Insert dynamically generated form inputs to MySQL database using Laravel

Question

I have a form for adding cheques to a system. The form has two parts; part one is for the common info i.e the customer details and part two is for the unique details i.e cheque details. Multiple cheques can be added for the same user by clicking an Add button which generates duplicate inputs for the form. While saving the data on mysql database using Laravel, i got the error

"Undefined offset: 1"

Here is my laravel controller :

public function store(Request $request)
{
  $customer_name = $request->customer_name;
  $cheque_number=$request->cheque_number;
  $count = count($cheque_number);

    for($i = 0; $i < $count; $i++){
        $objModel = new Cheque();
        $objModel->customer_name = $customer_name[$i];
        $objModel->cheque_number = $cheque_number[$i];
        $objModel->save();
    }
}

My main challenge is each cheque number should be saved on a new row, but with the same customer name which is provided only once.

the customer name is a single value while the cheque number can have more than one value. different cheque numbers should be stored in different db rows but with the same customer name — Vic, Feb 20 '19 at 12:16
https://stackoverflow.com/questions/43036183/array-check-undefined-offset-php/43036264 — Shibon, Feb 20 '19 at 12:30

score 1 · Accepted Answer · answered Feb 20 '19 at 12:18

try this one

public function store(Request $request)
{
  $customer_name = $request->customer_name;
  $cheque_number = $request->cheque_number;
  $count = count($cheque_number);

    for($i = 0; $i < $count; $i++){
       if(isset($cheque_number[$i])) {  //for check value is set or not..
          $objModel = new Cheque();
          $objModel->customer_name = $customer_name;  //same name
          $objModel->cheque_number = $cheque_number[$i]; //different number
          $objModel->save();  
       }
    }
}

score 0 · Answer 2 · answered Feb 20 '19 at 12:24

public function store(Request $request){
 $customer_name = $request->get('customer_name');
 $cheque_number = $request->get('cheque_number');
 $count = count($cheque_number);

 for($i = 0; $i < $count; $i++){
  if(isset($cheque_number[$i])){
   $objModel = new Cheque();
          $objModel->customer_name = $customer_name;  //same name
          $objModel->cheque_number = $cheque_number[$i]; //different number
          $objModel->save(); 
  }
 }
}

score -1 · Answer 3 · answered Feb 20 '19 at 12:24

You should try this:

public function store(Request $request)
{
  $customer_name = $request->customer_name;
  $cheque_number = $request->cheque_number;
  $count = count($cheque_number);

    for($i = 0; $i < $count; $i++){
       if(isset($cheque_number[$i])) {  
          $objModel = new Cheque();
          $objModel->customer_name = $customer_name;  //same name
          $objModel->cheque_number = $cheque_number[$i]; //different number
          $objModel->save();  
       }
    }
}

@Vic: Glad to help :) – AddWeb Solution Pvt Ltd Feb 20 '19 at 12:28 — AddWeb Solution Pvt Ltd, Feb 20 '19 at 12:28

Insert dynamically generated form inputs to MySQL database using Laravel

3 Answers3