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Image Rotator In Ajax

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18 Jun 2011 1  
Image Rotator In Ajax Step1:  In this Example, there arean Image (img1) and a TextBox(TextBox1).<br

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Image Rotator In Ajax

 

Step1:  In this Example, there arean Image (img1) and a TextBox(TextBox1).

<img id="img1" src="Image/1.jpg" /><br />

 

<asp:TextBox ID="TextBox1"runat="server">1</asp:TextBox>

 

 

Step 2: We call afunction ChangeImage() on the onloadevent of the <body> tag. So when the page loads the function callautomatiacally.

varxmlHttp

        vararr;

        functionChangeImage() {

            xmlHttp = GetXmlHttpObject()

            varurl = "Default.aspx"

            url = url + "?iname=" + document.getElementById('TextBox1').value

            xmlHttp.onreadystatechange =stateChanged

            xmlHttp.open("GET", url, true)

            xmlHttp.send(null)

            returnfalse;

        }

 

We set the Text property of the TextBox1

<asp:TextBox ID="TextBox1"runat="server">1</asp:TextBox>

 

                       

Now we write the functionChangeImage()

varxmlHttp

vararr;

 

functionChangeImage() {

xmlHttp =GetXmlHttpObject()

var url= "Default.aspx"

url = url + "?iname=" + document.getElementById('TextBox1').value

xmlHttp.onreadystatechange= stateChanged

xmlHttp.open("GET", url, true)

xmlHttp.send(null)

return false;

        }

 

 

Now we can write the  GetXmlHttpObject()to check the browser

 

function GetXmlHttpObject() {

var objXMLHttp = null

if (window.XMLHttpRequest) {

objXMLHttp = new XMLHttpRequest()

}

else if(window.ActiveXObject) {

objXMLHttp = new ActiveXObject("Microsoft.XMLHTTP")

}

return objXMLHttp

}

 

 Step 3:In the Page_Load Event

 

if(Request.QueryString["iname"] != null)

        {

            stringi = Request.QueryString["iname"].ToString();

            stringstr = "1";

            if(i == "2")

            {

 

                str = "2";

            }

 

            Response.Clear();

 

            Response.Write(str);

            Response.End();

        }

 

 

Complete Program

Default.aspx

 

<head runat="server">

<title></title>

<script language="JavaScript" type="text/javascript" >

varxmlHttp

vararr;

functionChangeImage() {

xmlHttp =GetXmlHttpObject()

var url= "Default.aspx"

url = url + "?iname=" + document.getElementById('TextBox1').value

xmlHttp.onreadystatechange= stateChanged

xmlHttp.open("GET", url, true)

xmlHttp.send(null)

return false;

}

functionstateChanged() {

 

if(xmlHttp.readyState == 4 || xmlHttp.readyState == "complete"){

 

varstr;

 

str =xmlHttp.responseText;

 

if(str == "1") {

document.getElementById("img1").src = "Image/211825_100000984069347_8367124_q.jpg";

document.getElementById("TextBox1").value = "2";

setTimeout("ChangeImage()", 1000);

}

if(str == "2") {

document.getElementById("img1").src = "Image/17Dec20100846PMfinal33.jpg";

setTimeout("ChangeImage()", 1000);

document.getElementById("TextBox1").value = "1";

}

 

}

}

functionGetXmlHttpObject() {

 

varobjXMLHttp = null

if(window.XMLHttpRequest) {

objXMLHttp = new XMLHttpRequest()

}

else if (window.ActiveXObject) {

objXMLHttp = new ActiveXObject("Microsoft.XMLHTTP")

}

returnobjXMLHttp

}

 

</script>

 

 

</head>

<body onload="ChangeImage()">

<form id="form1" runat="server">

<div>

 

<img alt=""  id="img1"

style="height: 156px; width: 233px" src="Image/1.jpg"/><br />

<br />

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<asp:TextBox ID="TextBox1" runat="server">1</asp:TextBox>

<br />

 

</div>

</form>

</body>

</html>

 

Default.aspx.cs

 

protected void Page_Load(objectsender, EventArgs e)

{

if (Request.QueryString["iname"] != null)

{

string i = Request.QueryString["iname"].ToString();

string str = "1";

if (i == "2")

{

 

str= "2";

}

 

Response.Clear();

 

Response.Write(str);

Response.End();

}

}

 

 

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