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LinkButton.PostBackUrl Property

Definition

Gets or sets the URL of the page to post to from the current page when the LinkButton control is clicked.

public:
 virtual property System::String ^ PostBackUrl { System::String ^ get(); void set(System::String ^ value); };
[System.Web.UI.Themeable(false)]
public virtual string PostBackUrl { get; set; }
[<System.Web.UI.Themeable(false)>]
member this.PostBackUrl : string with get, set
Public Overridable Property PostBackUrl As String

Property Value

The URL of the Web page to post to from the current page when the LinkButton control is clicked. The default value is an empty string (""), which causes the page to post back to itself.

Implements

Attributes

Examples

The following code example demonstrates how to use the PostBackUrl property to perform a cross-page post. When the user clicks the LinkButton control, the page posts the value entered in the text box to the target page specified by the PostBackUrl property. To run this sample, you must also create a file for the target page in the same directory as this code example. The code for the target page is provided in the next example.

Important

This example has a text box that accepts user input, which is a potential security threat. By default, ASP.NET Web pages validate that user input does not include script or HTML elements. For more information, see Script Exploits Overview.

<%@ page language="C#" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" >
<head id="head1" runat="server">
  <title>LinkButton.PostBackUrl Example</title>
</head>
<body>    
  <form id="form1" runat="server">
    
    <h3>LinkButton.PostBackUrl Example</h3>

    Enter a value to post:
    <asp:textbox id="TextBox1" 
      runat="Server">
    </asp:textbox>

    <br /><br />

    <asp:linkbutton id="LinkButton1" 
      text="Post back to this page"
      runat="Server">
    </asp:linkbutton>

    <br /><br />

    <asp:linkbutton id="LinkButton2"
      text="Post value to another page" 
      postbackurl="LinkButton.PostBackUrlPage2cs.aspx" 
      runat="Server">
    </asp:linkbutton>

  </form>
</body>
</html>
<%@ page language="VB" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" >
<head id="head1" runat="server">
  <title>LinkButton.PostBackUrl Example</title>
</head>
<body>    
  <form id="form1" runat="server">
    
    <h3>LinkButton.PostBackUrl Example</h3>

    Enter a value to post:
    <asp:textbox id="TextBox1" 
      runat="Server">
    </asp:textbox>

    <br /><br />

    <asp:linkbutton id="LinkButton1" 
      text="Post back to this page"
      runat="Server">
    </asp:linkbutton>

    <br /><br />

    <asp:linkbutton id="LinkButton2"
      text="Post value to another page" 
      postbackurl="LinkButton.PostBackUrlPage2vb.aspx" 
      runat="Server">
    </asp:linkbutton>

  </form>
</body>
</html>

The following code example demonstrates how to use the Page.PreviousPage property to access a value that was posted from another page using the PostBackUrl property. This page gets the string that was posted from the previous page and displays it to the user. If you attempt to run this code example directly, you will get an error because the value of the text field will be null. Instead, use this code to create a target page and place the file in the same directory as the code for the previous example. The name of the file must correspond to the value specified for the PostBackUrl property in the previous example. When you run the code for the previous example, this page will execute automatically when the cross-page post occurs.

Important

This example has a text box that accepts user input, which is a potential security threat. By default, ASP.NET Web pages validate that user input does not include script or HTML elements. For more information, see Script Exploits Overview.

<%@ page language="C#" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<script runat="server">
  
  void Page_Load (object sender, System.EventArgs e)
  {
    string text;
    
    // Get the value of TextBox1 from the page that 
    // posted to this page.
    text = ((TextBox)PreviousPage.FindControl("TextBox1")).Text;
    
    // Check for an empty string.
    if (text != "")
      PostedLabel.Text = "The string posted from the previous page is "
                         + text + ".";
    else
      PostedLabel.Text = "An empty string was posted from the previous page.";
  }

</script>

<html xmlns="http://www.w3.org/1999/xhtml" >
<head id="head1" runat="server">
  <title>LinkButton.PostBackUrl Target Page Example</title>
</head>
<body>
  <form id="form1" runat="server">
    
    <h3>LinkButton.PostBackUrl Target Page Example</h3>
      
    <br />
    
    <asp:label id="PostedLabel"
       runat="Server">
    </asp:label>

    </form>
</body>
</html>
<%@ page language="VB" %>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<script runat="server">
  Sub Page_Load(ByVal sender As System.Object, ByVal e As System.EventArgs)
    
    Dim text As String
    
    ' Get the value of TextBox1 from the page that posted
    ' to this page.
    text = CType((PreviousPage.FindControl("TextBox1")), TextBox).Text
       
    ' Check for an empty string.
    If Not (text = "") Then
      PostedLabel.Text = "The string posted from the previous page is " _
                         & text & "."
    Else
      PostedLabel.Text = "An empty string was posted from the previous page."
    End If
    
  End Sub

</script>

<html xmlns="http://www.w3.org/1999/xhtml" >
<head id="head1" runat="server">
  <title>LinkButton.PostBackUrl Target Page Example</title>
</head>
<body>
  <form id="form1" runat="server">
    
    <h3>LinkButton.PostBackUrl Target Page Example</h3>
       
    <br />
    
    <asp:label id="PostedLabel"
       runat="Server">
    </asp:label>

    </form>
</body>
</html>

Remarks

The PostBackUrl property allows you to perform a cross-page post using the LinkButton control. Set the PostBackUrl property to the URL of the Web page to post to when the LinkButton control is clicked. For example, specifying Page2.aspx causes the page that contains the LinkButton control to post to Page2.aspx. If you do not specify a value for the PostBackUrl property, the page posts back to itself.

Important

When performing a cross-page postback with controls with server-side validation, you should check that the page's IsValid property is true before processing the postback. In the case of a cross-page postback, the page to check is the PreviousPage. The following Visual Basic code shows how this is done:

Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load
        If Page.PreviousPage.IsValid Then
            ' Handle the post back
        Else
            Response.Write("Invalid")
        End If
End Sub

For more information on cross-page posting techniques, see Cross-Page Posting in ASP.NET Web Forms.

This property cannot be set by themes or style sheet themes. For more information, see ThemeableAttribute and ASP.NET Themes and Skins.

Applies to

See also