如何在gridview中打开一个指向链接的弹出窗口

Question

我正在尝试在 java 脚本中为我的网格视图构建一个弹出窗口函数，但我无法打开该窗口。

var oldgridcolor;
function SetMouseOver(element) {
    oldgridcolor = element.style.backgroundColor;
    element.style.backgroundColor = '#ffeb95';
    element.style.cursor = 'pointer';
    element.style.textDecoration = 'underline';
}
function SetMouseOut(element) {
    element.style.backgroundColor = oldgridcolor;
    element.style.textDecoration = 'none';

}
function SetMouseDown(element) {
    var r = confirm('Are you sure?');
    var url = window.location.pathname;
    var pathArray = url.split('/');
    var host = pathArray[1];
    var newHost = '/About.aspx';

    if (r == true) {
        window.location = host + newHost;
    }
    else {
        alert('it didnt work');
    }
    return false;
}

后面的代码

 protected void gvrecords_RowDataBound(object sender, GridViewRowEventArgs e)
  {
    if (e.Row.RowType == DataControlRowType.DataRow)
    {
        e.Row.Attributes["onmouseover"] = "javascript:SetMouseOver(this)";
        e.Row.Attributes["onmouseout"] = "javascript:SetMouseOut(this)";
        e.Row.Attributes["onmousedown"] = "javascript:SetMouseDown(this)";
    }

任何帮助将不胜感激。

Answer 1

请尝试一下，它可能对你有帮助。

function SetMouseDown(element) {
    var r = confirm('Are you sure?');
    var url = window.location.pathname;
    var pathArray = url.split('/');
    var host = pathArray[1];
    var newHost = '/About.aspx';

    if (r == true) {
        //window.location = host + newHost;

        window.open(host + newHost,'name','width=200,height=200');
    }
    else {
        alert('it didnt work');
    }
    return false;
}