#Laravel如何使用开关按钮存储值

Question

我想从开关按钮存储值，但当开关关闭时我无法保存值在我的刀刃上。

<input name="serial_status[]" type="checkbox"
data-toggle="switchbutton" checked data-onstyle="success"
data-size="xs" data-onlabel="ใช้งาน" data-offlabel="ไม่ใช้งาน">

In my controller:

if ($product->save()) {
$product->productSerials()->delete();
for ($i = 0; $i < count($request->serial_name); $i++) {
if ($request->get('serial_status')) {
$status = 'active';
} else {
$status = 'inactive';
}
$items[] = [
'product_id' => $product->id,
'serial_name' => $request->serial_name[$i],
'serial_status' => $status,
];
}
ProductSerial::insert($items);

Answer 1

我认为您可以在ajax的帮助下检查复选框是否处于活动状态。然后将数据传递给控制器，像这样:

$("#checkbox").click(function(){

// check your checkbox if active or not
if($(this).prop("checked") == true){
var checked = true;
}else if($(this).prop("checked") == false){
var checked = false;
}
// csrf token
var headers = { 'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content') };
// passing data to laravel controller
var data = {

'id': "1",
"serial_num":$("#checkbox").attr("name"),
"status": checked
};
// ajax post request
$.ajax({
type:'POST',
url:'/YourControllerUrl',
headers: headers,
data:data,
success:function(data) {
console.log(data.msg);
}

});
});