I'm trying to use query builder in controller using laravel, and i don't understand but the query didn't find the data.
Here's the code:
public function index()
{
$data = downloads::all();
if(request('searchName')){
$data = $data->where('fileName','like','%'.request('searchName').'%'); //Here's the problem
return view('download', compact('data'));
}
else{
return view('download', compact('data'));
}
}
i already tried dd(request('searchName')) and it display the input that i give, so there's no problem here
when I'm using $data->where('fileName','like','%'.request('searchName').'%') there's no data shown
i don't think that i misspell the fileName because when I'm using $data->where('fileName','like',request('searchName')) and it worked and display the file, but the fileName must be exactly the same as the inputed searchName, and of course what i wanted is not this
even when I'm using dd('%'.request('searchName').'%'); it will display "%*searchName*%" that's why i so confused when it didn't work when I'm using $data->where('fileName','like','%'.request('searchName').'%');
I even using SELECT * FROM *tables* WHERE fileName LIKE '%p%'; in SQL Workbench and it worked perfectly fine
Any suggestion of what should i do? Thank you
This looks odd. Why are you filtering the collection instead of adding the where conditional in your query?
Imagine you have thousands of download records but the where condition just match with a few ones, you will be fetching everything just for showing some of them.
IMO, a better approach should be
public function index(Request $request)
{
$data = downloads::
when($request->has('searchName'), function($query) use ($request){
$query->where('fileName','like','%'.$request->searchName.'%');
})
->get();
return view('download', compact('data'));
}
all() is static method not query builder.If you see internal of all() code then its calling get method
/**
* Get all of the models from the database.
*
* #param array|mixed $columns
* #return \Illuminate\Database\Eloquent\Collection|static[]
*/
public static function all($columns = ['*'])
{
return static::query()->get(
is_array($columns) ? $columns : func_get_args()
);
}
There are few ways to solve this .
public function index()
{
$downloads = downloads::query();
if(!empty(request('searchName'))){
$downloads->where('fileName','like','%'.request('searchName').'%');
}
$data=$downloads->get();
return view('download', compact('data'));
}
or
public function index()
{
$data = downloads::when(!empty(request('searchName')),function($query){
$query->where('fileName','like','%'.request('searchName').'%');
})->get();
return view('download', compact('data'));
}
You are trying to apply your querystring with like in a collection. In a collection, you can use the filter($callback_function) method to select elements in the collection. Pass a callback function that returns true for each element to be returned.
In your case, you can use the stristr() function to emulate a LIKE operator, something like this:
collect($data)->filter(function ($item) use ($searchName) {
return false !== stristr($item->fileName, $searchName);
});
Related
enter image description hereI am trying to implement a many to many relationship search with 2 models.
i get input from multiple checkbox values and want to search for items that match A or B when there is an input of data.
I read this url and wrote the same logic.
https://laracasts.com/discuss/channels/laravel/many-to-many-relationship-with-2-pivot-table-data-search
public function search(Request $request)
{
$languages = $request->lang;
$fields = $request->field;
$agencies = Agency::with('languages')->with('specialized_fields')
->orWhereHas('languages', function($query) use ($languages) {
$query->whereIn('language_id', $languages);
})
->orWhereHas('specialized_fields', function($query) use ($fields) {
$query->whereIn('specialized_field_id', $fields);
})
->get();
dd($agencies);
}
i expected to achieve A or B search but instead I got this error.
Argument 1 passed to Illuminate\Database\Query\Builder::cleanBindings() must be of the type array, null given, called in /var/www/jtf/vendor/laravel/framework/src/Illuminate/Database/Query/Builder.php on line 907
it seems that it causes this error if either A or B is null, but why??? Does the OrWhereHas method work only when theres an input??
/added info/
my error message
my agency model
class Agency extends Model {
protected $guarded = [
'id'
];
public function languages(){
return $this->belongsToMany('App\Language');
}
public function specialized_fields(){
return $this->belongsToMany('App\SpecializedField');
}
public function region(){
return $this->hasOne('App\Region');
} }
I believe it's because either $languages or $fields is null.
Since ->whereIn() is expecting an array, but you're passing null.
You just need to make sure you're passing an array.
