Home / Questions / 22198
Accepted
Antonio
Asked: 2020-09-01 10:44:27 +0800 CST 2020-09-01 10:44:27 +0800 CST

Assign value to an input returned from a controller

  • 772

i have this scriptin my sight

    $(document).ready(function () {
        $("#PlanVenta").change(function () {
            $("#Precio").empty();
            $.ajax({
                type: 'POST',
                url:'@Url.Action("PreciosPorTipoVenta")',
                datatype:'json',
                data:{Id_Plan:$("#PlanVenta").val(),Id_Articulo:$("#Id_Articulo").val()},
                success:function(precios){
                    $("#Precio").val(?????);
                },
            });
        });
    });
which sends data to this controller

public JsonResult PreciosPorTipoVenta(int Id_Plan, int Id_Articulo)
        {
            return Json(querty, JsonRequestBehavior.AllowGet);
        }

and I have this input to which I want to assign the value that the controller returns from ajax

<input type="text" id="Precio" name="Precio" />

what should I put there because I already tried to put precios.Valueas I had seen in an example but nothing comes out
enter image description here

jquery

2 Answers

  1. You need to go through the json returned, in this case you catch it as price, you have to go through the json price depending on the attributes that it has and that you want to show in the example input:

    $(document).ready(function () {
        $("#PlanVenta").change(function () {
            $("#Precio").empty();
            $.ajax({
                type: 'POST',
                url:'@Url.Action("PreciosPorTipoVenta")',
                datatype:'json',
                data:{Id_Plan:$("#PlanVenta").val(),Id_Articulo:$("#Id_Articulo").val()},
                success:function(precios){
                    //precios es la respueta de tu contralador
                    $('#Precio').val(precios.propiedad);

                },
            });
        });
    });
    • 3
  2. Best Answer

    oscar 's answer should be valid since you get a jsonthis way {precios:[{precio:600}]}.

    Try $('#Precio').val(precios[0].precio);as it is possible that you are getting an array with more data

    • 2