asp.net-mvc – ASP.NET MVC.如何创建接受和multipart / form-data的Action方法

前端之家收集整理的这篇文章主要介绍了asp.net-mvc – ASP.NET MVC.如何创建接受和multipart / form-data的Action方法前端之家小编觉得挺不错的,现在分享给大家,也给大家做个参考。
我有一个Controller方法,需要接受客户端发送的多部分/表单数据作为POST请求.表单数据有2个部分.一个是序列化为application / json的对象,另一个是作为application / octet-stream发送的照片文件.我的控制器上有一个方法,如下所示:
[AcceptVerbs(HttpVerbs.Post)]
void ActionResult Photos(PostItem post)
{
}

我可以在这里通过Request.File获取文件没有问题.但是PostItem为null.
不知道为什么?有任何想法吗

控制器代码

/// <summary>
/// FeedsController
/// </summary>
public class FeedsController : FeedsBaseController
{
    [AcceptVerbs(HttpVerbs.Post)]
    public ActionResult Photos(FeedItem FeedItem)
    {
        //Here the FeedItem is always null. However Request.Files[0] gives me the file I need  
        var processor = new ActivityFeedsProcessor();
        processor.ProcessFeed(FeedItem,Request.Files[0]);

        SetResponseCode(System.Net.HttpStatusCode.OK);
        return new EmptyResult();
    }

}

电线上的客户端请求如下所示:

{User Agent stuff}
Content-Type: multipart/form-data; boundary=8cdb3c15d07d36a

--8cdb3c15d07d36a
Content-Disposition: form-data; name="FeedItem"
Content-Type: text/xml

{"UserId":1234567,"GroupId":123456,"PostType":"photos","PublishTo":"store","CreatedTime":"2011-03-19 03:22:39Z"}

--8cdb3c15d07d36a
Content-Disposition: file; filename="testFile.txt"
ContentType: application/octet-stream

{bytes here. Removed for brevity}
--8cdb3c15d07d36a--

解决方法

FeedItem类是什么样的?对于我在帖子信息中看到的内容,它应该类似于:
public class FeedItem
{
    public int UserId { get; set; }
    public int GroupId { get; set; }
    public string PublishTo { get; set; }
    public string PostType { get; set; }
    public DateTime CreatedTime { get; set; }
}

否则它将不受约束.您可以尝试更改操作签名,看看是否有效:

[HttpPost] //AcceptVerbs(HttpVerbs.Post) is a thing of "the olden days"
public ActionResult Photos(int UserId,int GroupId,string PublishTo
    string PostType,DateTime CreatedTime)
{
    // do some work here
}

您甚至可以尝试在操作中添加HttpPostedFileBase参数:

[HttpPost]
public ActionResult Photos(int UserId,DateTime CreatedTime,HttpPostedFileBase file)
{
    // the last param eliminates the need for Request.Files[0]
    var processor = new ActivityFeedsProcessor();
    processor.ProcessFeed(FeedItem,file);

}

如果你真的感觉疯狂和顽皮,请将HttpPostedFileBase添加FeedItem:

public class FeedItem
{
    public int UserId { get; set; }
    public int GroupId { get; set; }
    public string PublishTo { get; set; }
    public string PostType { get; set; }
    public DateTime CreatedTime { get; set; }
    public HttpPostedFileBase File { get; set; }
}

最后一段代码片段可能就是您想要的结果,但逐步细分可能对您有所帮助.

这个答案也可以帮助你朝着正确的方向前进:ASP.NET MVC passing Model *together* with files back to controller

原文链接:https://www.f2er.com/aspnet/248401.html

猜你在找的asp.Net相关文章