Question

请问如何把ajax获取到的json数据以链接传参带值的方式带到另一个页面获取并解析？

$.ajax({
type: "post",
url: "../../expan/expan/getExpanInfo",
data: reqObj,
dataType: "text",
success: function(data) {
var rtnObj = eval('(' + data + ')');
var rtnFlag = rtnObj.rtnFlag;
load.close();
if (rtnFlag == '1' ) {
location.href = "service.html?orgNo=rtnObj.orgNo";
//window.open('service.html?rtnObj='+jsonString);
});
}else{
window.location.href = "../../annualassets/page/404.html?t=" + new Date().getTime();
}
}
});

这样对不对： location.href = "service.html?orgNo=rtnObj.orgNo"
在service.html页面如何获取传来的json呢

Answer 1

printPreviewDialog1.PrintPreviewControl.StartPage = 0;
class Composite : Component
{
private List children = new List();
public Composite(string name) : base(name) { }
public override void Add(Component c)
{
children.Add(c);
}
public override void Remove(Component c)
{
children.Remove(c);
}

追问

你写的没看懂啊，不是前台接收的js代码吧