我正在创建一个基于CMS的项目,在这个CMS中,我为管理员创建了一个页面来添加另一个管理员。所以我将其编码为:
<form role="form" method="POST" action="">
<div class="box-body">
<div class="form-group">
<label>User name</label>
<input type="text" class="form-control" placeholder="Enter username" name="uname" required>
</div>
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" id="exampleInputEmail1" placeholder="Enter email" name="email" required>
</div>
<div class="form-group">
<label for="exampleInputPassword1">Temporary password</label>
<input type="password" class="form-control" id="exampleInputPassword1" placeholder="Enter password" name="pass" required>
</div>
<div class="form-group">
<label>Group admin</label>
<select class="form-control" name="groups">
<option value="MainAdmin">Main Admin</option>
<option value="Administrator">Administrator</option>
<option value="ContentCreator1">Shop Content Creator</option>
<option value="ContentCreator2">Blog Content Creator</option>
<option value="SocialMediaManager">Social Media Manager</option>
<option value="Analyst">Analyst</option>
</select>
</div>
</div>
<div class="box-footer">
<button name="submit" type="submit" class="btn btn-primary">Submit</button>
</div>
</form>
如果提交的表单将以操作方式运行:
<?php
if (isset($_POST['submit'])){
$username = $_POST['uname'];
$email = $_POST['email'];
$password = $_POST['pass'];
$groups = $_POST['groups'];
if($groups == "MainAdmin"){
$level = 0;
}else if($groups == "Administrator"){
$level = 1;
}else if($groups == "ContentCreator1"){
$level = 2;
}else if($groups == "ContentCreator2"){
$level = 3;
}else if($groups == "SocialMediaManager"){
$level = 4;
}else if($groups == "Analyst"){
$level = 5;
}
else{
$level = Null;
}
if (filter_var($email, FILTER_VALIDATE_EMAIL) === false) {
$notice['email_validation'] = "The email that you have entered is not a valid one";
}else{
$registration = new Register();
$notice[] = $registration->CheckUname($username,$email,$password,$groups,$level);
}
}
?>
这是名为Register.class.php
的类:
<?php
class Register
{
protected $notice = array();
private $db;
public function __construct()
{
$this->db = new Connection();
$this->db = $this->db->dbConnect();
}
public function CheckUname($username,$email,$password,$groups,$level)
{
if(!empty($username)&&!empty($email)&&!empty($password)&&!empty($groups)&&!empty($level))
{
$chk1 = $this->db->prepare("SELECT user_name FROM admins WHERE user_name = ?");
$chk1->bindParam(1,$username);
$chk1->execute();
if($chk1->rowCount() == 1)
{
$notice['username_exists'] = "Try different username";
return $this->notice;
}else{
$chk2 = $this->db->prepare("SELECT email_address FROM admins WHERE email_address = ?");
$chk2->bindParam(1,$email);
$chk2->execute();
if($chk2->rowCount() == 1)
{
$notice['email_exists'] = "The email address that you have entered is already exists in database";
return $this->notice;
}else{
$this->NewAdmin($username,$email,$password,$groups,$level);
$notice['success_message'] = "New admin was successfully added";
return $this->notice;
}
}
}
}
public function NewAdmin($username,$email,$password,$groups,$level)
{
if(!empty($username)&&!empty($email)&&!empty($password)&&!empty($groups)&&!empty($level))
{
$reg = $this->db->prepare("INSERT INTO admins (user_name, email_address, password_hash, group_admin, date_joined, admin_level) VALUES ( ?, ?, ?, ?, NOW(), ?)");
$reg->bindParam(1,$username);
$reg->bindParam(2,$email);
$reg->bindParam(3,$password);
$reg->bindParam(4,$groups);
$reg->bindParam(5,$level);
$reg->execute();
}
}
public function getNotice()
{
return $this->notice;
}
}
?>
如您所知,如果管理员添加了一个重复用户名或电子邮件地址的新管理员,则会调用通知消息,如果插入成功,则会显示成功消息:
$ notice [&#39; username_exists&#39;] =&#34;尝试使用不同的用户名&#34 ;; $ notice [&#39; email_exists&#39;] =&#34;您输入的电子邮件地址已存在于数据库&#34 ;; $ notice [&#39; success_message&#39;] =&#34;新管理员已成功添加&#34;;
在课程结束时,我也做了一个这样的方法:
public function getNotice()
{
return $this->notice;
}
然后我尝试通过这种方式调用通知警告,如果页面上发生任何事情:
<?php
if(isset($notice['email_validation'])) {
echo "
<div class='alert alert-danger'>
<strong>Hey!</strong> ".$notice['email_validation'].".
</div>
";
}
if(isset($notice['username_exists'])) {
echo "
<div class='alert alert-danger'>
<strong>Hey!</strong> ".$notice['username_exists'].".
</div>
";
}
if(isset($notice['email_exists'])) {
echo "
<div class='alert alert-danger'>
<strong>Hey!</strong> ".$notice['email_exists'].".
</div>
";
}
if(isset($notice['success_message'])) {
echo "
<div class='alert alert-success'>
<strong>Hey!</strong> ".$notice['success_message'].".
</div>
";
}
?>
但问题是,没有任何消息在页面上根本不出现!
因此,如果您知道如何解决此问题,请告诉我,谢谢......
答案 0 :(得分:0)
在CheckUname()
功能中,您可以
$notice['success_message'] = "New admin was successfully added";
return $this->notice;
实际上有两个变量名为&#34; notice&#34;这里涉及的一个是局部变量$notice
,它被设置为你的消息,另一个是类成员变量$this->notice
,它永远不会被设置但会在这里返回。
您应该使用成员变量并调用getter getNotice()
来获取消息字符串,或者使用局部变量并从函数返回它。