Re: Dynamic control add in form and finally store in to database - ASP.NET Discussion Boards

I think it would be best to Zip the files yourself, then upload them to the server thats hosting your site. Display then on a page which will then allow you to then download the files. You would have to unzip them yourself though.

Thank you very much for your reply but I don't know how to display the unziped file in a web browser, that is where my problem is. The files I need to display are different extentions from tif to excel files and I have no idea how to display them.

inside a form
1)i have a report viewer inside a div tag,which has an id,
2)a button for print

when the button is pressed a script works as given below

function printPartOfPage(elementId)
{
var printContent = document.getElementById(elementId);
var windowUrl = 'about:blank';
var uniqueName = new Date();
var windowName = 'Print' + uniqueName.getTime();
var printWindow = window.open(windowUrl, windowName, 'left=50000,top=50000,width=750,height=800');
printWindow.document.write(printContent.innerHTML);
printWindow.document.close();
printWindow.focus();
alert(printWindow);
printWindow.print();
printWindow.close();
}

when it executes a new window gets opened in which the report viewer becomes visible

MY PROBLEM IS

While taking the printout of the values in this report viewer,the values are not printed out as output.Please help me.

Hi All,

I have a paging enabled datagrid, and am using a button to show/hide the editable columns (Template fields).

Lets say the user has navigated to second page, now I want to enable only the rows of the second page on click of edit button (which is outside of grid) instead of making all the rows editable.

NOTE: I cannot use Edit/Update Template fields in datagrid.

Any help is greatly appreciated.

Sean

You could try something like this.

//Loops through each row in the DataGrid
foreach (GridViewRow Row in GridView)
{
    //Checks that we are working within the Datarows
    if (Row.RowType == DataControlRowType.DataRow)
    {
        //Finds the Textbox (txtBoxMakeEnable) in each row and sets its
        //attribute enable to true.
        ((TextBox)Row.FindControl("txtBoxMakeEnable")).Enabled = true;
    }
}

Hope this helps.

Hi guys i have posted this one already,
But i didnt get exact o/p.

So once again i am posting here,

Actually in one link button click i have to add 3 textbox in my form...,

After enter in that textbox if my enduser need to fill one more set of records,

if he click that link button again, it have to add again 3 more textbox like this i need...........,

Finally when he click submit button the datas have to store into my database...,

Some guys said to use gridview...,

Suppose if i go for gridview in footer if i tell three textbox in template columns...,

In page load there is no values in my grid...,

So it wont show the footer template column...,

So i cant go for gridview also now,

Give me some ideas to solve this...,

Thanks & Regards,
NeW OnE,
please don't forget to vote on the post

You can still show a gridview if it is empty.
http://www.google.com/search?hl=en&client=opera&rls=en&hs=hiy&q=show+empty+gridview&btnG=Search

There quite a few examples here, for how to show a gridview when there no data.

For adding textboxes dynamicaly into a form you can either add textbox controls to a form directly or to a placeholder control inside the form.For retriving values correctly i suggest you to place controls inside a placeholder.

You want to create 3 Textboxes in each link button click.Here what i had done is I placed a dropdownlist with items 1,2,3 ..etc.According to the selected item value of the dropdownlist that much set of 3 Textboxes is generated.

For that first I place a DropDownList1,PlaceHolder1 and Button1 in the form.

