Question

这是我的profile.php代码，我想在网页中显示它

<!-- Bridegroom Profile Details -->
<div class="w3ls-list">

    <div class="container">

                <h2>Bridegroom Profile Details</h2>

    <div class="col-md-9 profiles-list-agileits">
        <div class="single_w3_profile">
            <div class="agileits_profile_image">
    <?php foreach ($profile_details as $row) { ?>
                                <img src="<?php echo $row['upload'];?>" alt="profile image" />
                                           <?php } ?>

            </div>

            <div class="clearfix"></div>
        </div>


        <div class="profile_w3layouts_details">

            <div class="agileits_aboutme">
                <h4>Profile</h4>
                <h5>Personal Details:</h5>
                <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Profile for : </label>
                    <div class="col-sm-9 w3_details">

                                               <!--Brother--> 
                    </div>
                    <div class="clearfix"> </div>
                </div>
                <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Name : </label>
                    <div class="col-sm-9 w3_details">

                    </div>
                    <div class="clearfix"> </div>
                </div>
                <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Gender : </label>
                    <div class="col-sm-9 w3_details">

                    </div>
                    <div class="clearfix"> </div>
                </div>
                                    <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Dob : </label>
                    <div class="col-sm-9 w3_details">

                    </div>
                    <div class="clearfix"> </div>
                </div>
                                    <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Religion : </label>
                    <div class="col-sm-9 w3_details">

                    </div>
                    <div class="clearfix"> </div>
                </div>
                                    <div class="form_but1">
                    <label class="col-sm-3 control-label1" for="Relation">Mobile : </label>
                    <div class="col-sm-9 w3_details">

                    </div>
                    <div class="clearfix"> </div>
                                    </div>


                            </div>

        </div>


    </div>
           </div>    

    </div>

这是我的控制器home.php 请帮助您了解如何显示此个人资料

 public function profile($param1='')
 {  

     //$data['user_detail'] = $this->db->get_where('admin',array('id'=> $this->session->userdata('admin_id')))->result_array();
     $data['page_name']='profile';
     $data['profile_details']=$this->db->get_where('profile',array('id' => $param1))->result_array();

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

Answer 1

因为你已经知道这样做了

<?php foreach ($profile_details as $row) { ?>
      <img src="<?php echo $row['upload'];?>" alt="profile image" />
<?php } ?>

我觉得你想做的就像下面的那样？

 <div class="form_but1">
    <label class="col-sm-3 control-label1" for="Relation">Name : </label>
                    <div class="col-sm-9 w3_details">
                         <?php echo $row['Name'];?> // or the name of the column
                    </div>
     <div class="clearfix"> </div>
  </div>

加

在数据库查询后放置print_r($data)并检查是否确实得到了结果。

public function profile($param1='')
 {  

     //$data['user_detail'] = $this->db->get_where('admin',array('id'=> $this->session->userdata('admin_id')))->result_array();
     $data['page_name']='profile';
     $data['profile_details']=$this->db->get_where('profile',array('id' => $param1))->result_array();

    print_r($data);

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


 }

Answer 2

由于您提取single record所以此处不需要foreach这样简单

<?php if(isset($profile_details[0]['upload'])) { ?>
      <img src="<?php echo $profile_details[0]['upload'];?>" alt="profile image" >
<?php } ?>

和

 <div class="form_but1">
     <label class="col-sm-3 control-label1" for="Relation">Name : </label>
         <div class="col-sm-9 w3_details">
               <?php if(isset($profile_details[0]['Name'])) { 
                            echo $profile_details[0]['Name'];
                 } ?>
          </div>
          <div class="clearfix"> </div>
        </div>

Answer 3

正确获取视图中的数据或在CodeIgniter中的页面中显示数据：

如果您知道要提取单个记录，请使用row_array而不是result_array。仅当您知道从数据库中获取多行时才使用result_array。

你的模特：

public function get_single_profile($param1)
{
        $query = $this->db
                ->select('*')
                ->where('id', $param1)      
                ->get('profile');
        $profile_data = $query->row_array();
        return $profile_data;
}

在控制器中：在__construct

中加载模型

    public function profile($param1)
    {
        $this->load->model ('your_model');
        $profile_data = $this->your_model->get_single_profile ($param1);

        $this->load->view ('index',['profile_data' => $profile_data]);
    }

在视图页面上：

<?php if(isset($profile_data) && !empty($profile_data)){ ?>

    <img src="<?php echo $profile_data['upload'];?>" alt="profile image" />
    //And So on...

<?php } ?>

如何在视图中获取数据或在codeigniter中的页面中显示数据

3 个答案:

加