[英]I can't convert my ajax $.get to $.ajax
I try to convert an ajax function that I do not understand. 我尝试转换一个我不理解的ajax函数。 Here the original function :
这里是原始功能:
$.get(page, function (data) {
$('.former-students__list').append(data.student
$('#load-more').data('next-page', data.next_page);
})
I'd like to convert it to make it more readable (with $ .ajax), So I could clearly see the events success, and so on. 我想将其转换为更具可读性(使用$ .ajax),因此我可以清楚地看到事件成功,依此类推。
Here is what I tried but it doesn't work 这是我尝试过的方法,但是不起作用
$.ajax({
url: page,
data: function (data) {
$('.former-students__list').append(data.students);
$('#load-more').data('next-page', data.next_page);
}
'success': 'Cool, it's work')
},
Thank you for your help and your explanations 感谢您的帮助和您的解释
The issue is because the data
parameter should be an object, a string or an array which you use to send data to the server. 问题是因为
data
参数应该是用于将数据发送到服务器的对象,字符串或数组。 However, given your $.get
sample, you don't actually need to use it in $.ajax
at all. 但是,给定
$.get
示例,实际上根本不需要在$.ajax
中使用它。
Instead, the success
property should be a function which receives the data back from the request and then acts on it. 相反,
success
属性应该是一个函数,该函数从请求中接收数据,然后对其执行操作。 Try this: 尝试这个:
$.ajax({
url: page,
success: function(data) {
$('.former-students__list').append(data.students);
$('#load-more').data('next-page', data.next_page);
})
});
For more information, please read the jQuery $.ajax documentation . 有关更多信息,请阅读jQuery $ .ajax文档 。
Try this: 尝试这个:
$.ajax({
url: page,
success: function (data) {
$('.former-students__list').append(data.students);
$('#load-more').data('next-page', data.next_page);
}
})
I think the data
property is for POST request payloads. 我认为
data
属性用于POST请求有效负载。
you should fire the success callback event so try this : 您应该触发成功回调事件,因此请尝试以下操作:
$.ajax({
url: page,
success: function(data) {
$('.former-students__list').append(data.students);
$('#load`enter code here`-more').data('next-page', data.next_page);
}
});
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