Ich möchte Daten aus der Datenbank anzeigen.Es gibt drei Tabelle zu verbinden.In zwei Tabellen haben den gleichen Feldnamen. wenn ich versuche, Daten anzuzeigen als follows.but gab es einen Fehler genanntDaten anzeigen von Codezeichner Join Tabelle
A PHP Error was encountered
Severity: Notice
Message: Undefined property: stdClass::$project_name1
Filename: views/boq_doc.php
Line Number: 12
-Controller
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);
}
}
Modell
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 $result;
}
}
Ansicht
<?php foreach ($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 $project->project_address; ?>">
<input type="submit" id="submit" name="dsubmit" value="Update">
</form>
<?php endforeach; ?>
</div>
Was ist, wenn Sie ein Argument der Methode 'Funktion show_project_id ($ id) {' passieren? – Tpojka
Funktion show_project_id() { $ id = $ this-> uri-> Segment (3); $ data ['projekte'] = $ this-> project_list_model-> show_projects(); $ data ['single_project'] = $ this-> projektliste_modell-> show_project_id ($ id); $ this-> load-> view ('boq_doc', $ data); } – ashik
Versuchen Sie damit, ich habe nur vorgeschlagen. Setzen Sie '$ id' Variable als Parameter der Methode. – Tpojka