我正在使用上传文件<asp:FileUpload>
and <asp:button>
控件,但我想在没有回发的情况下完成它。单击按钮后,我执行以下代码。
protected void btnUpload_Click(object sender, EventArgs e)
{
string strFileName = Path.GetFileName(FileUpload1.FileName); //fileupload1 is the <asp:fileupload ID
FileUpload1.SaveAs(Server.MapPath("~/UploadFile/" + strFileName + ""));
imgUpload.ImageUrl = "../UploadFile/" + strFileName + ""; //imgupload is the <img ID on which I am showing the image after upload
imgUpload.Visible = true;
}
上传文件后,我将显示项目解决方案中指定文件夹中保存的图像,但单击上传按钮时会加载整个页面,并且我不希望单击上传按钮时进行回发。
just add the script
<script language="javascript">
function Change(obj) {
__doPostBack("<%= btnUpload.ClientID %>", "");
}
</script>
Call the script in your button click event like this
<pre>
onchange="Change(this);"
</pre>
Your image control added in the update panel with contentTemplate
<asp:FileUpload ID="imgFileUploader" runat="server" />
<asp:Button ID="btnUpload" runat="server"
Text="Upload" onclick="btnUpload_Click" onchange="Change(this);" />
<br />
<br />
<asp:ScriptManager ID="ScriptManager1" runat="server">
</asp:ScriptManager>
<asp:UpdatePanel ID="UpdatePanel1" runat="server">
<ContentTemplate>
<img id="imgOrginal" runat="server" style="height: 200px; width: 200px;" />
</ContentTemplate>
</asp:UpdatePanel>
更多细节 http://www.codeproject.com/Tips/558327/Show-uploaded-image-without-postback
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)