随机图像名称图像上传MVC 4

dk_*_*032 0 c# asp.net-mvc asp.net-mvc-4

我希望为我在MVC 4 Web App中上传的图像生成一个随机名称.

我的控制器:

[HttpPost]
[ValidateAntiForgeryToken]
[ValidateInput(false)]
public ActionResult Create(Article article, HttpPostedFileBase file)
{
    if (ModelState.IsValid)
    {
        if (file != null && file.ContentLength > 0)
        {
            // extract only the filename
            var fileName = System.IO.Path.GetFileName(file.FileName);
            // store the file inside ~/App_Data/uploads folder
            var path = System.IO.Path.Combine(Server.MapPath("~/UploadedImages/Articles"), fileName);
            file.SaveAs(path);
            article.ArticleImage = file.FileName;
            ViewBag.Path = String.Format("~/UploadedImages/Events", fileName);
        }
        db.Articles.Add(article);
        db.SaveChanges();
        return RedirectToAction("Index");
    }

    ViewBag.SportID = new SelectList(db.Sports, "SportID", "Name", article.SportID);
    return View(article);
}
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我尝试过使用GetRandomFileName方法,但没有运气.不确定这是否是正确的方法.

提前致谢!

Zoh*_*led 5

最简单的方法可能是Guid.NewGuid()用作文件名.对于大多数目的而言,它是伪随机的,并且足够独特,可以Guid创建之前创建的非常低的机会.
您需要使用Path的GetExtension方法从原始文件名中提取文件扩展名,并且对于前面提到的非常低的机会,我建议编写如下方法:

string GenerateFileName(string TergetPath, HttpPostedFileBase file)
{
    string ReturnValue;
    string extension = Path.GetExtension(file.FileName);
    string FileName = Guid.NewGuid().ToString();
    ReturnValue = FileName + extension;
    if(!File.Exists(Path.Combine(TergetPath, ReturnValue)) 
    {
        return ReturnValue;
    }
    // This part creates a recursive pattern to ensure that you will not overwrite an existing file
    return GenerateFileName(TergetPath, file); 
}
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然后你可以从你现有的代码中调用它,如下所示:

var DirectoryPath = Server.MapPath("~/UploadedImages/Articles");
var path = GenerateFileName(DirectoryPath, file);
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