I am trying to pass variable.
I want to show all the sponsor ids from the id. i.e. if 203035051 passed in the URL then in the view i can see all ids for which 203035051 is sponsor.
Here is my anchor link:
<a href="<?=base_url('v3/member_income?id='.$id);?>">
Here is my Model: MemberIncome
public function display_records($id)
{
$this->db->select('*');
$this->db->from('users');
$this->db->where("id",$id);
$query = $this->db->get();
if ($query) {
return $query->row_array();
} else {
return false;
}
}
Here is my Controller
public function displaydata()
{
/*load Model*/
$this->load->model('MemberIncome');
$result['data']=$this->MemberIncome->display_records($id);
$this->load->view('customer/member_income', $result);
}
here is my route
$route['v3/member_income'] = 'IncomeData/displaydata';
Here the reason:
$route['v3/member_income'] = 'IncomeData/displaydata';
You need to send the ID through routes. Add the (:any) and $1 those are the parameters you need to send. Without it, the "displaydata" won't receive the ID.
$route['v3/member_income/(:any)'] = 'IncomeData/displaydata/$1';
In the anchor section, actually you don't need "?id=", just send it like this
<a href="<?= base_url('v3/member_income/'.$id); ?>">
Missing something in route to catch your ID. Try this way
$route['v3/member_income/(:any)'] = 'IncomeData/displaydata/$1';
Related
I'm sending an URL hashed and when i get it i have to show a view on Laravel, so i have those functions on the controller and also some routes:
This are my routes:
Route::post('/sendLink', 'Payment\PaymentController#getPaymentLink');
Route::get('/payment?hash={link}', 'Payment\PaymentController#show');
And this are the functions i have on my controller:
public function getPaymentLink (Request $request){
$budgetId = $request['url.com/payment/payment?hash'];
$link = Crypt::decryptString($budgetId);
Log::debug($link);
//here to the show view i wanna send the link with the id hashed, thats why i dont call show($link)
$view = $this->show($budgetId);
}
public function show($link) {
$config = [
'base_uri' => config('payment.base_uri'), ];
$client = new Client($config);
$banking_entity = $client->get('url')->getBody()->getContents();
$array = json_decode($banking_entity, true);
return view('payment.payment-data')->with('banking_entity', $array);
}
And this is getting a "Page not found" message error.
What i want to to is that when i the client clicks on the link i send him that has this format "url.com/payment/payment?hash=fjadshkfjahsdkfhasdkjha", trigger the getPaymentLink function so i can get de decrypt from that hash and also show him the view .
there is no need to ?hash={link} in get route
it's query params and it will received with $request
like:
$request->hash
// or
$request->get('hash')
You need to define route like this:
Route::get('/payment/{hash}', 'Payment\PaymentController#show');
You can now simply use it in your Controller method like below:
<?php
public function getPaymentLink (Request $request,$hash){
$budgetId = $hash;
// further code goes here
}
I would like to recover the slug of 2 categories from my routes but can’t write the Controller.
My Route
Route::get('technicians/o/{occupation}/c/{city}', 'User\TechnicianController#viewoccupationcity');
My Controller
public function viewoccupationcity($slug)
{
$technicians = TechnicianResource::collection(occupation::where('slug',$slug)->firstOrFail()->technicians()
->with('city','occupation')
->latest()->get());
return $technicians;
}
Route::get('technicians/o/{occupation}/c/{city}', 'User\TechnicianController#viewoccupationcity');
Your controller will accept the parameters from your route as variables by order
public function viewoccupationcity($ocupation, $city)
{
...
}
Example:
URL: technicians/o/foo/c/bar
public function viewoccupationcity($ocupation, $city)
{
// $ocupation will be 'foo'
// $city will be 'bar
}
Ok, you would need to retrieve 2 variables as that is what you are passing
public function viewoccupationcity($occupation, $city)
If you want the whole slug to do another search then you would use the $request object. So like so
public function viewoccupationcity(Request $request, $occupation, $city){ // You also need to include the Request decleration
$slug = $request->path();
$technicians = TechnicianResource::collection(occupation::where('slug',$slug)->firstOrFail()->technicians()
->with('city','occupation')
->latest()->get());
return $technicians;
}
EDIT: We are having to do a lot of guesswork as your question isn't very clear. I think what you are trying to achieve is probably this
public function viewoccupationcity($occupation, $city){
$technicians = TechnicianResource::collection(occupation::where('city',$city)->where('occupation',$occupation)->firstOrFail()->technicians()
->with('city','occupation')
->latest()->get());
return $technicians;
}
If you need something more then you need to give more details
This is My Controller.
public function show(Posts $posts)
{
$page = Posts::find($posts->id);
//dd($page);
return view('web_views.index',['page' => $page]);
}
This is my view page
<h4>{{$page->post_titile}}</h4>
It's better to use Route-Model Binding I think.
