在使用点击主页面时,使用以下代码显示一个弹出窗口:
queryString = "web.aspx?ds=" + Base64ForUrlEncode(ds.ToString())
newWin = "var Mleft = (screen.width/2)-(1200/2);var Mtop = (screen.height/2)-(700/2);window.open('" + queryString + ','_blank','height=700,width=1200,status=yes,toolbar=no,scrollbars=yes,menubar=no,location=no,top=\'+Mtop+\', left=\'+Mleft+\';'); window.opener.location.reload();";
ClientScript.RegisterStartupScript(this.GetType(), "pop", newWin, true);
如何在打开的PopUp窗口上重新加载父页面?
我试过
window.opener.location.reload()
没有成功:
newWin = "var Mleft = (screen.width/2)-(1200/2);var Mtop = (screen.height/2)-(700/2);window.open('" + queryString + ','_blank','height=700,width=1200,status=yes,toolbar=no,scrollbars=yes,menubar=no,location=no,top=\'+Mtop+\', left=\'+Mleft+\';'); window.opener.location.reload();";
提前感谢您的帮助。
答案 0 :(得分:1)
您是否尝试在打开弹出窗口后重新绑定父页面?