I want get id value from eloquent laravel and passing to variable to insert process. this is my eloquent to get data from siswa table
$siswa = Siswa::where('nis', $row['nis'])->first();
but when accessing id value like $siswa->id I'm getting error
enter image description here
Use this code
if($siswa){
$id=$siswa->id;
}
$siswa variable can be null. Look at its contents with the dd($siswa) command.
To escape the error you can use it like this:
if($siswa)
{
// $siswa not null.
}
You have two ways to fix this.
1/ You can use ->firstOrFail() to display a 404 page in case of your data is not found :
$siswa = Siswa::where('nis', $row['nis'])->firstOrFail();
// if $siswa is NULL, then it will display a 404 page, else it will continue
2/ You can add a simple condition to continue with your data :
$siswa = Siswa::where('nis', $row['nis'])->first();
if ($siswa) {
// then continue
}
First you can $id pass on your public function like
public function show($id)
{
$siswa = Siswa::where('id', $id)->first();
}
You can pass your id on your route
Related
I am writing a recursive function to call the child record from the parent record. It seems not to be working. i am getting this error; "Trying to get property 'refid' of non-object". Where am i getting it wrong. Please any idea? below is the code Snippet.
the function controller
public function DisplayDetail($id)
{
$displayDetail = DB::table('tblmembers')
->where('refid',$id)
->get();
return $this->DisplayDetail($displayDetail->refid);
}
main controller where the function is called
public function dashboard()
{
$profile = DB::table('tblmembers')->where('username',$userid)->first();
$data['userdetail'] = $this->DisplayDetail($profile->memid);
return view('main.userArea',$data);
}
the blade where the record fetched is displayed
#foreach($userdetail as $userd)
{{ $userd->memid }}
#endforeach
my sample data
refid | memid
-------------------
12345 | 123456
123456 | 1234567
123456 | 1234568
123456 | 1234569
1234567 | 1234570
from the above table; refid: 123456 brought memid: 1234567,1234568,1234569. then refid: 1234567 brought memid: 12345670
i want to display all the memid after login in as a user with memid 123456
You are doing one thing wrong in your function DisplayDetail. Here the the correction in your function
If you want to get single item then here is the correct code.
public function displayDetail($id)
{
$displayDetail = DB::table('tblmembers')
->where('refid',$id)
->first();
if($displayDetail) {
$displayDetail['userdetail'] = $this->displayDetail($displayDetail->refid);
}
return $displayDetail;
}
And dashboard function will be look like this
public function dashboard()
{
$profile=DB::table('tblmembers')->where('username',$userid)->first();
$userDetail = $this->DisplayDetail($profile->memid);
return view('main.userArea',[
'userdetail' => $userDetail
]);
}
This is the correct code. Try this and let me know if you have another query on this.
The error:
Trying to get property 'refid' of non-object
is occuring because your database query is using ->get(), which returns a Collection, rather than an object. You cannot get the property ->refid on a Collection, you can only get it from an object that resides in the collection.
As Lakhwinder Singh shows in his code, you need to use ->first(), as this will return one object. I would suggest using ->firstOrFail(), that way you will either get back an object which matches your ID, or it will fail if it cannot find it.
If you do:
$displayDetail = DB::table('tblmembers')
->where('refid',$id)
->firstOrFail();
You will now be able to call:
$displayDetail->refid
You can use that in your function call to displayDetail.
try this function
public function DisplayDetail($id,$data=[])
{
$displayDetail = DB::table('tblmembers')
->where('refid',$id)
->get();
if($displayDetail && isset($displayDetail->refid))// your condition for last child
{
$data = $this->DisplayDetail($displayDetail->refid,$data);
}
$data[] = array('refid' =>$displayDetail->id ,
'memid' => $displayDetail->secondid );
return $data;
}
i'll explain you later first modify according to your requirement and run
I am using following parameter value to be posted into controller but it shows null
Route::get('JobApplication/{JobId}','JobApplicationController#view');
Route::post('application/','JobApplicationController#Post');
In view method of JobApplicationController value is null when I try to get it as:
$JobId = Input::get('JobId') ;
$JobId= $request->JobId; or view($JobId){echo $JobId}
Try this:
public function Post($JobId) {
return $JobID;
}
You have set JobID as route segment, so you can access it directly as a parameter to the method.
<?php
class JobApplicationController
{
public function view($jobId)
{
dd($jobId);
}
}
The code $JobId = Input::get('JobId'); that you are trying is used in situations where JobId is sent as a query parameter, such as in url https://example.com/JobApplication?JobId=12321
I am writing an update methoda for my api. I neede to update requested fields.
