我正在使用
JQuery的$.getJSON函数来下载一些图像的URL,我试图在div中输出它们.我试图让输出看起来像这样:
<a href="the image url (shot.short_url)"><img src="the direct image url (shot.image_teaser_url)" /></a>
但是,它输出的是:
<div id="body-wrapper">
<a href="http://drbl.in/300896">[object Object]</a>
<a href="http://drbl.in/298080">[object Object]</a>
<a href="http://drbl.in/290395">[object Object]</a>
<a href="http://drbl.in/290324">[object Object]</a>
<a href="http://drbl.in/268595">[object Object]</a>
<a href="http://drbl.in/265197">[object Object]</a>
<a href="http://drbl.in/256368">[object Object]</a>
<a href="http://drbl.in/252519">[object Object]</a>
<a href="http://drbl.in/242235">[object Object]</a>
<a href="http://drbl.in/241676">[object Object]</a>
</div>
在输出图像的情况下,请告诉我出错的地方?
这是我的代码:
function work() {
$('#body-wrapper').empty();
$.getJSON("http://dribbble.com/jakekrehel/shots.json?callback=?", function(data){
$.each(data.shots, function(i,shot){
var image = $('<img/>').attr('src', shot.image_teaser_url);
var title = '<a href=\"' + shot.short_url + '\">';
var string = title;
string = string + image;
string = string + '</a>';
$('#body-wrapper').append(string);
});
});
}
最佳答案 在你的.each()回调中试试这个
// create image
var image = $('<img>').attr('src', shot.image_teaser_url);
// create anchor and append image
var anchor = $('<a>').attr('href', shot.short_url).append(image);
// append anchor to container
$('#body-wrapper').append(anchor);