ASP.NET

Marcus J. Smith19-Mar-07 4:40

Marcus J. Smith

19-Mar-07 4:40

You need to create a new textbox everytime you plan to add one so before the Txt.ID portion of each section call for a new textbox Txt = New Textbox.

_Cle^aKO

"I think you'll be okay here, they have a thin candy shell. 'Surprised you didn't know that.'" - Tommy (Tommy Boy)
"Fill it up again! Fill it up again! Once it hits your lips, it's so good!" - Frank the Tank (Old School)

joon _vh.19-Mar-07 5:08

Re: Add Control [modified]

19-Mar-07 5:08

I think even that won't do the trick.
Shouldn't you actually declare an entirely new variable everytime?

TextBox txt1 = new TextBox();<br />
TextBox txt2 = new TextBox();<br />
TextBox txt3 = new TextBox();<br />
TextBox txt4 = new TextBox();<br />
TextBox txt5 = new TextBox();

...

public object BufferOverFlow<br />
        {<br />
__get { return BufferOverFlow; }<br />
__set { BufferOverFlow = value; }<br />
}

Marcus J. Smith19-Mar-07 5:18

Marcus J. Smith

19-Mar-07 5:18

Once it's added no. I do this often when populating a datatable with a datarow. You use the same variable, next loop set it to new again then populate and add.

-- modified at 11:43 Monday 19th March, 2007

joon _vh.19-Mar-07 5:28

19-Mar-07 5:28

I learned something today!

public object BufferOverFlow<br />
        {<br />
__get { return BufferOverFlow; }<br />
__set { BufferOverFlow = value; }<br />
}

Re: Add Control [modified]

Marcus J. Smith19-Mar-07 5:42

Marcus J. Smith

19-Mar-07 5:42

Although it may be easier to follow in the case of the OP if different names are used that match the ID given but it isnt required.

joon _vh.19-Mar-07 5:53

19-Mar-07 5:53

I usually use this in a loop, and then I put the declaration in the loop too.
Is this good practice? Or is there no performance difference.

-- modified at 12:08 Monday 19th March, 2007

I threw together this code:

<br />
    class Program<br />
    {<br />
        static void Main(string[] args)<br />
        {<br />
            Stopwatch sw = new Stopwatch();<br />
            sw.Start();<br />
            test1();<br />
            sw.Stop();<br />
            Console.WriteLine(sw.Elapsed);<br />
            sw.Reset();<br />
            sw.Start();<br />
            test2();<br />
            sw.Stop();<br />
            Console.WriteLine(sw.Elapsed);<br />
            Console.Read();<br />
        }<br />
<br />
        static void test1()<br />
        {<br />
            <br />
            String testString;<br />
            for (int i = 0; i < 50000000; i++)<br />
            {<br />
                testString = i.ToString();<br />
            }<br />
        }<br />
<br />
        static void test2()<br />
        {<br />
            for (int i = 0; i < 50000000; i++)<br />
            {<br />
                String testString;<br />
                testString = i.ToString();<br />
            }            <br />
        }<br />
    }<br />

pretty fun (clogs for about 17 secs on my pc)... but not really conclusive. I asked a MSIL guru about it. You probably don't care even half as much as I do, but I like these kinds of perf. things.

public object BufferOverFlow<br />
        {<br />
__get { return BufferOverFlow; }<br />
__set { BufferOverFlow = value; }<br />
}

vengaqua19-Mar-07 20:28

vengaqua

19-Mar-07 20:28

if i add new control at runtime suppose three buttons how can i get the click event of these three buttons.

Mohinder Singh

vengaqua19-Mar-07 20:29

vengaqua

19-Mar-07 20:29

if i do this way for the button then i get three buttons but how can i have the click event of these three buttons .

Mohinder Singh

joon _vh.19-Mar-07 20:43

ASP.NET AJAX deployment error

19-Mar-07 20:43

By setting the event in code.

LinkButton.Click += new System.EventHandler(method);

in VS2005 Intellisense does most of the thinking for you.

public object BufferOverFlow<br />
        {<br />
__get { return BufferOverFlow; }<br />
__set { BufferOverFlow = value; }<br />
}

gus_br19-Mar-07 3:00

gus_br

19-Mar-07 3:00

Hi,

I`ve done an ASP.NET AJAX website and it works perfectly on my development machine. Ok! But when I try to deploy my application on a webfarm scenario my website doesn`t work properly.
It comes out with the following error message sometimes:

"
Sys.WebForms.PageRequestManagerParserErrorException: The message received from the server could not be parsed. Common causes for this error are when the response is modified by calls to Response.Write(), response filters, HttpModules, or server trace is enabled.

Details: Error parsing near '

page load fire twice ,why?

harryforum19-Mar-07 2:35

harryforum

19-Mar-07 2:35

when i set AutoEventWireup="false" page load doesn,t fire when i set
AutoEventWireup="true" page load fire twice, and seconbd time value of
page.ispostback=false.
can any one help me

Re: page load fire twice ,why?

Sandeep Akhare24-Mar-07 1:18

Sandeep Akhare

24-Mar-07 1:18

Thanks and Regards
Sandeep

If you want something you never had,
do something you have never done!

