Question

我有一张向系统添加支票的表格。表格分为两部分：第一部分用于常见信息（即客户详细信息），第二部分用于唯一信息（即支票详细信息）。通过单击“添加”按钮可以为同一用户添加多个检查，该按钮会为表单生成重复的输入。使用Laravel将数据保存在mysql数据库上时，出现错误

“未定义的偏移量：1”

这是我的laravel控制器：

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();
    }
}

我的主要挑战是每个支票号码应保存在新行中，但要使用相同的客户名（仅提供一次）。

Answer 1

尝试这个

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();  
       }
    }
}

Answer 2

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(); 
  }
 }
}

Answer 3

您应该尝试以下操作：

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();  
       }
    }
}

使用Laravel将动态生成的表单输入插入MySQL数据库

3 个答案: