Open aspx page after posback

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Dear All,

I am using the following code to open random number of pages based on GridView CheckBox Checked Status but It only opens the last page.

MIDL

string script = "<script>";
            foreach (GridViewRow row in gvM16Deductions.Rows)
            {
                chk = (CheckBox)row.FindControl("chkDeductionType");
                if (chk.Checked)
                {
                    string url = "../../Default5.aspx?m=" + cmbSDURN.SelectedValue + "&d=" + chk.ToolTip;
                    script += "window.open('"+url+"','window','toolbar=no,menubar=no,resizable=no,directories=no');";
                    //ClientScript.RegisterClientScriptBlock(typeof(string), "openPage" + chk.Text, "javascrit: openForm('" + cmbSDURN.SelectedValue + "','" + chk.ToolTip + "');");
                }
            }
            script += "</script>";
            ClientScript.RegisterClientScriptBlock(this.GetType(), "openPage", script);

What am I doing wrong?
I can't open all pages, the script from page source shows every thing is ok, but I dont know why only the last page opens?
One more interesting thing, if I give window.open('pageadddress+querystring') then it opens all pages, but I want to open it with param.

Posted 4-Jan-11 21:28pm

Abdul Rahman Hamidy

Updated 4-Jan-11 21:52pm

Dalek Dave

v2

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Dalek Dave 5-Jan-11 3:52am

Edited for Grammar and Readability.

2 solutions

Solution 2

I don't have an specific idea, but you can try this by changing to your code.

            foreach (GridViewRow row in gvM16Deductions.Rows)
            {
                chk = (CheckBox)row.FindControl("chkDeductionType");
                if (chk.Checked)
                {
string script = "<script>";                    
string url = "../../Default5.aspx?m=" + cmbSDURN.SelectedValue + "&d=" + chk.ToolTip;
                    script += "window.open('"+url+"','window','toolbar=no,menubar=no,resizable=no,directories=no');";
                    //ClientScript.RegisterClientScriptBlock(typeof(string), "openPage" + chk.Text, "javascrit: openForm('" + cmbSDURN.SelectedValue + "','" + chk.ToolTip + "');");

            script += "</script>";
            ClientScript.RegisterClientScriptBlock(this.GetType(), "openPage", script);
                }
            }

Posted 4-Jan-11 21:35pm

Hiren solanki

Comments

Sandeep Mewara 5-Jan-11 3:37am

Check my answer Hiren!

Abdul Rahman Hamidy 5-Jan-11 3:41am

thanks for your reply,as Sandeep Mewara stated the 2nd parameter in window.open was the problem, I am changing the param in loop and that solved my problem.

Hiren solanki 5-Jan-11 3:42am

Great.

Dalek Dave 5-Jan-11 3:53am

Good call, I think Sandeep beat you by seconds!

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Sandeep Mewara · Accepted Answer · 2011-01-04T21:35:00

what am i doing wrong which i can't open all pages, the script from page source shows every thing is ok, but I dont know why only the last page opens?
This is because of the 2nd parameter in the Window.Open method.
Second parameter is 'name' of the window. Since you are using same name for all the windows (fixed name here), all open into same one and thus the last one persists.

Look at the details here: MSDN: Window.Open[^]

Resolution: Change the name of the window everytime you form the script. Append an integer or any other way you find easy.