嗨,我正在尝试在发生错误时捕获错误,但我的Json遇到了问题,当音轨加载完美时,我收到了一个成功的答案,但在崩溃中我什么也没有
$.getJSON(apiUrl, function(data, error) {
index += 1;
if(data.tracks){
playerObj.tracks = playerObj.tracks.concat(data.tracks);
}else if(data.duration){
data.permalink_url = link.url;
playerObj.tracks.push(data);
}else if(data.creator){
links.push({url:data.uri + '/tracks'});
}else if(data.username){
if(/favorites/.test(link.url)){
links.push({url:data.uri + '/favorites'});
}else{
links.push({url:data.uri + '/tracks'});
}
}else if($.isArray(data)){
playerObj.tracks = playerObj.tracks.concat(data);
}
if(links[index] && (index % 18) != 0){
var mod = index % 18;
loadUrl(links[index]);
}else{
playerObj.node.trigger({type:'onTrackDataLoaded', playerObj: playerObj, url: apiUrl});
if (links[index]) {
loadMoreTracksData($player, links, key, index);
}
}
}).success(function() { console.log("second success"); }).error(function() { console.log("error"); });有谁知道如何正确地监听这个Api中的错误。
发布于 2013-06-24 18:39:23
假设你使用的是jQuery 1.5以上的版本,下面的结构更适合你。
仅当从服务器返回适用的HTTP代码(例如,500 / 404 / 400)时才调用失败方法,而不是简单地返回空的数据集,而是使用成功的HTTP 200响应。
var index = 0;
var processResponse = function(data) {
index += 1;
if(data.tracks){
playerObj.tracks = playerObj.tracks.concat(data.tracks);
}else if(data.duration){
data.permalink_url = link.url;
playerObj.tracks.push(data);
}else if(data.creator){
links.push({url:data.uri + '/tracks'});
}else if(data.username){
if(/favorites/.test(link.url)){
links.push({url:data.uri + '/favorites'});
}else{
links.push({url:data.uri + '/tracks'});
}
}else if($.isArray(data)){
playerObj.tracks = playerObj.tracks.concat(data);
}
if(links[index] && (index % 18) != 0){
var mod = index % 18;
loadUrl(links[index]);
}else{
playerObj.node.trigger({type:'onTrackDataLoaded', playerObj: playerObj, url: apiUrl});
if (links[index]) {
loadMoreTracksData($player, links, key, index);
}
}
var processError = function(error){
console.log("Error:" + error);
}
$.getJSON(apiUrl)
.done(function(data){ processResponse(data); })
.fail(function(error){ processError(error); })
.complete(function(xhr, status) {console.log(status);};https://stackoverflow.com/questions/17272941
复制相似问题