Laravel, When database is empty I'm getting error otherwise not but I have to use this











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0
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        $employee = Employee::where('user_id', Auth::user()->id)->get();

foreach (Company::all() as $company)
{
if ($company->id == $employee[0]->company_id && $company->employee_active === 1)
{
$event->menu->add([
'text' => 'Contracten',
'url' => 'dashboard/contracts',
'icon' => 'file-text',
'submenu' => [
[
'text' => 'Contract opzetten',
'url' => 'dashboard/contracts/create',
'icon_color' => 'red',
]
]
]);
}
}


When I use this code I'm getting undefined offset: 0, if the database is empty. How can get this write? Should I use an if or something like that










share|improve this question






















  • You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
    – Farooq Khan
    Nov 9 at 8:06












  • Should I put if(!empty($employee)){ inside the foreach or outside?
    – Bram Swinkels
    Nov 9 at 8:09










  • what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
    – N69S
    Nov 9 at 8:31















up vote
0
down vote

favorite












        $employee = Employee::where('user_id', Auth::user()->id)->get();

foreach (Company::all() as $company)
{
if ($company->id == $employee[0]->company_id && $company->employee_active === 1)
{
$event->menu->add([
'text' => 'Contracten',
'url' => 'dashboard/contracts',
'icon' => 'file-text',
'submenu' => [
[
'text' => 'Contract opzetten',
'url' => 'dashboard/contracts/create',
'icon_color' => 'red',
]
]
]);
}
}


When I use this code I'm getting undefined offset: 0, if the database is empty. How can get this write? Should I use an if or something like that










share|improve this question






















  • You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
    – Farooq Khan
    Nov 9 at 8:06












  • Should I put if(!empty($employee)){ inside the foreach or outside?
    – Bram Swinkels
    Nov 9 at 8:09










  • what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
    – N69S
    Nov 9 at 8:31













up vote
0
down vote

favorite









up vote
0
down vote

favorite











        $employee = Employee::where('user_id', Auth::user()->id)->get();

foreach (Company::all() as $company)
{
if ($company->id == $employee[0]->company_id && $company->employee_active === 1)
{
$event->menu->add([
'text' => 'Contracten',
'url' => 'dashboard/contracts',
'icon' => 'file-text',
'submenu' => [
[
'text' => 'Contract opzetten',
'url' => 'dashboard/contracts/create',
'icon_color' => 'red',
]
]
]);
}
}


When I use this code I'm getting undefined offset: 0, if the database is empty. How can get this write? Should I use an if or something like that










share|improve this question













        $employee = Employee::where('user_id', Auth::user()->id)->get();

foreach (Company::all() as $company)
{
if ($company->id == $employee[0]->company_id && $company->employee_active === 1)
{
$event->menu->add([
'text' => 'Contracten',
'url' => 'dashboard/contracts',
'icon' => 'file-text',
'submenu' => [
[
'text' => 'Contract opzetten',
'url' => 'dashboard/contracts/create',
'icon_color' => 'red',
]
]
]);
}
}


When I use this code I'm getting undefined offset: 0, if the database is empty. How can get this write? Should I use an if or something like that







laravel undefined offset






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 9 at 8:02









Bram Swinkels

32




32












  • You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
    – Farooq Khan
    Nov 9 at 8:06












  • Should I put if(!empty($employee)){ inside the foreach or outside?
    – Bram Swinkels
    Nov 9 at 8:09










  • what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
    – N69S
    Nov 9 at 8:31


















  • You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
    – Farooq Khan
    Nov 9 at 8:06












  • Should I put if(!empty($employee)){ inside the foreach or outside?
    – Bram Swinkels
    Nov 9 at 8:09










  • what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
    – N69S
    Nov 9 at 8:31
















You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
– Farooq Khan
Nov 9 at 8:06






You should put all of your $company logic in a simple empty check if(!empty($employee)){ // then your logic of $employee[0] will make sense }
– Farooq Khan
Nov 9 at 8:06














Should I put if(!empty($employee)){ inside the foreach or outside?
– Bram Swinkels
Nov 9 at 8:09




Should I put if(!empty($employee)){ inside the foreach or outside?
– Bram Swinkels
Nov 9 at 8:09












what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
– N69S
Nov 9 at 8:31




what is you logic in here? if the company of the connected user has employee_active attribute to 1 => add in menu ?
– N69S
Nov 9 at 8:31












3 Answers
3






active

oldest

votes

















up vote
-1
down vote



accepted










Really you should configure your relationships correctly so that you can do



$companies = Company::all();

foreach($companies as $company){

foreach($company->employees as $employee){

if($employee->active){
....
}

}

}


But in your case you can change your first line to



 $employee = Employee::where('user_id', Auth::user()->id)->first();


Which will return an Employee object collection rather than an array so you don't need to use an [index] to get the first object.






share|improve this answer

















  • 1




    it's a bad solution to cycle through all the companies and then get all there employee.
    – N69S
    Nov 9 at 8:28


















up vote
0
down vote













@Joe solution is correct. You can make an eager loading of companies and employees to avoid N+1 query problem. There is an official example of this case:
https://laravel.com/docs/5.7/eloquent-relationships#eager-loading






share|improve this answer




























    up vote
    0
    down vote













    You should leverage constrained eager loading to limit your results. This will return only the companies where the user is an employee.



