在 php java 脚本中打开新登陆页面自动填写表单值

Question

我在 php html 中有一个小表单，有五个字段。名字、姓氏、班级、年龄、分数。我想通过单击记录登录页面上的添加按钮来获取前三个字段（名字、姓氏、班级）的值。 （登陆页面有这些字段 FirstName、Lastname、class、Add 按钮）

表格代码

<form method=post, action="">
  Firstname <Input type=text name=fname, id=fname, value="">
  Lasttname <Input type=text name=lastname, id=lastname, value="">
  Class <Input type=text name=class, id=class, value="">
  Age <Input type=text name=Age, id=Age, value="">
  Marks <Input type=text name=marks, id=marks, value="">
</form>

登陆页面代码

// Attempt select query execution
$sql = "SELECT * FROM mis1";
if ($result = mysqli_query($link, $sql)) {
  if (mysqli_num_rows($result) > 0) {
    echo '<table class="styled-table">';
    echo "<thead>";
    echo "<tr>";

    echo "<th>FirstName</th>";
    echo "<th>LastName</th>";
    echo "<th>Class</th>";
    echo "<th>Action</th>";
    echo "</tr>";
    echo "</thead>";
    echo "<tbody>";
    while ($row = mysqli_fetch_array($result)) {
      echo "<tr>";
      echo "<td>".$row['fname'].
      "</td>";
      echo "<td>".$row['lastname'].
      "</td>";
      echo "<td>".$row['class'].
      "</td>";
      echo "<td>";
      echo '<a href="ua1.php?id='.$row['id'].
      '" class="mr-3" title="Update Record" data-toggle="tooltip">Add Button</a>';

      echo "</td>";
      echo "</tr>";
    }
    echo "</tbody>";
    echo "</table>";
    // Free result set
    mysqli_free_result($result);
  } else {
    echo '<div class="alert alert-danger"><em>No records were found.</em></div>';
  }
} else {
  echo "Oops! Something went wrong. Please try again later.";
}

// Close connection
mysqli_close($link); 
?>