Problem with gridview checkbox [modified]

[pinna_hari]

Hello, I have a gridview with checkboxes(inside Gridview) and a Button contol(outside of Gridview)with BtnDeleteSelected My aim is to get the value of 1st row 2nd column in gridview(1st Column is CheckBox control) and 2nd column is pid value.I want 'Pid' Value i wrote this code int pID = Convert.ToInt32(gv.DataKeys[row.RowIndex].Value); it is showing this error

Index was out of range. Must be non-negative and less than the size of the collection.
Parameter name: index

Pinna

modified on Thursday, May 22, 2008 9:15 AM

[NetBot]

haripinna wrote:

GridViewRow row = gv.Rows[i]; CheckBox chkbox = (CheckBox)row.FindControl("chkSelect")

Oh what code have u posted. I can`t get head or tails out of it..Pls post the entire code so that we can understand your logic implementation.

[eyeseetee]

at the moment you are only finding the checkbox Wheres the code which should be setting the checkbox to true or false Can you explain a bit more?

[pinna_hari]

Hello, Thanks for your response. I got answer for that question.. now my problem is I have a gridview with checkboxes(inside Gridview) and a Button control(outside of Gridview)with BtnDeleteSelected My aim is to get the value of 1st row 2nd column of gridview(1st Column is CheckBox control) and 2nd column is pid value.I want 'Pid' Value i wrote this code

for(i=0;i
{
GridViewRow row = gv.Rows[i];
CheckBox chkbox = (CheckBox)row.FindControl("chkSelect")
if(chkbox.checked)
{
int pID = Convert.ToInt32(gv.DataKeys[row.RowIndex].Value);
}
}

it is showing this error Index was out of range. Must be non-negative and less than the size of the collection. Parameter name: index

Pinna

[sukkureddy]

Try this string strContactId = "" + gvContacts.SelectedDataKey["ContactId"]; or string strContactId = "" + gvContacts.SelectedValue; You have to write these Code in OnSelectedIndexChanged event of gridview.Initially you have to set Datakeynames="ContactId" in gridview property. i hope it wil work for u

[pinna_hari]

Thanks..got it..

Pinna