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@ayuLiao
Last active January 19, 2019 11:08
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使用 $.ajax 做ajax请求
var onLoginSuccess = function (data) {
location.reload();
}
$("#changemaintain").click(function(){
var p = $('#changemaintain').val();
if(p == '0'){
p = '1';
}else{
p = '0';
}
params = {'maintain': p}
var ajaxObj = {
type: 'POST',
url: 'http://127.0.0.1:5002/lucky', //URL上一定要带HTTP,不然报错
data: params,
success: onLoginSuccess, //ajax成功,前端的回调函数
dataType: "json"
};
$.ajax(ajaxObj);
})
//通过方法封装AJAX
$.ajaxPost = function(url, data, callback) {
$.post(url, data, callback, "json");
};
$.ajaxPost('/post/ajaxUrl',{},function(data){});
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