    $companies = Company::where('employee_active', 1)->whereHas('employees', function($query) {
    $query->where('user_id', auth()->id());
    })->get();


    Now you dont need the conditional checks for generating the menu items.






    share|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      -1
      down vote



      accepted










      Really you should configure your relationships correctly so that you can do



      $companies = Company::all();

      foreach($companies as $company){

      foreach($company->employees as $employee){

      if($employee->active){
      ....
      }

      }

      }


      But in your case you can change your first line to



       $employee = Employee::where('user_id', Auth::user()->id)->first();


      Which will return an Employee object collection rather than an array so you don't need to use an [index] to get the first object.






      share|improve this answer

















      • 1




        it's a bad solution to cycle through all the companies and then get all there employee.
        – N69S
        Nov 9 at 8:28















      up vote
      -1
      down vote



      accepted










      Really you should configure your relationships correctly so that you can do



      $companies = Company::all();

      foreach($companies as $company){

      foreach($company->employees as $employee){

      if($employee->active){
      ....
      }

      }

      }


      But in your case you can change your first line to



       $employee = Employee::where('user_id', Auth::user()->id)->first();


      Which will return an Employee object collection rather than an array so you don't need to use an [index] to get the first object.






      share|improve this answer

















      • 1




        it's a bad solution to cycle through all the companies and then get all there employee.
        – N69S
        Nov 9 at 8:28













      up vote
      -1
      down vote



      accepted







      up vote
      -1
      down vote



      accepted






      Really you should configure your relationships correctly so that you can do



      $companies = Company::all();

      foreach($companies as $company){

      foreach($company->employees as $employee){

      if($employee->active){
      ....
      }

      }

      }


      But in your case you can change your first line to



       $employee = Employee::where('user_id', Auth::user()->id)->first();


      Which will return an Employee object collection rather than an array so you don't need to use an [index] to get the first object.






      share|improve this answer












      Really you should configure your relationships correctly so that you can do



      $companies = Company::all();

      foreach($companies as $company){

      foreach($company->employees as $employee){

      if($employee->active){
      ....
      }

      }

      }


      But in your case you can change your first line to



       $employee = Employee::where('user_id', Auth::user()->id)->first();


      Which will return an Employee object collection rather than an array so you don't need to use an [index] to get the first object.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Nov 9 at 8:14









      Joe

      2,55621627




      2,55621627








      • 1




        it's a bad solution to cycle through all the companies and then get all there employee.
        – N69S
        Nov 9 at 8:28














      • 1




        it's a bad solution to cycle through all the companies and then get all there employee.
        – N69S
        Nov 9 at 8:28








      1




      1




      it's a bad solution to cycle through all the companies and then get all there employee.
      – N69S
      Nov 9 at 8:28




      it's a bad solution to cycle through all the companies and then get all there employee.
      – N69S
      Nov 9 at 8:28












      up vote
      0
      down vote













      @Joe solution is correct. You can make an eager loading of companies and employees to avoid N+1 query problem. There is an official example of this case:
      https://laravel.com/docs/5.7/eloquent-relationships#eager-loading






      share|improve this answer

























        up vote
        0
        down vote













        @Joe solution is correct. You can make an eager loading of companies and employees to avoid N+1 query problem. There is an official example of this case:
        https://laravel.com/docs/5.7/eloquent-relationships#eager-loading






        share|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          @Joe solution is correct. You can make an eager loading of companies and employees to avoid N+1 query problem. There is an official example of this case:
          https://laravel.com/docs/5.7/eloquent-relationships#eager-loading






          share|improve this answer












          @Joe solution is correct. You can make an eager loading of companies and employees to avoid N+1 query problem. There is an official example of this case:
          https://laravel.com/docs/5.7/eloquent-relationships#eager-loading







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 9 at 8:45









          Ivan

          11




          11






















              up vote
              0
              down vote













              You should leverage constrained eager loading to limit your results. This will return only the companies where the user is an employee.



              $companies = Company::where('employee_active', 1)->whereHas('employees', function($query) {
              $query->where('user_id', auth()->id());
              })->get();


              Now you dont need the conditional checks for generating the menu items.






              share|improve this answer

























                up vote
                0
                down vote













                You should leverage constrained eager loading to limit your results. This will return only the companies where the user is an employee.



                $companies = Company::where('employee_active', 1)->whereHas('employees', function($query) {
                $query->where('user_id', auth()->id());
                })->get();


                Now you dont need the conditional checks for generating the menu items.






                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You should leverage constrained eager loading to limit your results. This will return only the companies where the user is an employee.



                  $companies = Company::where('employee_active', 1)->whereHas('employees', function($query) {
                  $query->where('user_id', auth()->id());
                  })->get();


                  Now you dont need the conditional checks for generating the menu items.






                  share|improve this answer












                  You should leverage constrained eager loading to limit your results. This will return only the companies where the user is an employee.



                  $companies = Company::where('employee_active', 1)->whereHas('employees', function($query) {
                  $query->where('user_id', auth()->id());
                  })->get();


                  Now you dont need the conditional checks for generating the menu items.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 9 at 9:04









                  DigitalDrifter

                  6,5472423




                  6,5472423






























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