I am using codeigniter framework. I have a table named client. I need to get the first name from the table and send it to my view page. I have a function in model file to retrieve the client data.
This is my model- function:
public function get_cli($id)
{
$this->db->select('*');
$this->db->from('client');
$this->db->where('id', $id);
$res = $query->result();
$row = $res[0];
return $row->first_name;
}
Here I get my client first name.
This is my controller file. Here I have a function to send the data from the client table to my view page.
public function ticket($id)
{
$id=$this->uri->segment(4);
$data['id']=$this->uri->segment(4);
$data['client']=$this->billing_model->get_cli($id);
$data['employee']=$this->employee_model->emp_name();
$data['main_content'] = 'admin/billing/ticket_page';
$this->load->view('includes/template', $data);
}
My view page.
<label>Client Id:<?=$id?></label>
<input type="text" value="<?=$client['first_name'];?>"/>
I am getting a white screen. Why is this happenening? Can someone help me with the code?
Change your model function to this:
public function get_cli($id)
{
$this->db->select('*');
$this->db->from('client');
$this->db->where('id', $id);
$query = $this->db->get();
$result = $query->result_array();
return $result[0]['first_name']; //returning only the first name key value of first row(0th row) of the result;
}
Then On the view page:
<input type="text" value="<?=$client;?>"/>
Which will print only the first name;
Another approach:
Change your model function to this:
public function get_cli($id)
{
$this->db->select('*');
$this->db->from('client');
$this->db->where('id', $id);
$query = $this->db->get();
$result = $query->result_array();
return $result; //returning the whole array;
}
Then On the view page:
<input type="text" value="<?=$client[0]['first_name'];?>"/>
Which will print only the first name value from the 0th(first row) of the array;
You should change your view page to this
<input type="text" value="<?=$client;?>" />
This will work as you are returning the firstname in your model not whole array.
So, your $client variable will contain the firstname. You dont need to refrence it.
Simply echoing $client will work.
When you have only single row in result i suggest you to use row_object() instead of result()
public function get_cli($id)
{
$this->db->select('*');
$this->db->from('client');
$this->db->where('id', $id);
$res = $query->row_object();
//$row = $res[0]; no need to write this line
return $res->first_name;
}
for your problem on your view
You will get first_name in $client
<input type="text" value="<?=$client?>"/>
if you are not getting anything please check your result from database might be null.
USE
$row['first_name'] instead of $row->first_name
Related
I am facing problem to join these tables and also know how to make controller code to view data in view file with text boxes.
public function get_order_return_info()
{
$this->db->select('tbl_order_details.*', false);
$this->db->select('tbl_order_details.order_details_id', false);
$this->db->select('tbl_order.order_id', false);
$this->db->select('tbl_product.product_id', false);
$this->db->select('tbl_inventory.product_quantity', false);
$this->db->from('tbl_order_details');
$this->db->join('tbl_order', 'tbl_order_details.order_id = tbl_order.order_id ', 'left');
$this->db->join('tbl_product', 'tbl_order_details.product_code = tbl_product.product_id ', 'left');
$this->db->join('tbl_inventory', 'tbl_product.product_id = tbl_inventory.product_id ', 'left');
$query_result = $this->db->get();
$result = $query_result->result();
return $result;
}
You can use MVC to solve this
//Controller
function get_data() {
$data['list'] = $this->your_model->get_data();
$this->load->view('your_view/location',$data);
}
//Model
function get_data() {
$sql = "your_query";
$list = $this->db->query($sql)->result_array();
return $list;
}
//View
foreach($list as $data) {
echo $data['your_selected_field']; // it could be table,text field,or just text
}
API Reference : https://www.codeigniter.com/user_guide/database/results.html#result-arrays
Hope this helps
First u can create OrderModel to work with order table. Then create function get_order_data.
//Order Model
public function get_orders() {
// your db query to get orders
return $result;
}
Then u can load model in controller using $this->load->model('OrderModel'); Best place is __construct to load model
And call your get_orders() function
// Your controller
// Create data array to store front data
$data = [];
$data['orders'] = $this->OrderModel->get_orders();
After getting $orders pass it to view using following codes
$this->load->view('path/to/view', $data);
Finally u can call your order data in view using $orders
I hope this will help u
You need to set alias to the table to function it properly like this.
$this->db->select('a.*,a.order_details_id, b.order_id ', false);
$this->db->from('tbl_order_details as a');
$this->db->join('tbl_order as b', 'a.order_id = b.order_id ', 'left');
If you want select 1 row only you need to use this function.
