使用jquery表单插件上传ASP.NET MVC Ajax文件？

2023年6月28日 159次阅读

我使用
Jquery Ajax Form Plugin上传文件.代码：

AuthorViewModel

public class AuthorViewModel
{
    public int Id { get; set; }

    [Required(ErrorMessage = "{0} alanı boş bırakılmamalıdır!")]
    [Display(Name = "Yazar Adı")]
    public string Name { get; set; }

    [Display(Name = "Kısa Özgeçmiş")]
    public string Description { get; set; }

    [Display(Name = "E-Posta")]
    public string Email { get; set; }

    public string OrginalImageUrl { get; set; }

    public string SmallImageUrl { get; set; }
}

形成

@using (Html.BeginForm("_AddAuthor", "Authors", FormMethod.Post, new { id = "form_author", enctype = "multipart/form-data" }))
{
    <div class="editor-label">
        <input type="file" name="file" id="file" />
    </div>
    <div class="editor-label">
        @Html.LabelFor(model => model.Name)
    </div>
    <div class="editor-field">
        @Html.EditorFor(model => model.Name)
    </div>
    <div class="editor-field">
        @Html.ValidationMessageFor(model => model.Name)
    </div>

    ...

    <div class="submit-field">
        <input type="submit" value="Ekle" class="button_gray" />
    </div>
}

脚本

<script>
$(function () {
    $('#form_author').ajaxForm({
        beforeSubmit: ShowRequest,
        success: SubmitSuccesful,
        error: AjaxError
    });
});

function ShowRequest(formData, jqForm, options) {
    $(".loading_container img").show();
}

function AjaxError() {
    alert("An AJAX error occured.");
}

function SubmitSuccesful(result, statusText) {
    // Veritabanı işlemleri başarılı ise Index sayfasına
    // geri dön, değilse partial-view sayfasını yenile
    if (result.url) {
        window.location.href = result.url;
    } else {
        $(".authors_content_container").html(result);
    }
}
</script>

调节器

[HttpPost]
public ActionResult _AddAuthor(AuthorViewModel viewModel, HttpPostedFileBase file)
{
    ...
    viewModel.OrginalImageUrl = file.FileName;
    ...
}

以上代码工作正常

题

如您所见,我从ViewModel发布文件分离.有没有办法将HttpPostedFileBase文件属性添加到ViewModel并将其绑定到视图中的viewModel,并将其发布到ViewModel中的控制器？

我希望,我可以解释一下.

编辑：

这个代码工作正常.我不想分别发布post,viewModel和HttpPostedFile.我想要这样的事情:(如果可能的话.)

模型

public class AuthorViewModel
{
    public int Id { get; set; }

    [Required(ErrorMessage = "{0} alanı boş bırakılmamalıdır!")]
    [Display(Name = "Yazar Adı")]

    HttpPostedFileBase file{ get; set; }
    ...
}

调节器

[HttpPost]
public ActionResult _AddAuthor(AuthorViewModel viewModel)
{
    var file = viewModel.file;
    ...
}

谢谢.

最佳答案 是的你可以添加AliRızaAdıyahşi.

这是属性：

public HttpPostedFileBase File { get; set; }

现在在你的形式你应该添加enctype作为小川马说：

@using (Html.BeginForm("_AddAuthor", "Authors", FormMethod.Post, new { id = "form_author", enctype="multipart/form-data" }))
{
    <div class="editor-label">
        <input type="file" name="file" id="file" />
    </div>
    <div class="editor-label">
        @Html.LabelFor(model => model.Name)
    </div>
    <div class="editor-field">
        @Html.EditorFor(model => model.Name)
    </div>
    <div class="editor-field">
        @Html.ValidationMessageFor(model => model.Name)
    </div>

    ...

    <div class="submit-field">
        <input type="submit" value="Ekle" class="button_gray" />
    </div>
}

在你的控制器动作：

[HttpPost]
public ActionResult _AddAuthor(AuthorViewModel viewModel, HttpPostedFileBase file)
{
    if(file!=null)
    {
        viewModel.File=file; //Binding your file to viewModel property
    }
    //Now you can check for model state is valid or not.
    if(ModelState.IsValid)
    {
        //do something
    }
    else
    {
        return View(viewModel);
    }
}

希望能帮助到你.这是你需要的吗？

编辑

没有别的办法了.它自动绑定到viewModel.

[HttpPost]
public ActionResult _AddAuthor(AuthorViewModel viewModel)
{
      var uploadedfile = viewModel.File;// Here you can get the uploaded file.
      //Now you can check for model state is valid or not.
    if(ModelState.IsValid)
    {
        //do something
    }
    else
    {
        return View(viewModel);
    }
}

这对我有用！

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