Develop Reference

Update gridview according to checkbox

I have a gridview. It contains checkboxes against all the values. Once I click the checkbox, and press submit, the selected items are inserted into another databse. I want that those selected items shouldn't appear on the current gridview if loaded next time.
For this, what I have planned is that there must be a column in current database to indicate whether the checkbox has been checked or not. The column will be filled with yes/no accordingly. And then display the rows which have only "No" in them.
However, this seems to be lengthy. Is there any other means to achieve this.

Since I believe this is a continuation of your problem from here,
I'll just copy my answer from there.
Sorry if it's not written in c#, I just noticed that tag just now.
You need to include the Primary Key or the column that makes your record Unique in its table. I'll just assume a "ImageID" column.
Place it on a hidden field on your first column.
You'll need it to issue a SQL DELETE or UPDATE statement on those rows after insertion to the destination DB. Although by using the latter, you'll need to have a new column on the table let's say - "TRANSFER_STATUS" tinyint, a switch which you can update 1 or 0.
<Columns>
<asp:TemplateField>
<HeaderTemplate>
<asp:CheckBox ID="CheckBox1" runat="server" AutoPostBack="true"
OnCheckedChanged="CheckUncheckAll"/>
</HeaderTemplate>
<ItemTemplate>
<asp:CheckBox ID ="CheckBox2" runat="server" />
<asp:HiddenField ID="hfImgID" runat="server"
Value=<%# Eval("ImageID") %> />
</ItemTemplate>
</asp:TemplateField>
Code behind:
Dim cbSelect As CheckBox, imgToInsert As Image
Dim ddlStatus As DropDownList, hfImgID As HiddenField
For Each r As GridViewRow In gvDetails.Rows
cbSelect = r.Cells(0).FindControl("CheckBox2")
If cbSelect.Checked Then
hfImgID = r.Cells(0).FindControl("hfImgID")
imgToInsert = r.Cells(1).FindControl("imgPreview")
ddlStatus = r.Cells(2).FindControl("dpdListEstatus")
/*Insert statement goes here...
DELETE or UPDATE statement goes here,
passing the ID of the image from hfImgID.Value*/
End If
Next r
**EDIT**
/*Call your SELECT statement again here to refresh your GridView.
If you used UPDATE, you need to include
your toggle switch in the WHERE clause. TRANSFER_STATUS = 0*/

I don't know what DB you are using, so this is limited, but it will get the value of a cell based on the checkbox--this assumes the chkbox is in the 3rd cell(index 2) and that there is other text in cells 0 and 1 (I don't know what your DGV looks like). it then puts the cell values of the checked rows into a dataset--from there you could dump the dataset into a db of your choice (not sure what you have). In terms of not showing the row on next load, you could change the text of cell 4 (a y/n cell, for example), then update the current database. Next time you load the DGV, include in your call to show only rows with 'y' in column X (are you using SQL?)
'get checked items and add to dataset
Dim i As Integer = 0
For i = 0 To Outdgv.RowCount - 1
If Outdgv.Rows(i).Cells(2).Value = True Then
Dim dsNewRow As DataRow
dsNewRow = dataset.Tables("report").NewRow()
Dim val1 As String = Outdgv.Rows(i).Cells(0).Value
Dim val2 As String = Outdgv.Rows(i).Cells(1).Value
dsNewRow.Item(1) = val1
dsNewRow.Item(2) = val2
dataset.Tables("report").Rows.Add(dsNewRow)
Outdgv.Rows(i).Cells(3).Value = 'n'
End If
Next

From your reply to my comment, it seems a restriction during the current session is suffient. Therefore you could keep track in a session variable.
For example, if your rows have an integer id, you could store a List<int> in session, and add ids to the list each time a checked row is submitted. Then refer to the session variable when refreshing the grid.

I'm not entirely sure what your goal appears to be, but if your trying to hide items that are checked directly from the grid that is possible.
Query: SELECT * FROM [db] WHERE ([Checked] = 0);
So the premise is that if the value is zero, it appears on the grid. If it becomes any other value, it won't appear on the grid.
Grid
<asp:TemplateField HeaderText="Active">
<ItemTemplate>
<asp:CheckBox Id="CustomerCheck" runat="server"
Checked='<%# Eval("Active") %>'
Enabled="false" />
<asp:LinkButton Id="lbCustomerActive" runat="server"
Text="Activate" CommandName="OnActivate"
ToolTip='<%# "Activate or Deactivate Course: " + Eval("CourseId") %>'
OnClick="CustomerCheck_Click" CssClass="Activator" />
</ItemTemplate>
</asp:TemplateField>
You'll now be able to actually write a method for server, called CustomerCheck_Click. That server postback will call your method which can write to the database, once that value in that column is changed the grid will no longer display it.
protected void CustomerCheck_Click(object sender, EventArgs e)
{
// Logic to modify column in database.
}

How to create editable GridView with dynamic controls

How do I create editable gridviews with dynamic controls. The following image gives an idea.
On page load, a gridview will be shown with only a single row other than the header. The row contains two dropdownlist, two textboxes and an image button. Once I enter data in the first row, and then press the image button, a new row with these controls will be created, and so on.
How is this possible?

