我在地图上显示某些内容时在 laravel 中收到错误 419，即使我的路径正确

Question

我使用谷歌地图API创建一个应用程序，允许用户在特定位置添加标记 但是当我单击地图添加位置时，我返回的所有数据都是正确的，但我得到错误 19 的响应代码......这是我保存数据的代码

function saveData(lat,lng) 
{
console.log("LAT:"+lat+" LNG:"+lng);
var description = document.getElementById('manual_description').value;
var formData = new FormData();
console.log("DESC:"+description);
formData.append("latitude", lat); //outputs correctly
formData.append("longitude", lng);//outputs correctly
formData.append("description", description); //outputs correctly
console.log(formData.get("latitude")+" "+formData.get("longitude")+" "+formData.get("description"));
downloadUrl(formData, function(data, responseCode) {
console.log("RESP:"+responseCode);
console.log(data.length);
if (responseCode === 200  && data.length > 1) {
var markerId = getMarkerUniqueId(lat,lng); // get marker id by using clicked point's coordinate
var manual_marker = markers[markerId]; // find marker
manual_marker.setIcon(purple_icon);
infowindow.close();
infowindow.setContent("<div style=' color: purple; font-size: 25px;'> Waiting for admin confirm!!</div>");
infowindow.open(map, manual_marker);
}else{
console.log("RESPONSECODE: "+ responseCode); //419
console.log("DATA: "+data); //html data with title of Page Expired
infowindow.setContent("<div style='color: red; font-size: 25px;'>Inserting Errors</div>");
}
});
function downloadUrl(formData, callback) 
{
var xhttp = new XMLHttpRequest;
xhttp.onreadystatechange = function() {
console.log(xhttp.readyState);
if (xhttp.readyState == 4) {
callback(xhttp.responseText, xhttp.status);
}
};
xhttp.open("POST", "http://localhost/cacheFinder/public/save");
xhttp.setRequestHeader('X-CSRF-TOKEN', '{{ csrf_token() }}');
xhttp.send(formData);
}

这是我的路线Route::post('save', 'MarkersController@save');

这是保存功能：

public function save(Request $request){
$lat = $request->input("latitude");
$lng = $request->input("longitude");
$description = $request->input("description");
Marker::create([
'latitude'=>$lat,
'longitude'=>$lng,
'description'=>$description
]);

return redirect('usermap');
}

任何帮助将不胜感激

Answer 1

将其添加到您的VerifyCsrfToken中间件中

protected $except = [
'/cacheFinder/public/*'
];

或

xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

xmlhttp.setRequestHeader('X-CSRF-TOKEN',window.Cookies.get('_csrf'));

Answer 2

这是因为 https://laravel.com/docs/5.7/csrf 在 CsrfMiddleware 中禁用路由保护或发布有效值