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php - PHP 更新脚本出错,未定义索引 : employee_id

转载 作者:行者123 更新时间:2023-11-29 23:38:04 24 4
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好的,我正在使用 php 脚本从 html 进行更新。首先,我选择员工 ID,然后以 html 表单获取结果,然后更新它,但我每次都会收到此错误:/

这是我的代码:

首先选择员工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>

这是我的 update_employee.php 脚本

<?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();
?>

最佳答案

您的employee_id 输入被标记为禁用,因此不会发送到表单处理程序。您的意思是将其标记为 type="hidden"而不是禁用它吗?

关于php - PHP 更新脚本出错,未定义索引 : employee_id,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26311533/

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