$this->db->get('tbl_order_details')->row_array();
or if you want all result row you need to use this.
$this->db->get('tbl_order_details')->result_array();
On your controller you should add this lines to pass your result on your view page.
public function index()
{
$data['orderDetails'] = $this->your_model->get_order_return_info();
$this->load->view('pages/yourview/index',$data);
}
And finally to set your order details to the text boxes. You need to echo it inside the text boxes like this.
<input type="Text" name="orderID" value="<?php echo $orderDetails['order_id'];?>" >
If you use result_array() you need to use foreach loop to echo it.
I am getting the following error.
Message: Trying to get property of non-object
Table Structure:
name ='fb' value='https://www.facebook.com/mypage'
Model:
function get_fb_url(){
$query=$this->db->select('value');
$query=$this->db->from('preferences');
$query=$this->db->where('name','fb');
return $this->db->get()->row();
}
Controller:
$data['fb_url']=$this->Home_model->get_fb_url()->value;
View:
<li><img src="css/imgs/fb.png" /></li>
When i run my page it give me error
Message: Trying to get property of non-object
only when if value in the table is link like above, if its simple text then text properly show on page.
How Can i get rid off this error?
Instead of echo use return type
Change
echo $this->db->get()->row();
to
return $this->db->get()->row();
And add $ before variable name in view
<a href="<?php echo $fb_url;?>"
UPDATED
Change your model code to
function get_fb_url() {
$this->db->select('value');
$this->db->from('preferences');
$this->db->where('name', 'fb');
return $this->db->get()->row();
}
you must try like this:
function get_fb_url() {
$this->db->select('value');
$this->db->from('preferences');
$this->db->where('name', 'fb');
$query = $this->db->get();
return $query->row();
}
and in Controller:
$result = $this->Home_model->get_fb_url();
//try print $result, it must be an stdObject
$data['fb_url']=$result->value;
it is because in $data['fb_url']=$this->Home_model->get_fb_url()->value; / differentiate the link to the other function, that's why u r getting this error.
get_fb_url is returning object array not object. so you have to return value from method.
function get_fb_url()
{
$query=$this->db->select('value');
$query=$this->db->from('preferences');
$query=$this->db->where('name','fb');
return $this->db->get()->row()->value;
}
and than
$data['fb_url']=$this->Home_model->get_fb_url();
In Your model do this
public function get_fb_url() {
$this->db->select('value');
$this->db->from('preferences');
$this->db->where('name','fb');
$query = $this->db->get();
$row = $query->row();
return $row;
}
In Controller :
$data = array();
$result = $this->Home_model->get_fb_url();
$data['fb_url'] = $result->value;
$this->load->view('your_view',$data);
In View :
echo $fb_url;
Please help me. i can not use my dataTable properly. what I want to do is
select from table and use thewherefunction. but i cant do it properly.
here is my controller code
public function reporttable ()
{
$zz = array('empnumber' => $this->input->post('empnumber'));
//$this->db->order_by("surname", "asc");
$this->datatables->select('date_auto,particulars,earned_VL,aul_VL,balance_VL,aulx_VL,earned_SL,aul_SL,balance_SL,aulx_SL,date_leave,action_taken')
->from('tblreport')->where($zz);
echo $this->datatables->generate();
}
this is my supposed query:
select * from tblreport where empnumber = (the empnumber in my textbox.)
there, i get a value from a textbox to my view. but it didn't work. i know that is wrong. can you please help me with my problem? thank you.
<p align="center"> <?php echo $this->table->generate();?></p></div>
<?php foreach ($tblemployee as $row){?>
<input type="text" name="empnumber" readonly="readonly" value="<?php echo $row->empnumber;?>"/>
<input type="hidden" name="empnumber" value="<?php echo $row->empnumber;?>"/>
here is my view for guide. thank you.
as Simple you can use
In Controller
$data['tblemployee'] = $this->model_name->reporttable($id)//assign your data base value to variable
$this->load->view('your view name',$data )
in Model
public function reporttable($id)
{
$query = $this->db->query("select * from tblreport where empnumber = '$id'");
$result = $query->result_array();
return $result; // this will return your data as array
}
In view
<?php
foreach ($tblemployee as $row)
{
echo $row['id];
}?>
Try this :
To make it more simplier.
