访问ajax之外的变量
我有以下代码：
var src, flickrImages = [];
$.ajax({
type: "GET",
url: "http://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=bf771e95f2c259056de5c6364c0dbb62&text=" + xmlTitle.replace(' ', '%20') + "&safe_search=1&per_page=5&format=json",
dataType: "json",
statusCode: {
404: function() {
alert('page not found');
}
},
success: function(data) {
$.each(data.photos.photo, function(i,item){
src = "http://farm"+ item.farm +".static.flickr.com/" + item.server + "/" + item.id + "_" + item.secret + "_s.jpg";
flickrImages[i] = '<img src="' + src + '">';
});
}
});
// undefined returned here for flickrImages
map.setZoom(13);
map.setCenter(new google.maps.LatLng(xmlLat,xmlLng));
infowindow.setContent('<strong>' + xmlTitle + '</strong><br>' + xmlExcerpt + '<br><br>' + flickrImages.join(''));
infowindow.open(map,this);
没有找到相关结果
已邀请:
4 个回复
芦歉竭皑
的调用是异步的，这意味着在你的请求完成之前它不会被阻塞。 这就是为什么有一个函数在底层HTTP请求完成时被“回调”。所以，你需要从函数处理你的变量，或者从你的函数处理你的变量，将传递给你处理的其他函数。屑凉赦
嫩昧竞莫
var myData $.ajax({ type: "GET", url: "http://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=bf771e95f2c259056de5c6364c0dbb62&text=" + xmlTitle.replace(' ', '%20') + "&safe_search=1&per_page=5&format=json", dataType: "json", statusCode: { 404: function() { alert('page not found'); } }, success: function(data) { myData = data myFunction() } });枫湃揩乾纲
你的ajax请求。这是你添加这个代码后的代码：$.ajax({ type: "GET", url: "http://api.flickr.com/services/rest/?method=flickr.photos.search&api_key=bf771e95f2c259056de5c6364c0dbb62&text=" + xmlTitle.replace(' ', '%20') + "&safe_search=1&per_page=5&format=json", dataType: "json", async: false, statusCode: { 404: function() { alert('page not found'); } }, success: function(data) { $.each(data.photos.photo, function(i,item){ src = "http://farm"+ item.farm +".static.flickr.com/" + item.server + "/" + item.id + "_" + item.secret + "_s.jpg"; flickrImages[i] = '<img src="' + src + '">'; }); } });