Jho*_*rra 10 c# asp.net checkbox postback gridview
我有一个带有模板字段的gridview.在该模板字段中是一个复选框.我在gridview之外有一个提交按钮来分配已检查的记录.在回发时没有复选框注册为正在检查.这是我的代码:
<Columns>
<asp:TemplateField>
<ItemTemplate>
<asp:CheckBox ID="cb" Checked="false" runat="server" />
<asp:Label ID="lblCFID" runat="server" Visible="false" Text='<%# Eval("ID") %>' />
</ItemTemplate>
</asp:TemplateField>
<asp:BoundField HeaderStyle-HorizontalAlign="Center" DataField="Name" HeaderText="Name" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" DataField="DOB" HeaderText="Date of Birth" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" HeaderText="Gender" DataField="Gender" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" HeaderText="Status" DataField="Status" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" HeaderText="Plan Name" DataField="PlanName" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" HeaderText="Type" DataField="ControlType" />
<asp:BoundField HeaderStyle-HorizontalAlign="Center" HeaderText="Date of Service" dataformatstring="{0:MMMM d, yyyy}" htmlencode="false" DataField="DateofService" />
</Columns>
protected void AssignRecords(object sender, EventArgs e)
{
int Rows = gvASH.Rows.Count;
for (int i = 0; i < Rows; i++)
{
//CheckBoxField cb = ((CheckBoxField)gvASH.Rows[i].Cells[1]).;
CheckBox cb = (CheckBox)gvASH.Rows[i].Cells[0].FindControl("cb");
Label lblID = (Label)gvASH.Rows[i].Cells[0].FindControl("lblCFID");
if (cb.Checked == true)
{
string ID = lblID.Text;
//Assign Code
}
}
}
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我在字符串ID = lblID.Text上设置断点; 但它永远不会发现任何被检查.
Muh*_*tar 14
我认为你缺少的是,当你点击按钮并且你的页面是回发时,你重新绑定到gridview,你需要在这种情况下绑定
if (!Page.IsPostBack)
{
GridView1.DataSourceID = "yourDatasourceID";
GridView1.DataBind();
}
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