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我想在另一个jquery函数里面使用jquery设置var,但是它没有设置var。我尝试了几种方法。这里是最后的从jquery函数里面设置var jquery函数不运行
jQuery(function($){
var pipeURL = "xxxxxxxxx.json";
var feedSuccess = function(data, status){
//var html = "<ul>";
var html = "";
var desc = "";
var description = "";
var start = "";
$.each(data.value.items, function(i,item){
//desc = item.description.replace("<br />"," ");
desc = item.description;
desc = desc.substring(0,600);
desc += "...";
descr = desc.replace(/(?:(?!UTC).)*/i,"");
$.getJSON(
"https://graph.facebook.com/"+item.eid+"/?fields=start_time&access_token=xxxxxxxxxxxxxxxxxx",
function(data){
start = data.start_time;
});
html += "<h3 class='pipes'><a class='pipes' href="+item.link+" target='_blank'>" + item.title +"</a></h3>"+"<p class='pipes'>"+start+"</p>"+
"<br/><a class='pipes' href="+item.link+" target='_blank'><img class='pipes' src='"+item['media:thumbnail'].url+"' ></a><br/>" ;
});
//html += "</ul>";
//Add the feed to the page
$("#insertFeed").empty().append(html);
};
$.ajax({
dataType: "json",
url: pipeURL,
success: feedSuccess,
timeout: 6000
});
});
定义var'html'的行与功能数据“var start”中的信息无法正常运行。它显示标题正常,图像正常,但“开始”显示空白。
,如果您正确格式的代码,你很可能会自己解决这个问题,或者至少是更容易让别人帮你。 – 2015-04-06 02:48:56
将'html'代码UP移动到定义了'start'的函数中,在它定义之后 – 2015-04-06 02:49:41
http://blog.slaks.net/2015-01-04/async-method-patterns/ – SLaks 2015-04-06 02:51:40