JQuery如何处理url返回的json数据?

2025-01-28 10:53:59
推荐回答(1个)
回答1:

functionhandleJson()
{
varj={"name":"Michael","address":{"city":"Beijing","street":"ChaoyangRoad ","postcode":100025} };
alert(j.name);
alert(j.address.city);
}

自己改改`