Upload Image

MiaanG19-Mar-07 2:26

MiaanG

19-Mar-07 2:26

I am working on a Web Application and my code language is VB.Net. i want to upload image and store its path to data base after uploading. I am using following code for uploading file.

Private Function UpLoadImage(ByVal objHtmlInputFile As HtmlInputFile, ByVal FolderName As String) As String
Dim strFilePath As String = FolderName
Dim strFileName As String = ""
Dim strFileNamePath As String = objHtmlInputFile.PostedFile.FileName
If strFileNamePath.Length > 0 Then
Dim strArr() As String = strFileNamePath.Split("\\")
Dim i As Integer = strArr.Length - 1
Do While i = strArr.Length - 1
If strArr(i).IndexOf(".") > -1 Then
strFileName = strArr(i)
End If
i -= 1
Loop
If Not (objHtmlInputFile.PostedFile Is Nothing) Then
Try
objHtmlInputFile.PostedFile.SaveAs(strFilePath + "\" + strFileName)
Response.Write("Uploaded Successfully")
Catch ex As Exception
Response.Write("<br" +="" objhtmlinputfile.value="" "<="" b="">
" + ex.Message)
End Try
End If
Return strFileName
End If
End Function

I am calling this function in this line
Dim FileName As String = Me.UpLoadImage(bookimage, "book_images")
this code generates an exception when it comes to Try Catch.]
Plz help me out.

M. Azeem

Re: Upload Image

Arindam Tewary19-Mar-07 2:39

Arindam Tewary

19-Mar-07 2:39

Please mention what exception you are getting !!! I guess putting your code without error/exception message doesnot help much !!!

Thanks,
Arindam D Tewary

Re: Upload Image

MiaanG19-Mar-07 21:48

MiaanG

19-Mar-07 21:48

Sorry i forgot to mention the exception before. Following exception is generated

Could not find a part of the path "C:\WINNT\system32\book_images\filename

where as "book_images" folder is created in my peojects root directory.

M. Azeem

Re: Upload Image

Vasudevan Deepak Kumar19-Mar-07 5:31

Vasudevan Deepak Kumar

19-Mar-07 5:31

What is the Exception/Message?

Vasudevan Deepak Kumar
Personal Homepage
Tech Gossips

need help whit girdview

Lappalainen Antti19-Mar-07 1:41

Lappalainen Antti

19-Mar-07 1:41

Hi!
I have two gridview that are displaying sql queryes there is two collums and i'm comparing theys two girdview's and i need to display the differences between gridview's on third gridview this gridview has two collums calld 'name' 'amount'. So how i can add rows onthe third girdview on code.

Kschuler19-Mar-07 9:25

Kschuler

19-Mar-07 9:25

I would make a DataTable with two columns (name and amount) and put my differences in this table. Then simply set the third gridview's datasource to this new table.

Lappalainen Antti19-Mar-07 19:55

Lappalainen Antti

19-Mar-07 19:55

Thanks for the answer, but i have been trying that and not getting it working. here is my code where i'm doing wrong??

DataTable datatable = new DataTable();
DataColumn column;
// Create new DataColumn, set DataType, ColumnName
// and add to DataTable.
column = new DataColumn();
column.DataType = System.Type.GetType("System.String");
column.ColumnName = "Nimike";
datatable.Columns.Add(column);
// Create second column.
column = new DataColumn();
column.DataType = Type.GetType("System.String");
column.ColumnName = "kpl";
datatable.Columns.Add(column);
DataRow dr = datatable.NewRow();
dr["Nimike"] = "Scott";
dr["kpl"] = "25";
datatable.Rows.Add(dr);
dr = datatable.NewRow();
dr["Nimike"] = "Scott";
dr["kpl"] = "25";
datatable.Rows.Add(dr);
dr = datatable.NewRow();
dr["Nimike"] = "Scott";
dr["kpl"] = "25";
datatable.Rows.Add(dr);
dr = datatable.NewRow();
dr["Nimike"] = "Scott";
dr["kpl"] = "25";
datatable.Rows.Add(dr);
gvErot12.DataSource = datatable;
gvErot12.DataBind();

I know that the answer is simple but I cant get it working the gridview is empty but in gvErot12->Rows->Count is 4 but in the gridview is empty.
The code above is only example for the code that I'm using.

Thx

varshavmane19-Mar-07 20:01

varshavmane

19-Mar-07 20:01

I dont in C# but in VB u need to declare the Datatable as shared.It might be static in C#.

Kschuler20-Mar-07 2:51

Kschuler

20-Mar-07 2:51

Like the guy who posted before me, I also code in VB. So I'm not sure if this is a problem at all, but in VB you can't declare a column, add it to a table, and then use it to create another column and add it to the table. You have to create two instances of a column and add them separately. The same thing goes with the datarow object.

Where are you putting this code? Is it in the PageLoad event?

I would suggest putting a break point in and actually looking at the datatable to make sure it looks the way you want it to. This will narrow it down to find out if the problem is with setting up the datatable or with actually binding the data to the grid.

Sorry I couldn't be of more help.