我是laravel的新手,所以请考虑一下。当我使用jquery.post方法时,我总是得到结果失败,我不知道为什么这是路线的问题或不在这里是我的代码请看看
<script>
$(".fa-exchange").click(function(){
var doctor_id=$(this).attr('doctorid');
$("#doctor_id").val(doctor_id);
var status=$(this).attr('status');
$("#status").val(status);
});
$("#change-ok").click(function(){
var doctor_id=$("#doctor_id").val();
var status=$("#status").val();
var url='{{url('/change-doctor-status/')}}';
$.post(url, {doctor_id:doctor_id,status:status }).done(function (result) {
alert("success");
}).fail(function(result) {
alert("There was an error. Try again please!");
});
});
</script>
我的控制器看起来像这样
public function change_doctor_status(Request $request)
{
$id = $request['doctor_id'];
$status = $request['status'];
if($status==1)
{
$query=DB::table('doctors')->where('id',$id)->update(['status' => 0]);
if($query)
{
return true;
}
}
else if($status==0)
{
$query=DB::table('doctors')->where('id',$id)->update(['status' => 1]);
if($query)
{
return true;
}
}
}
我这样做了
Route::post('/change-doctor-status', 'Admin\AdminController@change_doctor_status');
在这里,当我找到结果时,我总是得到这样的警报(&#34;出现了错误。请再试一次!&#34;);它存储在失败部分
答案 0 :(得分:1)
如果您在刀片文件中写这个
$.post(url, {doctor_id:doctor_id,status:status }).done(function (result) {
alert("success");
}).fail(function(result) {
alert("There was an error. Try again please!");
再添加一行。
$.post(url, {doctor_id:doctor_id,status:status,_token:'{{ csrf_token() }}' }).done(function (result) {
alert("success");
}).fail(function(result) {
alert("There was an error. Try again please!");