So ı am trying to get only requested fields and update them. However, the code below returns me null even thoug I cast $fill to string.
foreach ($fillableFields as $fill){
$customerId->$fill = $request->get("$fill") ;
edit- 1
My code is like below ;
$fillableFields = array('lastname','firstname');
when I dd($fill) it returns me null .
You forgot to call the save() method to update your object :
foreach ($fillableFields as $fill){
$customerId->$fill = $request->get($fill);
$customerId->save();
}
Your method should be.
public function yourMethod(Request $request)
{
try {
foreach ($fillableFields as $fill){
$customerId->$fill = $request->get($fill) ;
}
}
} catch (\Exception $ex) {
}
}
Best way to get request parameters is provided at laravel's official documentation as below.
You can get request parameter based on your specific need.
By using below statment you can get only those perameter you have specified in only attribute as array eg : username, password.
$input = $request->only(['username', 'password']);
If you wantto get all fields excluding particular field than you can define as below with except parameter. eg: by using below it will allow to get all the fields except credit_card.
$input = $request->except('credit_card');
For more information on request parameter please refer official doc url : https://laravel.com/docs/5.4/requests
the following function is supposed to read the name of the given asset code from the database. but it triggers the error: "Trying to get property of non-object"
function sban_name($asset){
$this->db->select('name');
$this->db->from('asset_types');
$this->db->where('code',$asset);
return $this->db->get()->result()->row('name');
}
All I want is to have the name of the asset returned back to the controller! Your help is highly appreciated!
Use row() like,
return $this->db->get()->row()->name;
Use row() for a single row, and result() for multiple rows.
do like this, asset_types is your table name
function sban_name($asset){
$this->db->select('name');
$this->db->from('asset_types');
$this->db->where('code',$asset);
return $this->db->get('asset_types');
}
And in your controller acess it like
$result=$this->modelname->sban_name('$asset')->row();
$name=$result->name;
I think it's important to check if the record that satisfies the conditions even exists in the database. Code for the model:
function sban_name($asset){
$this->db->select('name');
$this->db->from('asset_types');
$this->db->where('code',$asset);
$row = $this->db->get()->row();
if (isset($row)) {
return $row->name;
} else {
return false;
}
}
Simply call the function from the controller like so:
$response = $this->model_name->sban_name($asset)
Try this code of block , I already checked and works fine:
function sban_name($asset)
{
$this->db->select('name');
$this->db->from('asset_types');
$this->db->where('code', $asset);
return $this->db->get()->row()->name;
}
//Anyone can help to create a view data with same id? it is a multiple viewing.
this is my Controller. i dont khow apply in Model and View
function Get_Pitch($id){
$this->load->model('users_model');
$data['query'] = $id;
$this->load->view('view_pitch', $data);
}
Example this is my url "http://localhost/SMS_System/home/sample/102"
in my database is
id=1 name=erwin user_id=102
id=2 name=flores user_id=102
id=3 name=sample user_id=202
how to view the same user_id?
First of all with what you've supplied your URL won't work, you aren't following the normal conventions for CI so it won't know where to look. I am assuming your controller is called sample then you need to tell the application which function you're calling in that controller, finally URL names should be lower case so I changed that, so your URL should read:
"http://localhost/SMS_System/home/sample/get_pitch/102"
Also you need to get your data from a model, you loaded the model then didn't use it. The line after loading the model calls a function from that model and passes it the id you got from your url. Notice the if not isset on the id, this ensures that if someone goes to that page without the id segment there are no errors thrown from the model having a missing parameter, it will just return nothing, that is handled in the view.
Controller:
function get_pitch($id){
//the following line gets the id based on the segment it's in in the URL
$id=$this->uri_segment(3);
if(!isset($id))
{
$id = 0;
}
$this->load->model('users_model');
$data['query'] = $this->users_model->getUserData($id);
$this->load->view('view_pitch', $data);
}
Your model takes the id passed from the controller and uses that to retrieve the data from the database. I normally create the array I am going to return as an empty array and handle that in the view, this makes sure you get no errors if the query fails. The data then returns to the controller in the last line and is passed to the view in your load view call.
Model:
function getUserData($id)
{
$this->db->where('id',$id);
$result = $this->db->get('users') //assuming the table is named users
$data = array(); //create empty array so we aren't returning nothing if the query fails
if ($result->num_rows()==1) //only return data if we get only one result
{
$data = $result->result_array();
}
return $data;
}
Your view then takes the data it received from the model via the controller and displays it if present, if the data is not present it displays an error stating the user does not exist.
View:
if(isset($query['id']))
{
echo $query['id']; //the variable is the array we created inside the $data variable in the controller.
echo $query['name'];
echo $query['user_id'];
} else {
echo 'That user does not exist';
}