Introduction
In case of ImageRotator
, we can use FileUpload
control to upload my images.
Here I use it in a simple way.
Step 1
We use a image control (Image1
), FileUpload
control, button (Button1
), HiddenField
(for storing the file name) and a Save button to call the JavaScript function.
Step 2
On the click of Button1
, we upload a file with the help of FileUpload1
and store the file name in HiddenField1
and show the image in the image control.
string filename = FileUpload1.PostedFile.FileName.ToString();
FileUpload1.SaveAs(Server.MapPath("~/") + filename);
Image1.ImageUrl = filename.ToString();
HiddenField1.Value = filename.ToString();
Step 3
On the click of Button2
, we upload a file with the help of FileUpload2
and store the file name in HiddenField2
and show the image in the image control:
string filename = FileUpload2.PostedFile.FileName.ToString();
FileUpload2.SaveAs(Server.MapPath("~/") + filename);
Image1.ImageUrl = filename.ToString();
HiddenField2.Value = filename.ToString();
Step 4
On the click of Button3
, we upload a file with the help of FileUpload3
and store the file name in HiddenField3
and show the image in the image control.
string filename = FileUpload3.PostedFile.FileName.ToString();
FileUpload3.SaveAs(Server.MapPath("~/") + filename);
Image1.ImageUrl = filename.ToString();
HiddenField3.Value = filename.ToString();
Suppose we add only three Images for ImageRotator
.
Step 5
In the click of the Save Button (HTML control) we call the JavaScript function (show()
) in the OnClick
event.
<script language="javascript" type="text/javascript" >
function show()
{
var a=document.getElementById('HiddenField1').value;
document.getElementById('Image1').src=a;
setTimeout("show2()",1000);
}
function show2()
{
var b=document.getElementById('HiddenField2').value;
document.getElementById('Image1').src=b;
setTimeout("show3()",1000);
}
function show3()
{
var b=document.getElementById('HiddenField3').value;
document.getElementById('Image1').src=b;
setTimeout("show()",1000);
}
</script>