Question

我是MVC Architechture的新手。我想将模型的数据传递给行动，但我不知道如何通过它。

当我执行以下代码时，我认为它会自动传递模型数据。

View of ShoppingController:

@using (Ajax.BeginForm("MakePayment", "Shopping", new AjaxOptions { HttpMethod = "POST", InsertionMode = InsertionMode.Replace, OnBegin = "VarifyParameters" }))
{
    <table>
        <tr>
            <td>
                First Name *
            </td>
            <td>
                @Html.TextBoxFor(model => model.firstName, new { @id = "txtFName", @class = "text-field" })
            </td>
        </tr>
        <tr>
            <td>
                Last Name *
            </td>
            <td>
                @Html.TextBoxFor(model => model.lastName, new { @id = "txtLName", @class = "text-field" })
            </td>
        </tr>
        <tr>
            <td>
                <input type="button" value="Make Payment" onclick="makePayment()" />
            </td>
        <tr>
    <table>
    //there are more button here with different functionalities
 }


//javascript
 function VarifyParameters() {
 try {
        return false;
  } catch (e) {

    }
  }

我没有获得任何数据。如何通过javascript将所有模型数据传递给操作

//checkoutController  
public ActionResult MakePayment(PaymentDetails paymentDetails)
{
     //paymentDetails are always null
}

有人可以建议我如何传递模型数据吗？

Answer 1

您可以使用Html.BeginForm并将按钮类型更改为submit：

@using (Html.BeginForm("MakePayment", "Shopping"))
    {
    <table>
        <tr>
            <td>
                First Name *
            </td>
            <td>
                @Html.TextBoxFor(model => model.firstName, new { @id = "txtFName", @class = "text-field" })
                @Html.ValidationMessageFor(model => model.firstName)
            </td>
        </tr>
        <tr>
            <td>
                Last Name *
            </td>
            <td>
                @Html.TextBoxFor(model => model.lastName, new { @id = "txtLName", @class = "text-field" })
                @Html.ValidationMessageFor(model => model.lastName)
            </td>
        </tr>
        <tr>
            <td>
                <input type="submit" value="Make Payment" />
            </td>
        <tr>
    <table>
    }

您必须将[HttpPost]添加到您的控制器：

//checkoutController
[HttpPost]  
public ActionResult MakePayment(PaymentDetails paymentDetails)
{
     //your code
}

您也可以使用Ajax.BeginForm，并且可以使用其OnBegin，OnSuccess，OnFailure功能。像这样：

@using (Ajax.BeginForm("MakePayment", "Shopping", new AjaxOptions { HttpMethod = "POST", InsertionMode = InsertionMode.Replace, OnBegin = "VarifyParameters" }))
            {

        <table>
            <tr>
                <td>
                    First Name *
                </td>
                <td>
                    @Html.TextBoxFor(model => model.firstName, new { @id = "txtFName", @class = "text-field" })
                    @Html.ValidationMessageFor(model => model.firstName)
                </td>
            </tr>
            <tr>
                <td>
                    Last Name *
                </td>
                <td>
                    @Html.TextBoxFor(model => model.lastName, new { @id = "txtLName", @class = "text-field" })
                    @Html.ValidationMessageFor(model => model.lastName)
                </td>
            </tr>
            <tr>
                <td>
                    <input type="submit" value="Make Payment" />
                </td>
            <tr>
        <table>

}

但是在使用ajax时，请不要忘记在视图中添加jquery.unobtrusive-ajax.min.js引用。

从视图传递模型数据但不使用MVC 4

1 个答案: