标题:怎么把链接返回的json放入gridview的指定位置
作者:星火零光
日期:2015-02-28 17:32
内容:
function load(successCallback) {
var xhr = new XMLHttpRequest();
xhr.open("GET","http://api.acfun.tv/apiserver/recommend/list");
xhr.onreadystatechange = function()
{
if ( xhr.readyState == xhr.DONE)
{
if ( xhr.status == 200)
{
var res = JSON.parse(xhr.responseText)
successCallback(res)
}
}
}
xhr.send();
}
我想把上面l链接返回的title和cover作为标题和封面在网格视图里显示，可是这个 successCallback(res)函数该怎么写呢？
链接返回
{"success":true,"msg":"ok","status":200,"data":{"list":[{"slideId":17954,"title":"第57届格莱美颁奖典礼","subtitle":"高清中字","description":"","cover":"http://static.acfun.mm111.net/dotnet/artemis/u/cms/www/201502/121746280ajo.jpg","releaseDate":1423794178000,"specialId":"1740084","config":6,"url":null,"type":0,"weekday":null,"time":null},{"slideId":17936,"title":"2015羊年美国春晚","subtitle":"大案力作","description":"","cover":"http://static.acfun.mm111.net/dotnet/artemis/u/cms/www/201502/11180422peru.png","rel ..
#1 [星火零光 03-01 16:40]
额，解决了