How to open file in browser with hyperlink

Question

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public string GetCheckListDocUrl(object DeliverablesID)
{

SOBO objScope = new SOBO();
objScope.DelivarablesID = Convert.ToInt32(DeliverablesID);
DataSet dsDelivarablesDoc = objScope.GetSODeliverablesDocument();
// scopeDoc = dsDelivarablesDoc.Tables[0].Rows[0]["PLC_Filepath"].ToString();
string siteurl = System.Configuration.ConfigurationManager.AppSettings["websiteroot"].ToString();
string path = DeliverablesID.ToString().Replace("~", "").TrimStart('/');
siteurl = siteurl + path;
return siteurl;
}

What I have tried:

i'm trying to open a file by clicking on filename on browser(Google Chrome) but when i'm clicking on file despite of having right address browser does not open the file

Posted 27-Jun-16 1:19am

Member 10549697

Updated 28-Jun-23 2:48am

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Comments

[no name] 27-Jun-16 7:28am

What kind of file is that ? As long as I know, it is a security problem, if you can open file from browser click. Browser can only automatically open HTM, HTML, TXT, Images.... and not EXE, ZIP, ...

Member 10549697 27-Jun-16 7:29am

even it does not open text files

2 solutions

Solution 2

How to open file in browser with hyperlink?

Posted 28-Jun-23 2:48am

vishal_jaimataji

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Add your solution here

Treat my content as plain text, not as HTML

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Madhu Nair · Accepted Answer · 2016-06-27T01:41:00

Solution 1

Try the given below code -

C#

WebClient req = new WebClient();
HttpResponse response = HttpContext.Current.Response;
response.Clear();
response.ClearContent();
response.ClearHeaders();
response.Buffer = true;

response.AddHeader("Content-Disposition", "attachment;filename=\"" + System.IO.Path.GetFileName(HttpContext.Current.Server.MapPath(FileName)) + "\"");
byte[] data = req.DownloadData(HttpContext.Current.Server.MapPath(FileName));
response.BinaryWrite(data);
response.End();

Posted 27-Jun-16 1:41am

Madhu Nair

Comments

Member 10549697 27-Jun-16 7:57am

how to add filename in this code?

Madhu Nair 27-Jun-16 8:48am

Create a function with the code and use FileName variable mentioned in the code as a parameter where in you can pass the full path of the file name. In the code i posted I was passing the file name as "~/upload/test.txt". Hence I used Server.MapPath function.

Member 10549697 29-Jun-16 3:02am

okay.thank you.it helped