【发布时间】:2018-06-27 13:18:42
【问题描述】:
HTML 表单的摘录:
<input type="text" id="1" name="name[1]" class="input-number" onchange="input_onchange(this.id)" value="39" min="1" max="25">
<input type="text" id="1" name="name[2]" class="input-number" onchange="input_onchange(this.id)" value="39" min="1" max="25">
我的 AJAX 调用:
$.ajax({
type: 'POST',
url: '/sepetim/ajax',
data: {
'_token': $('input[name="_token"]').val(),
//'name': $('input[name=name]').val(),
'name': $('input[name=name[' + id + ']]').val()
},
success: function(data) {
$('#fiyat').replaceWith("<span class='para fw6'>" + data.formattedPrice + " TL</span>");
}
});
从我的控制器中提取:
public function addStock(Request $request) {
$data = new ShoppingCartDetail();
$data->cart_id = $request->input("cart_id");
$data->stock_id = $request->input("stock_id");
$data->price = 1;
$data->save();
$data->name = $request->input("name");
$data->price = 1200;
$formattedPrice = number_format($data->name * $data->price, "2" , ",", ".");
return response ()->json ([ "data" => $data , "formattedPrice" => $formattedPrice ]);
}
我认为有问题的行是这一行:
$data->name = $request->input("name");
我已经尝试过,但我无法获得名称为“name”的输入?
【问题讨论】: