從codeigniter連接表查看數據

Question

我想查看數據庫中的數據。有三個表join.in兩個表中有相同的字段名稱。當我嘗試查看數據follows.but有一個叫

A PHP Error was encountered

Severity: Notice

Message: Undefined property: stdClass::$project_name1

Filename: views/boq_doc.php

Line Number: 12

控制器

class Project_list extends CI_Controller { 

function __construct(){ 
parent::__construct(); 
$this->load->model('project_list_model'); 
} 
function show_project_id() { 
$id = $this->uri->segment(3); 
$data['projects'] = $this->project_list_model->show_projects(); 
$data['single_project'] = $this->project_list_model->show_project_id($id); 
$this->load->view('boq_doc', $data); 
} 

} 

型號

class Project_list_model extends CI_Model { 


// Function To Fetch All Students Record 
function show_projects(){ 

$this->db->select("project.project_name AS project_name1 , project.id AS id, client.firstname AS firstname1, client.lastname AS lastname1,staff.firstname AS firstname2, staff.lastname AS lastname2, project.location,project.category, project.start_date, project.end_date"); 
$this->db->from('project'); 
$this->db->join('client', 'project.client_id = client.client_id'); 
$this->db->join('staff', 'staff.id = project.staff_id'); 
$query = $this->db->get(); 
return $query->result(); 
} 
// Function To Fetch Selected Student Record 
function show_project_id($data){ 
$this->db->select('*'); 
$this->db->from('project'); 
$this->db->where('id', $data); 
$query = $this->db->get(); 
$result = $query->result(); 
return $result; 
} 

} 

查看

錯誤210

<?php foreach ($projects as $project): ?> 
<li><a href="<?php echo base_url() . "index.php/project_list/show_project_id/" . $project->id; ?>"><?php echo $project->project_name1; ?></a></li> 
<?php endforeach; ?> 
</ol> 
</div> 
<div id="detail"> 
<!-- Fetching All Details of Selected Student From Database And Showing In a Form --> 
<?php foreach ($single_project as $project): ?> 

<form method="post" action="<?php echo base_url() . "index.php/update_ctrl/update_project_id1"?>"> 
<label id="hide">Id :</label> 
<?php echo $project->project_name1; ?> 
<label>Name :</label> 
<input type="text" name="dname" value="<?php echo $project->location; ?>"> 
<label>Email :</label> 
<input type="text" name="demail" value="<?php echo $project->start_date; ?>"> 
<label>Mobile :</label> 
<input type="text" name="dmobile" value="<?php echo $project->end_date; ?>"> 
<label>Address :</label> 
<input type="text" name="dmobile" value="<?php // echo $project->firstname1; ?>"> 
<input type="text" name="daddress" value="<?php // echo $project->project_address; ?>"> 
<input type="submit" id="submit" name="dsubmit" value="Update"> 
</form> 
<?php endforeach; ?> 
</div> 

Answer 1

你的foreach循環的語法似乎是錯誤的嘗試這個代替

<? 
    foreach ($projects as $project) 
    { 
    ?> 
     <li><a href="<?php echo base_url()."index.php/project_list/show_project_id/" . $project->id; ?>"><?php echo $project->project_name1; ?></a></li> 
<? 
} 
?> 
     </ol> 
     </div> 
     <div id="detail"> 
     <!-- Fetching All Details of Selected Student From Database And Showing In a Form --> 
     <?php foreach ($single_project as $project){ ?> 

     <form method="post" action="<?php echo base_url() . "index.php/update_ctrl/update_project_id1"?>"> 
     <label id="hide">Id :</label> 
     <?php echo $project->project_name1; ?> 
     <label>Name :</label> 
     <input type="text" name="dname" value="<?php echo $project->location; ?>"> 
     <label>Email :</label> 
     <input type="text" name="demail" value="<?php echo $project->start_date; ?>"> 
     <label>Mobile :</label> 
     <input type="text" name="dmobile" value="<?php echo $project->end_date; ?>"> 
     <label>Address :</label> 
     <input type="text" name="dmobile" value="<?php // echo $project->firstname1; ?>"> 
     <input type="text" name="daddress" value="<?php // echo $project->project_address; ?>"> 
     <input type="submit" id="submit" name="dsubmit" value="Update"> 
     </form> 
     <? 
     } 

     ?> 
     </div>