Your route (if you are using resource route, it's already done):
Route::get('posts/{post}', 'PostController#show); // domain.tld/posts/1
Your method should look like this:
public function show(Post $post)
{
return view('web_views.index',compact('post'));
}
In your view:
<h4>{{ $post->post_titile }}</h4>
May be $posts->id you are passing, has no result in database so you need to check it by using if-statement
public function show(Posts $posts)
{
$page = Posts::find($posts->id);
if($page == null){
$page = ['post_title' => 'Not Available'];
}
return view('web_views.index',['page' => $page]);
}
I believe this error is because of not finding data for an ID into database. So if as per above script will pass the fetched data to view otherwise it will push an item into $page array.
I am making a site in codeigniter, here is my controller, model and view:
controller:
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();// assign your value to CI variable
$this->load->view('home', $data); //passing your value to view
}
model:
public function email()
{
$query = $this->db->query("SELECT email FROM change_password");
$result = $query->result_array();
return $result;
}
view:
foreach ( $dbemail as $new_dbemail )
{
echo $new_dbemail['database_field'];//in here you can get your table header.
//Ex if your table has name field and you need to sho it you can use $new_dbemail['name']
}
but i want to get $new_dbemail['name'] in my controller, how i can get the value of $new_dbemail['name'] in controller rather than view ???
To do so... you have to fetch data into a row_array not in result_array in Model.
Controller
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();
$database_field = $data['dbemail']['database_field']; //to get result in controller
$date['database_field'] = $data['dbemail']['database_field']; //to get result in view
$this->load->view('home', $data);
}
Model
public function email()
{
$query = $this->db->query("SELECT email FROM change_password");
return $query->row_array();
}
View
<p> Database Field value : <?=$database_field;?></p>
You are Calling the Model In A Variable :-
$data['dbemail']=$this->login_model->email();
So in this you have the results which you are fetching from Database. Than you passing this variable data to your View.
$this->load->view('home', $data);
So you already have the value in controller which u fetched from model which is stored in $data.
You can View the Values of $data , using
echo "<pre>";
print_r($data);
and it will show what data you are receiving
Try like this...
In controller...
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$dbemail=$this->login_model->email();//array containing all emails
//For each loop here for displaying emails
foreach($dbemail as $email){
$mail = $email; //assigning to variable
echo $mail."<br/>";//displaying email
}
$data['dbemail'] = $dbemail;//for passing to view
$this->load->view('home', $data); //passing your value to view
}
First you need to fetch the name from the database in the model. Change the query line:
$query = $this->db->query("SELECT name, email FROM change_password");
Then to use the name data in the controller you could try this:
function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemail']=$this->login_model->email();
$name = $data['dbemail']['name']; // example usage to retrieve name in controller
$this->load->view('home', $data);
}
Now the name data is available as $name in controller.
Controller:
public function checking()
{
$email=$this->input->post('email');
$this->load->model('login_model');
$data['dbemails']=$this->login_model->email();
$this->load->view('home', $data); //passing your value to view
}
Model
public function email()
{
$this->db->select("email"); // the row of the table you want to select.
$this->db->get('change_password'); //table name. you can still use your query, its just much neat when you use a specific query builder.
$result = $this->result(); //giving you a result of type object
return $result;
}
View
foreach ( $dbemails as $dbemail )
{
echo $dbemail->email; // the '->' will be the identifier of which row you want to get or display, i assume its email beacause you specify it in the model.
}
Hope this what your looking for.
My link in view page
Detail
My controller
public function cari(){
//what i'm doing here ? please help ?
}
My model
function search_by_id($id){
$query = $this->db->where('id',$id);
$query = $this->db->get();
return $query->result();
}
How pass value from link to model and show the result to view again?
it's my first time using codeigniter, need some help
You have to pass argument into method:
public function cari($hasil_id){
if ((int)$hasil_id > 0)
{
$this->load->model('Name_of_model');
$data['search_by_id'] = $this->Name_of_model->search_by_id($hasil_id);
$this->load->view('hasil_view', $data);//assuming page for item
}
else
{
redirect('not_good_id_method', 'refresh');//in case of not valid id
}
}
In your controller ,receive the value passed through link by uri class.And then pass it into your model.
Controller :
public function cari()
{
$id= $this->uri->segment(3);
$this->model_name->search_by_id($id);
}
Take a look on this.