I have a single form which I am using to create and edit information.
I am using Form::model to show all the data in the edit form which are queried from the database. Now I had to add another form part whose data is being stored in a different table. But I need to show those data during edit in the same form. I have tried to put two parameter in the form:model which did not work or I am doing it wrong.
{!! Form::model([$employee_data,$edu_info],['route'=>['employee.update',$employee_data->id],'method'=>'put','files'=> true]) !!}
Then I tried to merge the queried data in my controller like this:
public function edit($id)
{
$edit_info['title'] = 'Edit User';
$edit_info['country'] = Country::all();
$employee_basic = Employee::withTrashed()->where('id',$id)->first();
$employee_edu = EmployeeEdu::withTrashed()->where('employee_id',$id)->first();
$employee_all_data = $employee_basic->merge($employee_edu);
$employee_all_data->all();
$edit_info['employee_data'] = $employee_all_data;
return view('admin.employee.edit',$edit_info);
}
This did not work either. I get the following error:
Call to undefined method Illuminate\Database\Query\Builder::merge()
How can I achieve my intended result?
EDIT: I tried to use ->get() instead of ->first(), In this case I do not get the error but my merge does not work as when I dd($employee_all_data) it gives me only the value of $employee_edu.
Try this:
$employee_all_data = $employee_basic->toArray() + $employee_edu->toArray();
Related
I have a page where you can create your workout plan. Second page contains "pre-saved" workouts and I want them to load by passing parameters from second page to first. If you directly access first page, you create your workout plan from scratch.
// first page = https://prnt.sc/y4q77z
// second page = https://prnt.sc/y4qfem ; where you can check which one you want to pass to first page
// final step looks like this: https://prnt.sc/y4qh2q - but my URL looks like this:
www.example.com/training/plan?sabloni%5B%5D=84&sabloni%5B%5D=85&sabloni%5B%5D=86
this 84,85,86 are IDS
Can I pass params without changing URL ? Like having only /training/plan without anything after ?
public function plan(Request $request){
$workout = false;
if($request->workout){
$workout = $request->workout;
$workout = SablonTrening::find($sabloni); // $workout = array [1,3,4,5,6]
}
return view('trener.dodaj_trening', compact('workout'));
}
If you are getting to the /training/plan page with GET request, you could simply change it to POST. That way the parameters would be hidden in the URL but would be present in the request body. You would need a new post route:
Route::post('/training/plan', 'YourController#plan')->name('training.plan');
And then, in the form where you are selecting these plans, change the method on submit:
<form action="{{route('training.plan')}}">
//Your inputs
</form>
Your method should still work if your inputs stay the same.
Note: Not sure you would still keep the functionalities that you need, since I can't see all the logic you have.
If you have any questions, let me know.
To pass data from on blade to another blade.
At the end of first post before redirect()-route('myroute') add $request->session()->put('data', $mydata);
At the begining of the route 'myroute', just get back your data with $data = $request->old('data');
I've checked the Q&A about this and can't find anything, so thought i'd ask.
I have a very simple Laravel controller returning all results from a table as below via the 'Name model'. There is then also a further call to my controller, via the model to count the rows and all works and sends to the result set fine...
// All results from my 'Name' model:
$results = $this->name->getAllResults(); // All works fine.
// I then use my controller again, count the rows via the model and add them to $results:
$results['count'] = $this->countNames(); // Again fine
BUT, when i try to add a string to the $results array before i pass it off to th view, as in:
$results['test'] = 'Test'; // This fails in the view
$results['test'] = 124; // But this passes in the view and renders.
It only seems to allow me to add an INT to my result set array. as $results['test'] = 124 also fails.
I then finally, have this sending to my view via:
return view('names', compact('results')); // And that works fine.
Can anyone see what it is I am missing and why integer added to $results works and not a string?. Many thanks in advance.
You are updating collection data. The following line will give collection of models.
$results = $this->name->getAllResults(); // This may give collection of the model
And below, you are updating the collection object.
$results['count'] = $this->countNames();
You can do the following to safely send data to view, without modifying any.
$results = $this->name->getAllResults();
$count = $this->countNames();
$test = 'Test';
$test2 = 124;
return view('names', compact('results','count','test','test2'));
I am working on search filter on checkbox click, with Laravel and Ajax call. So I get results when I click on a checkbox. my query is as follows:
$editors = User::with(['editor.credentials','editor.specialties','editor.ratings']);
$temp=$editors->whereHas('editor', function($q) use ($a_data){
$q->whereHas('specialties',function($sq) use($a_data){
$sq->whereIn('specialty',$a_data);
});
})->paginate(2);
This gives me all the data I need. however how should I get the links for pagination?
$temp->setBaseUrl('editors');
$links = $temp->links()->render();
I am currently doing this and with $links which I am sending over as response to ajax call, I set the pagination with $links data. Now, I need to append the query to next page like page=2?query="something". I don't know how should I go about appending the remaining query result links to next page links. i.e. I don;t know what should come in the query="something" part. Can someone guide me. thanks
{{ $users->appends($_GET)->links() }}
It will append all query string parameters into pagination link
As of Laravel 7, you can call the withQueryString() method on your Paginator instance.
Quote from the documentation:
If you wish to append all current query string values to the pagination links you may use the withQueryString method:
{{ $users->withQueryString()->links() }}
See "Appending To Pagination Links": https://laravel.com/docs/7.x/pagination#displaying-pagination-results
Check the answer from #Arda, as it's global solution. Below you can find how to do it manually.
