如何使用Ajax并在另一个URL上发布数据请求? 解决ajax问题
$.ajax({
type: 'POST',
url: 'https://kyleschaeffer.com/feed/',
data: { postVar1: 'theValue1', postVar2: 'theValue2' },
beforeSend:function(){
// this is where we append a loading image
$('#ajax-panel').html('<div class="loading"><img src="/images/loading.gif" alt="Loading..." /></div>');
},
success:function(data){});
});
答案 0 :(得分:0)
从某种意义上说,此jQuery功能非常完美,它可以处理您将需要进行的所有AJAX请求中的99%,它包含成功和失败功能,以确保用户获得所需的正确反馈,并且在处理请求以进行引导时,您会看到旋转的加载映像。
$.ajax({
type: 'POST',
url: 'https://kyleschaeffer.com/feed/',
data: { postVar1: 'theValue1', postVar2: 'theValue2' },
beforeSend:function(){
// this is where we append a loading image
$('#ajax-panel').html('<div class="loading"><img src="/images/loading.gif" alt="Loading..." /></div>');
},
success:function(data){
// successful request; do something with the data
$('#ajax-panel').empty();
$(data).find('item').each(function(i){
$('#ajax-panel').append('<h4>' + $(this).find('title').text() + '</h4><p>' + $(this).find('link').text() + '</p>');
});
},
error:function(){
// failed request; give feedback to user
$('#ajax-panel').html('<p class="error"><strong>Oops!</strong> Try that again in a few moments.</p>');
}
});