我正在laravel
中使用jquery ajax从数据库更新数据。我写了这段代码,而$ ajax()部分不起作用。但是在那之前的一切都很好。请让我知道我在哪里做错了?
1)AJAX功能:
$(".Editform").click(function() {var id=$(this).attr("id");
// alert(id);
$("#saveBtn").on ('click',function() { var name =
// alert(name);
$("input[name=name]").val();
$.ajax({type:'POST',url:'edit-records',data:{id:id,name:name },success:
function(result){alert.show(); $('#MyModal').modal('hide');
console.log(result);}});});});
2)控制器:
public function updaterec(Request $request, $id)
{
$name = $request->input('name');
DB::update('update m_employee set employee_name = ? where employee_id = ?',[$name,$id]);
echo "Record updated successfully";
}
3)web.php:
Route::post('edit-records','viewemployeeController@updaterec');
Route::get('view-records','viewemployeeController@index');
如果未在路径中定义,则不应使用$id
作为参数。
更改控制器:
public function updaterec(Request $request)
{
$name = $request->input('name');
$id = $request->input('id');
DB::update('update m_employee set employee_name = ? where employee_id = ?',[$name,$id]);
echo "Record updated successfully";
}
--------- OR ---------
更改您的路线和Ajax代码:
Route::post('edit-records/{id}','viewemployeeController@updaterec');
$.ajax({type:'POST',url:'edit-records/'+id,data:{id:id,name:name },success: function(result){alert.show(); $('#MyModal').modal('hide'); console.log(result);}});});
});