送回原始的html页面。我在外部文件中创建了php，它在浏览器中显示正确的值，但我的问题是尝试在html页面上显示php。

我启用了我的服务器来解析html文件并将php放在html文件中但仍然无法正常工作。当我输入一个值并提交页面重新加载但没有显示任何内容时。

如果有人知道这样做的最佳方式，我将不胜感激。我已经阅读了很多线程，但似乎都没有。

PHP：

<?php

define('DB_NAME', 'database');
define('DB_USER', 'user');
define('DB_PASSWORD', 'password');
define('DB_HOST', 'localhost');

$conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if (!$conn) {
     die('Could not connect: ' . mysqli_connect_error());
     }

$studentnum = $_POST['studentnum'];     

$sql = "SELECT * FROM test WHERE number LIKE '%$studentnum%'"; 
$result=mysqli_query($conn, $sql);

echo '<table class="table table-striped table-bordered table-hover">'; 
echo "<tr>
      <th>Name</th>
      <th>Number</th>
      <th>Floor</th>
      <th>Room</th>
      <th>Message</th>
      </tr>"; 
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
  echo "<tr><td>"; 
  echo $row['name'];
  echo "</td><td>"; 
  echo $row['number'];
  echo "</td><td>";   
  echo $row['floor'];
  echo "</td><td>"; 
  echo $row['room'];
  echo "</td><td>";   
  echo $row['message'];
  echo "</td></tr>";  

}
echo "</table>";  

mysqli_close($conn);
?>

HTML：

<form action="test.php" method="post">
            <label>Enter Student Number:</label>
            <input name="studentnum" type="number" placeholder="Type Here">
                <br>
                <br>
            <input type="submit" value="Enter">

    </form>

Answer 1

如果你想在同一页面上全部使用，你可以使用下面的代码，假设你的html文件中的解析php工作正常，你的页面是index.html（在表单操作中指定）。

<?php

// check if the form has been submitted and display the results
if (isset($_POST['studentnum'])) {

  define('DB_NAME', 'database');
  define('DB_USER', 'user');
  define('DB_PASSWORD', 'password');
  define('DB_HOST', 'localhost');

  $conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
  if (!$conn) {
    die('Could not connect: ' . mysqli_connect_error());
  }

  // escape the post data to prevent injection attacks
  $studentnum = mysqli_real_escape_string($conn, $_POST['studentnum']);

  $sql = "SELECT * FROM `test` WHERE `number` LIKE '%$studentnum%'"; 
  $result=mysqli_query($conn, $sql);

  // check if the query returned a result
  if (!$result) {
      echo 'There are no results for your search';
  } else {
    // result to output the table
    echo '<table class="table table-striped table-bordered table-hover">'; 
    echo "<tr>
          <th>Name</th>
          <th>Number</th>
          <th>Floor</th>
          <th>Room</th>
          <th>Message</th>
          </tr>"; 
    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
    {
      echo "<tr><td>"; 
      echo $row['name'];
      echo "</td><td>"; 
      echo $row['number'];
      echo "</td><td>";   
      echo $row['floor'];
      echo "</td><td>"; 
      echo $row['room'];
      echo "</td><td>";   
      echo $row['message'];
      echo "</td></tr>";  
    }
    echo "</table>";
  }

  mysqli_close($conn);
} // end submitted
else
{
// not submitted to output the form
?>
<form action="index.html" method="post">
  <label>Enter Student Number:</label>
  <input name="studentnum" type="number" placeholder="Type Here">
  <br>
  <br>
  <input type="submit" value="Enter">
</form>
<?php
} // end not submitted
?>

显示从数据库中提取的数据，基于html表单输入并在html页面中显示

1 个答案: