如何填充表单:使用onclick图像的数据库记录

<form action="submit.php" method="post">
  <input type="text" name="name" placeholder="姓名">
  <input type="email" name="email" placeholder="电子邮箱">
  <input type="tel" name="phone" placeholder="电话号码">
  <input type="hidden" id="imagePath" name="imagePath">
  <input type="submit" value="提交">
</form>
<input type="file" id="imageUpload">

<script>
  document.getElementById('imageUpload').addEventListener('change', function(e) {
    var file = e.target.files[0];
    var reader = new FileReader();
    reader.onload = function(e) {
      document.getElementById('imagePath').value = e.target.result;
    }
    reader.readAsDataURL(file);
  });
</script>

// 建立数据库连接
$servername = "数据库服务器地址";
$username = "数据库用户名";
$password = "数据库密码";
$dbname = "数据库名";

$conn = new mysqli($servername, $username, $password, $dbname);

// 处理表单提交
$name = $_POST['name'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$imagePath = $_POST['imagePath'];

// 获取onclick图像的数据记录
$sql = "SELECT * FROM form_data WHERE image='$imagePath'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
  // 填充表单字段
  $row = $result->fetch_assoc();
  $name = $row['name'];
  $email = $row['email'];
  $phone = $row['phone'];
}

// 在此处执行保存表单数据到数据库的代码

// 关闭数据库连接
$conn->close();