In model :
public function reporttable ($id){
$this->db->select('*');
$this->db->from('tblreport');
$this->db->where('empnumber', $id);
$query = $this->db->get();
return $query->return_array(); // use this if so want to return many query if not you can also use first_row('array')
}
In controller :
public function function_name (){
$data['variable_name'] = $this->model_name->reporttable($id); // change the model_name by your own model where your function reporttable is located and use the variable_name for your view,
$this->load->view('view_name' , $data); // load the view page you want to display.
}
In Controller
$data['tblemployee'] = $this->model_name->reporttable($id)//assign your data base value to variable
$this->load->view('your view name',$data )
in Model
public function reporttable($id)
{
$query = $this->db->query("select * from tblreport where empnumber = '$id'");
$result = $query->result_array();
return $result; // this will return your data as array
}
In view
<?php
foreach ($tblemployee as $row)
{
echo $row['id];
}?>
Good day all. I'm new to codeigniter and I'm trying to pass a variable from one view page to another. But I get error and after a lot of tries still couldn't figure it out.
This is the Model code:
function get_post($postID){
$this->db->select()->from('khanposts')->where(array('post_id'=>$postID))->order_by('fullname', 'desc');
$query=$this->db->get();
return $query->result_array();
}
function update_post($postID, $data)
{
$this->where('post_id', $postID);
$this->db->update('khanposts', $data);
}
This is the Controller code:
function editpost($postID)
{
$data['success']=0;
if($_POST){
$data_post=array(
'fullname'=>$_POST['fullname'],
'dob'=>$_POST['dob'],
'blood'=>$_POST['blood'],
'village'=>$_POST['village'],
'occupation'=>$_POST['occupation'],
'company'=>$_POST['company'],
'email'=>$_POST['email'],
'contact'=>$_POST['contact'],
'password'=>$_POST['pass'],
'marry'=>$_POST['marry']);
$this->khanpost->update_post($postID, $data);
$data['success']=1;
}
$data['post']=$this->khanpost->get_post($postID);
$this->load->view('edit_post', $data);
}
This is the code of View page which passes the value to edit_post view page:
foreach ($posts as $row){
<tr><td><i>Edit</i></td></tr>';
}
This is code of edit_post view page where it must get the value of $row['post_id']:
echo form_open(base_url().'khanposts/editpost/'.$row['post_id']);
echo '<b>Full Name: </b>';
$data_form=array('name'=>'fullname', 'size'=>30, 'id'=>'fullname', 'class'=>'inputstyle', 'value'=>$row['fullname'] );
echo form_input($data_form);
How do I assign the passsed variable($row['post_id']) in form_open()? Any solution will be really helpful. Tnx.
As far as I understand the problem;
Pass PostID to view in editpost function:
function editpost($postID)
{
...
$data['postID'] = $postID;
}
Get it in view page like:
echo form_open(base_url('khanposts/editpost/'.$postID));
You should load form helper:
$this->load->helper('form');
And your data_form input:
$data_form = array('name'=>'fullname', 'size'=>30, 'id'=>'fullname', 'class'=>'inputstyle', 'value' => $post['fullname']);
You should use row_array instead of result_array, because of you get single post.
function get_post($postID)
{
$this->db->select()->from('khanposts')->where(array('post_id'=>$postID))->order_by('fullname', 'desc');
$query=$this->db->get();
return $query->row_array();
}
I'm new in CodeIgniter. I'm having a problem fetching "id" from "project view" then i will pass it to another view which a call "project details view".
I've tried using ampersand (&) in URL and i guess that does not work in CodeIgniter.
Here are my codes:
In my controller:
public function project(){
$data["title"] = "CodeIgniter Projects";
$this->load->model("projects_model");
$data["result"] = $this->projects_model->getProjects();
$this->load->view("project", $data);
}
public function project_details($project_id){
$this->load->model("projects_model");
$data["result"] = $this->projects_model->getProjectDetails($project_id);
$this->load->view("project_details", $data);
}
for my model
function getProjects(){
$query = $this->db->query("SELECT * FROM projects");
return $query->result();
}
function getProjectDetails($project_id){
$this->db->where('project_id', $project_id);
$query = $this->db->get('projects');
}
in my view i only use < a href="controller/project_details">< / a> to call the project details view.
You can pass anything to controller from a view as a uri_segment. The general syntax is,
Controller/function/parameter
Here, in your case
controller/project_details/<?php echo $project_id; ?>