我的代码看起来像这样。问题是,PHP方面做的工作并返回正确的价值。但是ajax不执行success: function
内的事情。我错过了什么?
AnswerDiv.on("click", ".NotSelectedAnswer", function() {
var NotSelectedAnswerBtn = $(".NotSelectedAnswer"),
SelectedAnswerBtn = $(".SelectedAnswer"),
AnswerDiv = $("div.Answer"),
querystring="fromID="+SelectedAnswerBtn.data("id")+"&toID="+$(this).data("id")+"&op=SelectAsAnswer";
$.ajax({
url: 'processor.php',
type: "POST",
dataType: "json",
data: querystring,
success: function(data) {
if(data.status)
{
SelectedAnswerBtn.removeClass("SelectedAnswer").addClass("NotSelectedAnswer").button("enable");
$(this).removeClass(" NotSelectedAnswer").addClass("SelectedAnswer").button("disable");
$("div.Answer[data-id=" + SelectedAnswerBtn.data("id") + "]").toggleClass("SelectedDiv");
$("div.Answer[data-id=" + $(this).data("id") + "]").toggleClass("SelectedDiv");
}
}
});
return false;
});
答案 0 :(得分:1)
尝试在ajax调用之前缓存$(this)
AnswerDiv.on("click", ".NotSelectedAnswer", function() {
var NotSelectedAnswerBtn = $(".NotSelectedAnswer"),
SelectedAnswerBtn = $(".SelectedAnswer"),
AnswerDiv = $("div.Answer"),
thisElem=$(this),
querystring="fromID="+SelectedAnswerBtn.data("id")+"&toID="+$(this).data("id")+"&op=SelectAsAnswer";
$.ajax({
url: 'processor.php',
type: "POST",
dataType: "json",
data: querystring,
success: function(data) {
if(data.status)
{
SelectedAnswerBtn.removeClass("SelectedAnswer").addClass("NotSelectedAnswer").button("enable");
thisElem.removeClass(" NotSelectedAnswer").addClass("SelectedAnswer").button("disable");
$("div.Answer[data-id=" + SelectedAnswerBtn.data("id") + "]").toggleClass("SelectedDiv");
$("div.Answer[data-id=" +thisElem.data("id")+ "]").toggleClass("SelectedDiv");
return false;
}
}
});
});