Gridview by default does not provide a simple way of handling the Insert operations.
You will have to perform several steps to handle this.
First of all each gridview column has to be converted to TemplateColumn(TemplateField). Then in the footer template you have to insert your controls(dropdown, textbox etc.)
<asp:TemplateField HeaderText="Item Category">
<%--Define item template--%>
<FooterTemplate>
<asp:DropDownList ID="dropDownListCategory" runat="server"></asp:DropDownList>
</FooterTemplate>
</asp:TemplateField>
...... similarly other footer templates
You have to add the onclick Event for the ImageButton.
protected void imageButtonInsert_Click(object sender, EventArgs e)
{
//Get the gridview row
GridViewRow row = (sender as Control).NamingContainer as GridViewRow;
DropDownList dropDownListCategory = row.FindControl("dropDownListCategory") as DropDownList;
///Similarly you can access the other controls here
//If you are using SqlDataSource then you can add/assign these values to the insert parameters
//Note that, you need to have insertcommand defined for the sql data source with appropreate parameters
SqlDataSource1.InsertParameters.Add("category", dropDownListCategory.SelectedValue);
//Similarly assign the other parameter values
//Call the insert method of the sql data source.
SqlDataSource1.Insert();
//If you are not using data source approach, here you can insert the data to
// database by calling sql command or other ways
//Rebind the gridview.
GridView1.DataBind();
}
This should work, when you have some rows to display in the gridview.
BIG PROBLEM : EMPTY ROWS
The above approach works when you have at least one row in the gridview. But if you do not have any rows, that is for the first time, the gridview will be empty. If the gridview data source is empty, then it does not display anything, even header,footer, rows etc.
As a initial workaround, first you need to make the header is always visible
ShowHeaderWhenEmpty="true" ShowFooter="true"
This makes sure that you have the headers all the time.
Now footer will not at all display unless and until there is some row in the gridview.
As a work around you can use EmptyDataTemplate of the gridview.
Here you need to add all your controls again, dropdownlist, textboxes etc. Not that you have to use here to make the same layout. You can use the same control ids, so that you dont need to write the codebehind again
<EmptyDataTemplate>
<tr>
<td><asp:DropDownList ID="dropDownListCategory" runat="server"></asp:DropDownList></td>
<td><asp:Button runat="server" CommandName="Insert" ID="buttonInsert" OnClick="imageButtonInsert_Click" Text="Insert" /></td>
<%--Other controls go here--%>
</tr>
</EmptyDataTemplate>
This is now good for both first time and successive load.
The next question you may have is how to bind these dropdowns. For this you need to use the RowCreated event, in which you can access the dropdowns and fill them
protected void GridView1_RowCreated(object sender, GridViewRowEventArgs e)
{
if (e.Row.RowType == DataControlRowType.Footer || e.Row.RowType == DataControlRowType.EmptyDataRow)
{
DropDownList dropDownListCategory = e.Row.FindControl("dropDownListCategory") as DropDownList;
//similarly access the other controls and bind them
}
}

For Doing this use can use Dynamic table at client side. where you can create whole table dynamically and also apply styling that can look like grid view. for Styling use can use JQuery DataTables which provice much more controls.
And also by using JQuery DataTables or JQuery Code you can also add rows dynamically at client side.

ASP.NET Gridview with Link button column on Dynamic Grid

ASP.NET Gridview with Link button column on Dynamic Grid
The ASP.NET Gridview should have a linked column - 1st column, when clicked should take to another page with the clicked cell value. the Grid is a dynamic one, that is the columns are not fixed, no of columns/ column itself is dynamic. I added a asp control just for the first columns and remaining columns are dynamic, in the code behind and I add the first column "linkbutton" always.
<Columns>
<asp:TemplateField HeaderText="linkbutton">
<ItemTemplate>
<asp:LinkButton ID="linkbutton" runat="server" Text='<%#EVal("linkbutton") %>'
CommandName="ShowDetails" CommandArgument='<%#Eval("linkbutton") %>'>
</asp:LinkButton>
</ItemTemplate>
</asp:TemplateField>
</Columns>
linkbutton is part of the datatable, so when I bind the datatable to gridview it appears twice, once for the templatefield and once from the datatable bind. gridview.column().visible=false didnot work as it considers linkbutton as the only column not the datatable columns.
i tried to add the Linkbutton control from code behind that one also didnt work.