$languages = array_filter((array) $request->lang); // cast to array & remove null value
$fields = array_filter((array) $request->field);
$agencies = Agency::with('languages', 'specialized_fields')
->orWhereHas('languages', function($query) use ($languages) {
$query->whereIn('language_id', $languages);
})
->orWhereHas('specialized_fields', function($query) use ($fields) {
$query->whereIn('specialized_field_id', $fields);
})
->get();
I'm speculating that you started your where query chain with an orWhereHas() which may have caused the problem, try starting with whereHas() instead.
public function search(Request $request){
$languages = $request->lang;
$fields = $request->field;
$agencies = Agency::with('languages', 'specialized_fields') // you can get away by just using one with(), not needed but its cleaner this way
->whereHas('languages', function($query) use ($languages) { // previously orwherehas
$query->whereIn('language_id', $languages);
}) ->orWhereHas('specialized_fields', function($query) use ($fields) {
$query->whereIn('specialized_field_id', $fields);
})
->get();
dd($agencies);
}
I am using this EloquentFilter to filter Database on the frontend and the controller below works
public function indexArtisan(Request $request)
{
$workers = Worker::filter($request->all())->get();
$states = DB::table('states')->pluck("name", "id");
return view ('artisans', compact('states'))->with(['workers' => $workers]);
}
However, when i try to use the WHERE query to add conditions to the data been called with the controller, the WHERE condition is met and data returned but my filter no longer works. domain.com?name=xyz
public function indexArtisan(Request $request)
{
$work = DB::table('workers')->where('career', '=', 'artisan')->get();
$workers = Worker::filter($request->all())->get();
$states = DB::table('states')->pluck("name", "id");
return view ('artisans', compact('states','work'))->with(['workers' => $workers]);
}
Also tried
$work = DB::table('workers')->where('career', '=', 'artisan')->get();
$worker = $work->filter($request->all());
$states = DB::table('states')->pluck("name", "id");
return view ('artisans', compact('states'))->with(['worker' => $worker]);
I get
How do i execute the WHERE condition without breaking the filter
I've read https://laravel.com/docs/5.6/queries#conditional-clauses i still cant figure out what i'm doing wrong.
If you use $work->filter($request->all()); the filter() method is the method of collection class, because $work is a collection now.
So, you need to use filter() method of model class
DB::table('workers')->where('career', 'artisan')->filter($request->all())->get();
should work.
I've this method:
public function indexGuest($idAzienda, Request $request){
$companyId = $idAzienda;
$ddts = Ddt::where('company_id',$companyId);
$ddt_urls= Ddt_file_url::all();
if($request->start_date || $request->end_date){
$ddts->whereBetween('created_at',[new Carbon($request->start_date),new Carbon($request->end_date)]);
}
$ddts->paginate(10);
return view('guest.ddt-management')->with('ddts', $ddts)->with('ddt_urls',$ddt_urls)
->with('companyId',$companyId);
}
My start_date and end_date comes in strings like "yyyy-mm-dd".
I've tried to pass it straight to the query and like in the example like a carbon object with no hope!
After executing the query (now only the one without the wherebeeteween clause) i cant apply the method "paginate" to the collection, no error are raised but when i pass it to the view, the "link()" method not work and raise an error again.
where I wrong?
Laravel 5.4
Structure your wheres like this.
public function indexGuest($idAzienda, Request $request) {
[...]
$ddts = Ddt::where('company_id', $companyId)
->where(function($query) use ($request) {
if($s = $request->get("start_date") {
$s_date = Carbon::parse($s)->format("Y-m-d");
$query->whereDate("created_at", ">=", $s_date);
}
if($e = $request->get("end_date") {
$e_date = Carbon::parse($e)->format("Y-m-d");
$query->whereDate("created_at", "<=", $e_date);
}
})
->paginate(10);
[...]
}
stuck on a form that allows the user to enter a value into a choice of two fields. I can query the database using one field but want to add more range to database queries. With the following code below when i try to access the page to query it just shows me a white screen.
public function index()
{
$data = $request->all();
if(!empty($data['pstoreNum']))
{
$pstoreNum = $data['pstoreNum'];
$result = DB::table('perfumes')->where('StoreNumber','=',$pstoreNum)
->get();
return view('perfumes',compact('result'));
}
else if(!empty($data['pweekNum']))
{
$pweekNum = $data['pweekNum'];
$result = DB::table('perfumes')->where('WeekNumber','=',$pweekNum)
->get();
return view('perfumes',compact('result'));
}
}
My routes file simple calls the index function. Any help would be appreciated.
