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How to Maintain FileUpload Control’s State after PostBack

, 17 Aug 2010
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How to Maintain FileUpload Control’s State after PostBack

Introduction

Once I was trying to create a page it has two file upload controls and many other fields and around 30 post backs. Once user select files in file upload controls and in any selection when page’s post back event occurs, file upload control loose it’s value. It was very annoying situation for me to maintain the state of FileUpload Control. I search around so many articles and sites but didn't get any fixed solution. Then I decided to do it myself and publish for other developers. This is very common problem with many developers.


About ASP.Net's File Upload Control

File upload control has a text box and a browse button. When a file is selected, on PostBack PostedFile property gets initilized with HttpPostedFile object for the file. As we know that http request can't maintain state, and PostedFile is initilized with HttpPostedFile Object so it loose it's state. PostedFile property can be used to get information about file e.g. ContentLength, Content Type, FileName, InputStream. It has also a method SaveAs to store file on disk. You can read more about it on MSDN. To maintain the length of solution I will focus to code snippet.

How do we Achieve


I have seen so many options available, but many of them are not working or need too much effort. How can we do this with minimal efforts? Idea is Session Object. As we know session object can store any object and in case of OutProc/State Server Serializable Object only. So idea is very simple store fileupload object in session and in the post back get the values back from it. Use this code in Page_Load.

        //If first time page is submitted and we have file in FileUpload control but not in session 
        // Store the values to SEssion Object 
        if (Session["FileUpload1"] == null && FileUpload1.HasFile)  
        { 
            Session["FileUpload1"] = FileUpload1; 
            Label1.Text = FileUpload1.FileName; 
        } 
        // Next time submit and Session has values but FileUpload is Blank 
        // Return the values from session to FileUpload 
        else if (Session["FileUpload1"] != null && (! FileUpload1.HasFile)) 
        { 
            FileUpload1 = (FileUpload) Session["FileUpload1"]; 
            Label1.Text = FileUpload1.FileName; 
        } 
        // Now there could be another sictution when Session has File but user want to change the file 
        // In this case we have to change the file in session object 
        else if (FileUpload1.HasFile) 
        { 
            Session["FileUpload1"] = FileUpload1; 
            Label1.Text = FileUpload1.FileName; 
        }

Conclusion


This is very simple and effort less solution. Does not require any extra effort.

License

This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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About the Author

Chirantan Upadhyay
Team Leader Interglobe Technologies
India India
Hi I am chirantan upadhyay. I always like to do R&D on MS Technology and search newer and newer things. I always think something typical then try to make that possible in Code. I am B.Sc.,C-DAC,MCTS-Sharepoint,Brainbench ASP.Net and Brainbench C# Certified.
I like to have fun in Nature and Adventurous things like river rafting, rock climbing, and rippling.

Comments and Discussions

 
GeneralMy vote of 1 Pinmemberyogesh199016-Sep-14 17:59 
GeneralMy vote of 5 PinmemberRockstar_1-Aug-13 20:28 
SuggestionThis is not a good idea PinmemberTore Olav Kristiansen7-Jul-13 23:23 
GeneralRe: This is not a good idea PinmemberJuju Shen7-Aug-13 23:22 
QuestionHow to use the code. Please explain Pinmembermathewtinu20-Feb-13 23:05 
QuestionThaks But.....! PinmemberRobertDowney26926-Oct-12 21:50 
GeneralMy vote of 5 PinmemberMlsoun17-Sep-12 13:01 
GeneralRe: My vote of 5 PinmemberChirantan Upadhyay25-Sep-12 23:05 
GeneralRe: My vote of 5 Pinmembermathewtinu20-Feb-13 23:06 
GeneralMy vote of 5 PinmemberMember 787890413-Sep-12 6:18 

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