My code follows

protected void Page_Load(object sender, EventArgs e)
{
for (int i = 0; i < Convert.ToInt16(DropDownList1.SelectedItem.Text); i++)
{
PlaceHolder p2 = new PlaceHolder();
p2.ID = "p2" + i;
PlaceHolder1.Controls.Add(p2);
for (int j = 0; j < 3; j++)
{
TextBox tb = new TextBox();
tb.ID = "TextBoxg"+i + j;
p2.Controls.Add(tb);
RequiredFieldValidator f1 = new RequiredFieldValidator();
f1.ControlToValidate = "TextBoxg" + i+j; ;
f1.ErrorMessage ="*";
p2.Controls.Add(f1);
p2.Controls.Add(InsertBreaks(1));
}
p2.Controls.Add(InsertLineBreaks(2));
}
}
private static Label InsertBreaks(int breaks)
{
Label lblBreak = new Label();
for (int i = 0; i < breaks; i++)
{
lblBreak.Text += " ";
}
return lblBreak;
}
private static Label InsertLineBreaks(int breaks)
{
Label lblLineBreak = new Label();
for (int i = 0; i < breaks; i++)
{
lblLineBreak.Text += "
";
}
return lblLineBreak;
}
protected void Button1_Click(object sender, EventArgs e)
{
string[] data = new string[3];
foreach (Control c in PlaceHolder1.Controls)
{
int yu = 0;
foreach (Control c23 in c.Controls)
{
if (c23.GetType().ToString().Equals("System.Web.UI.WebControls.TextBox"))
{
data[yu] = ((TextBox)c23).Text;
yu = yu + 1;
}
}
//data[0], data[1], data[2] will contain the datas of each row.
//Here you can give your insert command to the database

}

}
}

Explanation:Here inside PalceHolder1 I dynamically created a placeholder with 3 Text Boxes for each row.According to the selected value of the dropdownlist, number of placeholders are dynamically created inside PlaceHolder1.For each TextBoxes I dynamically created required field validators also.If you dont want validator you can remove it and use.

Hope this code will clear your doubts

i want create templatefield in gridview
i wrote this code:
RadioButton rb = new RadioButton();
TemplateField select = new TemplateField();
select.ItemTemplate = (ITemplate)rb.TemplateControl;
select.HeaderText = "SELECT";
timeGrid.Columns.Add(select);

but dont show any radiobutton in each row. how should i do???? plz.

[Message Deleted]

Did you not pay attention to the answer given to you before?

i have 2 button and a custom control,which contains a grid view, on page.

i set datasource of gridview; sequentially rowdatabound and rowcreated events are raised. After the events are raised in custom control i need to hide buttons on page how could i do ?

Note : button are in a server side div control. So i can set the display:none with javascript in control.

Hi,

I have a problem when integrating payment process. I send a (url+query string of my site) to payment site in form of hidden field, when i do payment, payment site gives me a link in which href is same as url given by me.

Now when i click the url i actually update the payment details in my system.
Now i want to confirm that this url is coming from payment site not from my site as any one can update the details by hitting the url.

Please help me to get the address of site who send me the url.

Regards,
Anup

Try using the Request.ServerVariables Collection[^] - in particular, the HTTP_REFERER may be what you want.

Thanks for reply..

I am using the same but it gives me the url of current page instead of page from where url is opened up.

Have you any other trick??

Thanks

Hi,

I want to change value of textbox2 while entering value in textbox1.

Is it possible?

Thanks

sjs4u wrote:
Is it possible?

Yes. Hook onkeypressed or changed JS events. Get the current text from textbox1 and set to textbox2.

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R u really a programmer?

Anand Desai
Developer
Atharva Infotech

Why r u saying that?

cos I cant believe, an ASP.NET programmer dont know what u have asked

Anand Desai
Developer
Atharva Infotech

My question was simple.

Though I know the JS and the above solution given by navneet.

I was thinking, if any other solution was there or not instead of writing JS.

U can use events running on server also, but it'll cause postback n ur page will be refreshed on every change. so the best solution is using Javascript use "tb1.onlostfocus" event for best result if u want to change tb2 after completing writting in tb1. otherwise use tb1.onkeyup" for changing tb2 for every character pressed in tb1.

got it buddy?

Anand Desai
Developer
Atharva Infotech

Thanks

Is it possible? And if it is, then how? (InProc session with cookieless="false")

modified on Tuesday, August 5, 2008 3:38 AM

Sessions are not shared. Why do you want to do that ?

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