Use appends on Paginator:
$querystringArray = Input::only(['search','filter','order']); // sensible examples
// or:
$querystringArray = ['queryVar' => 'something', 'anotherVar' => 'something_else'];
$temp->appends($querystringArray);
Append all input except the actual page, form token and what you don't want to pass:
$paginatedCollection->appends($request->except(['page','_token']));
For the latest version of Laravel at the moment (5.2), you can just use the Request facade to retrieve the query string and pass that to your paginator's appends() method
$input = Request::input();
$myModelsPaginator = App\Models\MyModel::paginate();
$myModelsPaginator->appends($input);
Add this anywhere in your app (e.g routes.php, filters.php or anything that's autoloaded), no need to edit any pagination codes that is written already. This works flawlessly using view composers, and you don't need to know any query string parameters:
////////PAGINATION QUERY STRING APPEND
View::composer(Paginator::getViewName(), function($view) {
$queryString = array_except(Input::query(), Paginator::getPageName());
$view->paginator->appends($queryString);
});
//////////////////
Inspired from previous answers I ended up using the service container for both frontend + api support.
In your AppServiceProvider#boot() method:
$this->app->resolving(LengthAwarePaginator::class, function ($paginator) {
return $paginator->appends(array_except(Input::query(), $paginator->getPageName()));
});
you can used request helper in view as same
{{ $users->appends(request()->query())->links() }}
in your view where you display pagination...
{{ $results->appends(Request::except('page'))->links() }}
appends keeps the query string value except "page". not sure if it will work with POST request
{{ $users->appends(Request::only(['filter','search']))->links()}}
Updating #rasmus answer for Laravel 8.
In your AppServiceProvider boot method, add the following lines and your existing query string will be be used for all pagination links
$this->app->resolving(Paginator::class, function ($paginator) {
return $paginator->appends(Arr::except(Request::query(), $paginator->getPageName()));
});
$this->app->resolving(LengthAwarePaginator::class, function ($paginator) {
return $paginator->appends(Arr::except(Request::query(), $paginator->getPageName()));
});
And for completeness, use these classes:
use Illuminate\Pagination\LengthAwarePaginator;
use Illuminate\Pagination\Paginator;
use Illuminate\Support\Arr;
use Illuminate\Support\Facades\Request;
Im trying to load data from a model if the data from a previously loaded model request is_numeric. I have written my code like this, though it doesn't work and i sort of get why. But i dont know how to get the content from the data['schedule_item'].
$this->load->model('schedule_model');
$this->load->model('partners_model');
$data['schedule_item'] = $this->schedule_model->getAll();
if(is_numeric($schedule_item['featured_partner1']) {
$data['featured_partner1'] = $this->schedule_model->get_partner($schedule_item['featured_partner1']);
}
You are getting the wrong variable you have assigned the model values on the index of $data array not to $schedule_item so it is undefined $schedule_item will be accessible in view only when all the indexes of $data is converted to the variables you should try to do this like
$this->load->model('schedule_model');
$this->load->model('partners_model');
$schedule_item=$this->schedule_model->getAll();
$data['schedule_item'] =$schedule_item;
if(is_numeric($schedule_item['featured_partner1']) {
$data['featured_partner1'] = $this->schedule_model->get_partner($schedule_item['featured_partner1']);
}
OR
$this->load->model('schedule_model');
$this->load->model('partners_model');
$data['schedule_item'] =$this->schedule_model->getAll();
if(is_numeric( $data['schedule_item']['featured_partner1']) {
$data['featured_partner1'] = $this->schedule_model->get_partner($schedule_item['featured_partner1']);
}
Here is my function. It is very simple.
function load_data() {
$query = $this->db->query('SELECT * FROM configurations WHERE username="' . $this->session->userdata('username') . '"');
return $query;
}
My controller has this line of code:
$data['query'] = $this->configurations->load_data();
In my view, I tried:
foreach($query->result_array() as $row) {
echo $row->first;
}
But I get an error that I am trying to get a property of a non-object. Isn't the query being returned from the model as an object?
You should use $query->result_array, row_array, result, or row otherwise your returning the object intead get the results. Check the CI manual.
You are returning the results as array and using $row as object!
Try:
foreach($query->result() as $row) {
Refer.
Try changing $this->load->model('login/configurations', '', TRUE); to $this->load->model('login/configurations', 'configurations', TRUE); and see if it works. If it does, it is related to how you're extending your model class. That is, it would be related to what name you give inside configurations.php.
Hope this helps.
Your undefined variable error tells me that your query might not be running correctly. To diagnose...enable the profiler to check your query.
From the documentation:
$this->output->enable_profiler();
Permits you to enable/disable the
Profiler, which will display benchmark
and other data at the bottom of your
pages for debugging and optimization
purposes.
To enable the profiler place the
following function anywhere within
your Controller functions:
$this->output->enable_profiler(TRUE);
When enabled a report will be
generated and inserted at the bottom
of your pages.
Your query will be shown at the end of
the page
Double check your query syntax to make sure it is running properly, and
your code in it's current state is returning an object of objects and arrays:
print_r($query)
CI_DB_mysql_result Object
(
[conn_id] => Resource id #29
[result_id] => Resource id #39
[result_array] => Array
(
)
[result_object] => Array
(
)
[current_row] => 0
[num_rows] => 3
[row_data] =>
)
you need to access the individual properties to get to the actual data.
$result=$query->result();
print_r($result);
should do it
Had this issue before - basic problem is that if the Query returns no result set then $query->result_array() == NULL
NULL is not a list and can't be processed in a foreach.
I have found that checking for this condition solves this issue.
From your question, you are getting the result as a pure array (result_array) but printing the data as object ($row->first).
result() or row() is returning in object but result_array is returning in array.
change your view part like,
foreach($query->result_array() as $row) {
echo $row['first'];
}
or if you want to use an object then,
foreach($query->result() as $row) {
echo $row->first;
}
Generating Query Results in Codeigniter