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分享程序员开发的那些事...
且构网 - 分享程序员编程开发的那些事

你如何在MVC5中上传pdf文件?

更新时间:2023-02-24 22:32:45

命名空间Template.MVC5.Controllers

{

公共类FileLoadController:Controller

{



private readonly FileloadBusiness _db = new FileloadBusiness();



public ActionResult Index()

{

var x = new FileloadBusiness();

return View( x.GetAll());

}







public ActionResult创建()

{

返回查看();

}



[HttpGet]

public ActionResult Uploaded()

{

ViewBag.Message =doc;

返回查看();

}

[HttpPost]

public ActionResult Uploaded(Fileload查看型号)

{

返回RedirectToAction(CustHome,客户);

}




[HttpPost]

公共ActionResult创建(FileloadView doc)

{

试试

{

_db.Insert(doc);



返回RedirectToAction(Uploaded);
}

catch(例外e)

{



返回查看() ;

}

}



DataContext db = new DataContext();



公共ActionResult下载()

{

返回查看(_db.GetAll());

}



public FileContentResult FileDownload(int? id)

{

byte [] fileData;

string fileName;



var x = new FileloadBusiness();

var fileRecord = db.FileLoads.Find(id);





fileData =(byte [])fileRecord.File.ToArray();

fileName = fileRecord.FileName;





返回文件(fileData,text,fileName);





}



}

}
namespace Template.MVC5.Controllers
{
public class FileLoadController : Controller
{

private readonly FileloadBusiness _db = new FileloadBusiness();

public ActionResult Index()
{
var x = new FileloadBusiness();
return View(x.GetAll());
}



public ActionResult Creating()
{
return View();
}

[HttpGet]
public ActionResult Uploaded()
{
ViewBag.Message = "doc";
return View();
}
[HttpPost]
public ActionResult Uploaded(FileloadView model)
{
return RedirectToAction("CustHome", "Customer");
}


[HttpPost]
public ActionResult Creating(FileloadView doc)
{
try
{
_db.Insert(doc);

return RedirectToAction("Uploaded");
}
catch (Exception e)
{

return View();
}
}

DataContext db = new DataContext();

public ActionResult Download()
{
return View(_db.GetAll());
}

public FileContentResult FileDownload(int? id)
{
byte[] fileData;
string fileName;

var x = new FileloadBusiness();
var fileRecord = db.FileLoads.Find(id);


fileData = (byte[])fileRecord.File.ToArray();
fileName = fileRecord.FileName;


return File(fileData, "text", fileName);


}

}
}