You can add query functions within your query like so
public function index(Request $request)
{
$data = $request->all();
$result = \DB::table('perfumes')->where(function($query) use ($data) {
if(!empty($data['pstoreNum'])) {
$query->where('StoreNumber', '=', $data['pstoreNum']);
}
if(!empty($data['pweekNum'])) {
$query->where('WeekNumber', '=', $data['pweekNum']);
}
})->get();
return view('perfumes',compact('result'));
}
You can then use the one query and add multiple wheres on various conditions.
https://laravel.com/docs/5.2/queries#advanced-where-clauses
simple issue I presume.
My controller is getting the if to display from the url using $this->uri->segment(3). This will always be a single value. I am putting this in an array to pass to the model with:
$customerid = array(
'id' => $this->uri->segment(3)
);
The controller syntax is below:
function confirm_delete_customer()
{
$data['title']="Confirm Customer Deletion";
$customerid=array(
'id'=>$this->uri->segment(3)
);
//query model to get data results for form
$data=array();
if($query=$this->model_master_data->get_customer_records_to_delete()){
$data['records']=$query;
$this->load->view("master_data/view_master_data_header",$data);
$this->load->view("master_data/view_master_data_nav");
$this->load->view("master_data/view_content_master_data_confirm_customer_deletion",$data);
$this->load->view("master_data/view_master_data_footer");
}
I am then trying to access this array value and pass it to my model to process. If I hard code the array into the model it works as per below syntax:
Model - Manual Syntax is:
function get_customer_records_to_delete()
{
$query = $this->db->get_where('customers', array('id'=>43));
return $query->result();
}
if I try replace this with the array from my controller it fails with error:
Undefined variable: customerid
idea of model that I want to get working:
function get_customer_records_to_delete()
{
$query = $this->db->get_where('customers', $customerid);
return $query->result();
}
I have a feeling it is something small. however is this the best way to get a single record from the database in order to output to a view?
Thanks in advance for the assistance.
The best way to do that is:
function confirm_delete_customer()
{
$data=array();
$data['title']="Confirm Customer Deletion";
$customerId = $this->uri->segment(3);
//Prevent SQL injections
if(!is_numeric($customerId) || empty($customerId)) {
show_error("Bad Request");
}
$query = $this->model_master_data->get_customer_records_to_delete($customerId);
if ($query){
$data['records']=$query;
$this->load->view("master_data/view_master_data_header",$data);
$this->load->view("master_data/view_master_data_nav");
$this->load->view("master_data/view_content_master_data_confirm_customer_deletion",$data);
$this->load->view("master_data/view_master_data_footer");
}
}
and then you can simply call:
function get_customer_records_to_delete($customerId)
{
$query = $this->db->get_where('customers', array('id'=>$customerId));
return $query->result();
}
at your model.
You need to pass the value as an argument to the function so it can access it.
Ex:
get_customer_records_to_delete($customerid)
{
// now $customerid is accessible
$query = ....;
return $……;
}
You should heavily rely on function parameters. Grab the customer id from the controller and send it to the model. Moreover, you can use row() to get a single result from the database.
Controller:
function confirm_delete_customer(){
$data['title']="Confirm Customer Deletion";
$customerid=$this->uri->segment(3);
//query model to get data results for form
$data=array();
if($query=$this->model_master_data->get_customer_records_to_delete( $customerid)) //you are sending customer id as a parameter here
$data['records']=$query;
$this->load->view("master_data/view_master_data_header",$data);
$this->load->view("master_data/view_master_data_nav");
$this->load->view("master_data/view_content_master_data_confirm_customer_deletion",$data);
$this->load->view("master_data/view_master_data_footer");
}}
Model
function get_customer_records_to_delete($customerid)
{
$query = $this->db->get_where('customers', array("id"=>$customerid)); //you are using the customer id sent from the controller here
return $query->row(); //this will return a single row
}
Old thread but the answer is to declare the variable as "public" in the controller (i.e. public $customerid;), in which case it'll be available to your model. In some cases it's probably safer to explicitly pass as an argument. However, when you have several variables, it's useful to have the option to declare them instead.