本文介绍了如何使用mvc 4在数据库中保存上传的文件名?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
代码:
code in business layer :
FilesEntity files = new FilesEntity();
public void SaveFileName(FilesEntity filentity)
{
string cs = ConfigurationManager.ConnectionStrings["DBCS"].ConnectionString;
using (SqlConnection con = new SqlConnection(cs))
{
SqlCommand cmd = new SqlCommand("spnotaryfiles", con);
cmd.Parameters.AddWithValue("@Filename", files.filename);
cmd.CommandType = CommandType.StoredProcedure;
con.Open();
cmd.ExecuteNonQuery();
}
}
COntroller中的代码:
Code in COntroller :
[HttpPost]
public ActionResult Index()
{
{
FilesEntity entity = new FilesEntity();
string path = Path.Combine(Server.MapPath("~/Images"),
Path.GetFileName(entity.filename));
DataAccess da = new DataAccess();
da.SaveFileName(entity);
}
return View();
}
视图中的COde
COde in view
@using (Html.BeginForm("Index", "File", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
<input type="file" name="fileUpload" /><br />
<input type="submit" name="Submit" id="Submit" value="Upload" />
}
请帮我将文件名保存在数据库中。提前谢谢
please help me to save the file name in database. thanks in advance
推荐答案
FilesEntity files = new FilesEntity();
public void SaveFileName(FilesEntity filentity)
{
files = filentity;
string cs = ConfigurationManager.ConnectionStrings["DBCS"].ConnectionString;
using (SqlConnection con = new SqlConnection(cs))
{
SqlCommand cmd = new SqlCommand("spnotaryfiles", con);
cmd.Parameters.AddWithValue("@Filename", files.filename);
cmd.CommandType = CommandType.StoredProcedure;
con.Open();
cmd.ExecuteNonQuery();
}
}
-KR
-KR
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