I would set the autogeneratecolumns to false, and then use asp:BoundFields in order to show the columns from the datatable, other than the linkbutton, which you are handling with the asp:TemplateField.
You can then use GridView.Columns().Visible to hide/show different columns.
You were not able to use GridView.Columns().Visible with the auto generated columns because they are not added to Columns():
Explicitly declared column fields can be used in combination with
automatically generated column fields. When both are used, explicitly
declared column fields are rendered first, followed by the
automatically generated column fields. Automatically generated column
fields are not added to the Columns collection.

Show and Hiding information in ASP.NET data grid

i have a datagrid which contains a lot of information in one column.
I would like a way for the user to preview some of the content in that column and then click a "more" hyperlink in that column that takes them to another page that displays all the information in that column allowing them to add more info into it and other items in that row.
I have done the more hyper link on the end of all the information but i would i change for example:
"HI my name is stefan i like walking on the beach and other stuff...more"
to
"Hi my name is stefan... more"
And how would i pull up the selected row in an new datagrid on a new asp page after they click more?
When i click the more hyperlink it doesn't open the correct row but only blogid 2.

you can use HyperLink in GridView. You can use CommangArgument and CommandName Property to Hold the specic data. When u clicked the hyperlink the data will be passed as querystring to the next page. so that you can use the querytring
And to display " ... more" you can use substring method.. substring for a legnth and concat the ....more to your hyperlink text.
MSDN for CommandArgument
Sample for Passing data with LinkButton

Handle OnRowDataBound as so:
protected void GridView1_RowDataBound(Object sender, GridViewRowEventArgs e)
{
if(e.Row.RowType == DataControlRowType.DataRow)
{
// say for example that the column with long text is the 5th one
e.Row.Cells[5].Text = e.Row.Cells[5].Text.Length>20?e.Row.Cells[5].Substring(0,20)+" more...";
}
}
Where the first column (Cells[0]) has the blog ID. I think that's how you have your gridview now.

I think maybe you're looking for the text-overflow: ellipsis; CSS styling. You can then use jQuery or plain javascript to toggle the style and reveal all the text.
I'm not quite sure how this is handled by Firefox today, but IE is ok.

I managed to fix this by using an sql query,
Left(content,50)
Which displays the first 50 characters.

You can try following code:
<asp:TemplateField HeaderText="Description" HeaderStyle-HorizontalAlign="Left" ItemStyle-HorizontalAlign="Left">
<ItemTemplate>
<%# (Eval("strDescription").ToString().Length > 55 ? Eval("strDescription").ToString().Substring(0, 55) + "....." : Eval("strDescription"))%>
</ItemTemplate>
</asp:TemplateField>
and just add hyperlink in this item template and assign your blogid(in your case) to that hyperlink to open detail page that you want.

Adding checkbox dynamically to dataGrid in C# Visual studio 2005

I have a dataGrid(not dataGridView) in which i need to add Checkbox dynamically at the first column. How can i do this.?

In the ItemDataBound event, which fires for each row of the DataGrid, you could dynamically add a control to the first cell. Easier to use a TemplateColumn, but if you want to do it dynamically as you asked, this is how I'd do it.
private void DataGrid1_ItemDataBound(object sender,
System.Web.UI.WebControls.DataGridItemEventArgs e)
{
if ((e.Item.ItemType == ListItemType.AlternatingItem) ||
(e.Item.ItemType == ListItemType.Item))
{
CheckBox chk = new Checkbox();
e.Item.Cells[0].Controls.Add(chk);
}

The following woprks with a GridView, and I believe that it's also the same for a DataGrid.
Just add a Template Column. (You can do that in source or via the GUI). Then add a checkbod to the ItemTemplate:
<Columns>
<asp:TemplateField>
<HeaderTemplate>
Retry
</HeaderTemplate>
<ItemTemplate>
<asp:CheckBox ID="CheckBox1" runat="server" />
</ItemTemplate>
</asp:TemplateField>
</Columns>
ps. Might want to consider dropping in a GridView if you're on 2.0+

I don't believe you can add them to the first column dynamically if the datagrid already has columns, because it would just append the new column to a datagrid by using the Add method, making it the last column:
CheckBoxColumn checkCol = new CheckBoxColumn();
DataGrid1.Columns.Add(checkCol);
but to add them dynamically you could either follow thse steps at CodeProject Adding a CheckBox column to your DataGrid
or
If you still want the look of them being in the first column, then you can just create them in your client side code and set their visibile attribute to false and when you conditions are met in code behind then set the attribute to true, which gives the idea that it was crated dynamically