你好,我做了一个代码,要求在ajax调用后添加一个类。我确定代码正确,但是仍然没有添加该类。它真的很奇怪,因为代码中的其他所有内容都可以正常工作,而且我确定addClass的代码正确,而且我已经检查了控制台是否有任何错误,但是没有错误。这是我的代码
$(document).on('click', '.miclickks', function(event) {
event.preventDefault();
var for_uid = $(this).parents("li").attr('data');
var for_name = $(this).parents("li").attr('unme');
var for_pic = $(this).parents("li").attr('upic');
var owner_uid = $('.row-fluid').attr('uid');
var owner_name = $('.row-fluid').attr('usnm');
var owner_pic = $('.row-fluid').attr('usp');
var type = "kiss";
var dataString = "type=" + type + "&for_uid=" + for_uid + "&for_name=" + for_name + "&for_pic=" + for_pic + "&owner_uid=" + owner_uid + "&owner_pic=" + owner_pic + "&owner_name=" + owner_name;
$.ajax({
type: "POST",
url: "include/ajax.php",
data: dataString,
success: function (html) {
if(html=="300")
{
$('#myModal .modal-body p').html("Error Please Try Again.");
$('#myModal').modal('show');
}
else
{
$(this).addClass('active');
}
}
});
});
最佳答案
两种方式:
第一:
var that = this;
$.ajax({
type: "POST",
url: "include/ajax.php",
data: dataString,
success: function (html) {
if (html == "300") {
$('#myModal .modal-body p').html("Error Please Try Again.");
$('#myModal').modal('show');
} else {
$(that).addClass('active');
}
}
});
第二(使用
context
选项):$.ajax({
type: "POST",
url: "include/ajax.php",
data: dataString,
context: this,
success: function (html) {
if (html == "300") {
$('#myModal .modal-body p').html("Error Please Try Again.");
$('#myModal').modal('show');
} else {
$(this).addClass('active');
}
}
});