php – 如何在代码点火器中将值从控制器传递到视图？

发布时间：2021-08-28 03:45:18 所属栏目：大数据来源： https://www.jb51.cc

导读：控制器代码 public function store_service_provider() { $get = file_get_contents('PHP://input'); $json_decode = json_decode($get,true); foreach ($json_decode as $key = $value) { $data[$key] = $value; } $data['role'] ='provider'; // assign ro

控制器代码

  public function store_service_provider() {
            $get = file_get_contents('PHP://input');
            $json_decode = json_decode($get,true);
            foreach ($json_decode as $key => $value) {
                $data[$key] = $value;
            }        
            $data['role'] ='provider';  // assign role as provider
            $record = $this->serviceprovider_model->store_service_provider($data);                     
             if($record == '2'){
                $data['json'] = json_encode(array('flag' => 'false', 'message'=> 'you have account with this email id Thank you '));
                $this->load->view('jsonview', $data);
            } else {
            log_message("info",json_encode($record));
            $message['uname'] = '<html><h1>Hi '.$record['first_name']. " " .$record['last_name'].'</h1><br&;
            <h2><strong>Welcome to Look My Service.</strong></h2><br/>
            <h3>Your Username: '.$record['email'].'</h3><br&;
            <h2>To set your password
            <a href ="'.base_url().'serviceprovider#/change_password/'.$record['id'].'">Click here </a></h2><br&;
            <strong> Thank You </strong></html>';        
            $this->email->to($data['email']);
            $this->email->subject('Hall-book');
            $this->email->message(''.$message['uname'].'');
            $this->email->message($this->>view('hall_service/email_message',$record,true));
            $contact_number = $this->input->post("service_seeker_contact_number");        
            if(!$this->email->send()) {
                $data['json'] = json_encode(array('flag' => 'email_error', 'message'=> ' not correct login '));
                $this->load->view('jsonview', $data);
            }else{
                 $message = 'Welcome to Look My Service. Your Username  : '.$record['email'].' To set your password check you email Thank you.';
                  send($data['phone'], $message);
                echo json_encode(array('flag' => 'true', 'message'=> 'correct login '));
            }
        }
        }

在此处输入代码

 <html>
         <head>
              <title></title>
         </head>
    <body>
         <h1>sudarshan</h1>
         <?PHP foreach($record as $record):?>
              <span><?PHP echo $record; ?></span><br>
              <h1>Hi <?=$record->first_name?> " " <?=$record['last_name']?></h1><br/>
              <h2><strong>Welcome to Look My Service.</strong></h2><br/>
              <h3>Your Username: <?=$record->email?></h3><br/>
              <h2>To set your password
              <a href ="base_url()serviceprovider#/change_password/<?=$records['id']?>">Click here </a></h2><br/>
         <?PHP endforeach;?>
         <strong> Thank You </strong>
    </body>
    </html>

我将$记录传递给一个视图,在控制器中我可以获取所有值,但在我无法获取$record的值,如何解决此问题？如何点击链接后禁用链接

解决方法:

@aruna angadi

永远记住当您将变量中的任何数据传递给控制器??文件中的视图时,您无法在视图文件中获取相同的名称变量名称.

您只需向下一步即可获取数据.

例：-

如果你想通过

$data = array();

$data['record'] = $my_record_data;

$this->load->view('view_file_name',$data);

那么你将获得带有$record变量的数据.

你可以像这样使用

视图文件中的$记录.

谢谢

（编辑：北几岛）

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