在PHP更新腳本中出現錯誤，未定義索引：employee_id

Question

好的，我正在使用PHP腳本從html進行更新。首先我選擇僱員ID，然後在之後的HTML表單我更新它取結果，但每次我收到此錯誤：/

這裏是我的代碼：

首先選擇員工ID

 <?php 
    if (isset($_GET['submit'])) { 
$con=mysqli_connect("localhost","root","","hct_db"); 
    $employee_id = mysqli_real_escape_string($con,$_GET['employee_id_db']); 

// Check connection 
if (mysqli_connect_errno()) { 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$result = mysqli_query($con,"SELECT * FROM employee where employee_id = '".$employee_id."'"); 
while($row = mysqli_fetch_array($result)) { ?> 



        <div class="panel-body"> 
         <div class="row"> 
          <div class="col-lg-6"> 
           <form role="form" action="update_employee.php" method="post"> 
            <fieldset disabled> 
             <div class="form-group"> 
              <label for="disabledSelect">Employee ID</label> 
              <input class="form-control" id="disabledInput" name="employee_id" type="text" value="<?php echo(htmlspecialchars($row['employee_id'])); ?>" disabled> 
             </div></fieldset> 
            <div class="form-group"> 
             <label>First Name</label> 
             <input class="form-control" name="first_name" value="<?php echo(htmlspecialchars($row['first_name'])); ?>" /> 
            </div> 
            <div class="form-group"> 
             <label>Last Name</label> 
             <input class="form-control" name="last_name" value="<?php echo(htmlspecialchars($row['last_name'])); ?>"> 
            </div> 
            <div class="form-group"> 
             <label>Username</label> 
             <input class="form-control" name="username" value="<?php echo(htmlspecialchars($row['username'])); ?>"> 
            </div> 
            <div class="form-group"> 
             <label>Password</label> 
             <input class="form-control" name="password" value="<?php echo(htmlspecialchars($row['password'])); ?>"> 
            </div> 
            <div class="form-group"> 
             <label>Date Of Joining</label> 
             <input type="date" class="form-control" name="date_of_join" value="<?php echo(htmlspecialchars($row['date_of_join'])); ?>"> 
            </div> 
            <div class="form-group"> 
             <label>Email ID</label> 
             <input class="form-control" type="email" name="email_id" value="<?php echo(htmlspecialchars($row['email_id'])); ?>"> 
            </div> 
            <div class="form-group"> 
             <label>Address</label> 
             <input class="form-control" name="address" value="<?php echo(htmlspecialchars($row['address'])); ?>">          </div> 
            <div class="form-group"> 
             <label>Contact No.</label> 
             <input class="form-control" name="contact_no" value="<?php echo(htmlspecialchars($row['contact_no'])); ?>"> 
            </div> 


            <button type="submit" class="btn btn-default">Submit</button> 
            <button type="reset" class="btn btn-default">Reset</button> 
           </form> 
          </div> 

這裏是我的update_employee.php腳本

<?php 
session_start(); 
ob_start(); 

$con=mysqli_connect("localhost","root","","hct_db"); 
// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

// escape variables for security 
$first_name = mysqli_real_escape_string($con, $_POST['first_name']); 
$last_name = mysqli_real_escape_string($con, $_POST['last_name']); 
$address = mysqli_real_escape_string($con, $_POST['address']); 
$contact_no = mysqli_real_escape_string($con, $_POST['contact_no']); 
$employee_id = mysqli_real_escape_string($con, $_POST['employee_id']); 
$username = mysqli_real_escape_string($con, $_POST['username']); 
$password = mysqli_real_escape_string($con, $_POST['password']); 
$date_of_join = mysqli_real_escape_string($con, $_POST['date_of_join']); 
$email_id = mysqli_real_escape_string($con, $_POST['email_id']); 

$qq= mysqli_query($con,"UPDATE employee SET first_name='$first_name', last_name='$last_name', address='$address', contact_no='$contact_no', username='$username', password='$password', date_of_join='$date_of_join', email_id='$email_id' 
WHERE employee_id = '".$employee_id."'"); 
    if ($qq) { 
      echo 'Updated Profile!'; 
     }else 
      echo 'Failed to update your Profile.'; 

    mysqli_close($con); 
    ob_end_flush(); 
?> 

Answer 1

你的僱員標識輸入被標記爲禁用，因此不發送ŧ o表單處理程序。你的意思是把它標記爲type =「hidden」而不是禁用它？