Question

I have this hyperlink called “SEND” in a ASP page called Home and here it is:

<asp:TemplateField>
                    <ItemTemplate>
                        <asp:HyperLink ID="HyperLink1" runat="server" 
                            NavigateUrl='<%# Eval("Post_ID", "~/RCA.aspx?Post_ID={0}") %>' 
                            Text="SEND"></asp:HyperLink>
                    </ItemTemplate>

when the user clicks the hyperlink it goes to another page called RCA and in this page there is a Button and here it is the code:

<asp:Button ID="btnRCA" runat="server" onclick="Button1_Click" 
                  Text="Assign RCA" Width="147px" />

so I want this button to be visible only when clicked the hyperlink in the HOME page. I am planning to have another button or control in the RCA page that will make it invisible when clicked or before someone leaves the page they have to make it invisible the Button by clicking some other control. can someone help me with this? thanks

Answer 1

Use a QueryString parameter.

Home.aspx

//When linked to RCA.aspx from Home.aspx, a parameter called ShowButton=1 is included 
//in the URL.

<asp:HyperLink ID="HyperLink1" runat="server" 
    NavigateUrl='<%# Eval("Post_ID", "~/RCA.aspx?Post_ID={0}&ShowButton=1") %>' 
    Text="SEND"></asp:HyperLink>

RCA.aspx

//By default, since you want the button to NOT appear for all incoming traffic EXCEPT 
//that which came from Home.aspx, the button's Visible property is set to false.

<asp:Button ID="btnRCA" runat="server" onclick="Button1_Click" 
    Text="Assign RCA" Width="147px" Visible="false" />

RCA.aspx.cs

protected void Page_Load(object sender, EventArgs e)
{
    //If the querystring has a parameter called ShowButton and it's equal to "1",
    //then set the button to Visible = true.
    //Else, do nothing, keeping the button in it's default, Visible=false state.

    //By putting this in an "IsPostback == false" check, you can guarantee that this will
    //only  happen on first page_load, and won't be triggered again even if you do other 
    //actions in the page that cause Postback

    //For example, if you don't use this !IsPostback check, and you end up creating some 
    //new function that causes the button to be hidden again, but then you make a 
    //selection from a dropdown list that causes postback, you will trigger the call to 
    //make the button Visible again, even though that's probably what you don't want at 
    //this point, since your other new function set it to Visible = false.

    if (!IsPostback)
    {
        if (Request.QueryString["ShowButton"] == "1")
        {
            RCAbtn.Visible = true;
        }

        if (Request.QueryString["Post_ID"] != null)
        {
            //do whatever you need to with the post ID  
        }
    }
}

SomeOtherPage.aspx.cs

Response.Redirect("RCA.aspx?Post_ID=1234"); //button will be invisible

And then let’s say later that you want to re-direct from some other page and have the button be visible, like the redirect from Home:

Response.Redirect("RCA.aspx?Post_ID=1234&ShowButton=1"); //button will be visible

If you don’t like cluttering up your URL or you feel that it looks tacky to have what you are doing so plainly available to the user’s eyes, you don’t necessarily need to use “ShowButton”. You could say ?Post_ID=1234&fkai3jfkjhsadf=1, and then check your query string for “fkai3jfkjhsadf”. I like to do that sometimes because then from the users point of view, it makes me look like I’m doing something really technical and encrypted, and not just passing around a bunch of basic instructions in plain English 🙂 Downside there is you need keep track of your own query string parameters.

---

Edit:

If you want to get the URL with only the Post_ID and nothing else, you can do this:

string currenturl = Request.Url.ToString(); //get the current URL
string urlToSend = currenturl.Substring(0, currenturl.IndexOf("?")); //cut off the QueryString entirely
urlToSend += "?Post_ID=" + Request.QueryString["Post_ID"]; //re-append the Post_ID

Be aware that your call to Substring will cause an exception if the URL doesn’t have a QueryString, so please patch that up in whatever way works best for you (try/catch, etc.).

After that, you should just be able to use the “urlToSend” string in your mailMessage.Body.