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如何将变量“appName”从SerialPopUp函数传递到我创建的弹出窗口?预先感谢您提供的任何帮助。从父窗口Javascript函数传递变量到弹出窗口Div
function SerialPopUp(appInfo)
{
appInfo=appInfo.split("\n");
var appName = appInfo[1].trim();
var appType = appInfo[5].trim();
var url = appInfo[7].trim();
url = appInfo[7].substring(9,appInfo[7].indexOf(">")-17);
var serialNum = "";
if(appType == 'External Paid'){
x = window.open(url, "Get Serial Number","resizable=1,status=1, width=700, height=250");
x.document.getElementById("appName").innerHTML= appName;
}
}