CODEWITHSUNDEEP
c#asp.netasp.net-mvchtmlrazor
MrProgram
MrProgram

Reputation: 5242

Get values from HttpPostedFileBase in MVC

My controller:

    [HttpPost]
    public ActionResult ShowExcelFile(HttpPostedFileBase getFile)
    {
        //Make something with values of getFile
        return PartialView("ShowExcelFile");
    }

In my view:

@using (Html.BeginForm("ShowExcelFile", "ShowExcel", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
     <input type="file" id="getFile" name="getFile" /><br />
     <input type="submit" value="Upload file" />
}

How can I read the values from getFile?

Upvotes: 0

Views: 5923

Answers (3)

dom
dom

Reputation: 6832

An easy way to parse an Excel file is to use a library such as Excel Data Reader (available as a Nuget package as well). And once installed, reading your Excel file is quite simple.

using Excel;

[HttpPost]
public ActionResult ShowExcelFile(HttpPostedFileBase getFile)
{
    if (getFile != null && getFile.ContentLength > 0)
    {
        // .xlsx
        IExcelDataReader reader = ExcelReaderFactory.CreateOpenXmlReader(getFile.InputStream);

        // .xls
        IExcelDataReader reader = ExcelReaderFactory.CreateBinaryReader(getFile.InputStream);

        reader.IsFirstRowAsColumnNames = true; // if your first row contains column names
    }

    return PartialView("ShowExcelFile");
}

From that point it's hard to tell your exact needs without knowing the contents of your Excel file. You can convert the file to a System.Data.DataSet which will contain every sheet and data of your Excel file.

DataSet dataSet = reader.AsDataSet();

Upvotes: 4

Nitin Varpe
Nitin Varpe

Reputation: 10694

Dont know what do you mean by

Make something with values of getFile

To get extension of file use

string fileExtension = Path.GetExtension(getFile.FileName);

//save this file using



string path = Path.Combine(Server.MapPath(Url.Content("~/Content")), "Name"+fileExtension);
file.SaveAs(path);

Upvotes: 0

hutchonoid
hutchonoid

Reputation: 33306

You could use a ViewModel to do this by adding it as a property:

View

@using (Html.BeginForm("ShowExcelFile", "ShowExcel", FormMethod.Post, new { enctype = "multipart/form-data" }))
{
     @Html.TextBoxFor(m => m.Files, new { type = "file", name = "Files" })<br />
     <input type="submit" value="Upload file" />
}

Model

public class AModel 
{
    public AModel()
    {
        Files = new List<HttpPostedFileBase>();
    }

    public List<HttpPostedFileBase> Files { get; set; }
    // Rest of model details
}

You can the retrieve the files by removing the un-needed parameter i.e.

Controller

[HttpPost]
public ActionResult ShowExcelFile(AModel model) 
 {
    var file = model.Files[0];
    ...
 }

Upvotes: 1

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