我遇到了一个问题,即我对 $.ajax 的调用成功完成并返回了 Firebug 报告的 200OK 响应内容,但成功、完成和错误回调不执行。这仅发生在 Firefox 中,在 Chrome 中它工作正常(我正在运行 Firefox22)。
$.ajax(site_url+url+'/fetch_salt',{type:'POST',data:data,success:check_salt});
var group = '';
function check_salt(d)
{
console.log(d);
Firebug 报告的请求的实际响应是:
choose_login:{"admin":"Admin Zone"}
和响应类型:
Content-Type text/html
我尝试过强制设置像 dataType 和 contentType 这样的设置,以防 jquery 假设 json 或其他东西,并且我尝试了匿名函数来处理错误、成功和完整回调,但没有任何效果。
我发布了完整的函数代码,以防万一出现某种语法错误怪癖:
function prep_login_form(elem,url,challenge)
{
function show_error(msg)
{
$(elem).find('.ecms-error-for-password .ecms-error-text').html(msg).closest('.ecms-error-container').removeClass('ecms-error-hidden');
}
function submit()
{
var data = {email:$(elem).find('input[name="email"]').val()};
data[csfr_token_name] = csfr_hash;
$.ajax({type:'POST',url:site_url+url+'/attempt_login',data:data,success:check_salt});
var group = '';
function check_salt(d)
{
console.log(d);
if (d=='no_email')
{
show_error('Invalid Email address');
}
else if (d=='account_disabled')
{
show_error('This account has been disabled, please contact your administrator');
}
else if (d.substr(0,12)=='choose_login')
{
var cl;
eval('cl = '+d.substr(13));
var cou = 0;
for (p in cl)
{
cou++;
}
if (cou==1)
{
group = p;
var mydata = $.extend(data,{group:p});
$.ajax(site_url+url+'/fetch_salt',{type:'POST',data:mydata,success:check_salt})
}
else
{
var str = '<div class="login-selection-popup"><p>We have detected that your email address is linked to more than one account.<br />Please select which zone you would like to login to.</p><ul class="choose-login-popup">';
for (p in cl)
{
str+='<li><a rel="'+p+'">'+cl[p]+'</a></li>';
}
str+='</ul></div>';
open_modal({heading:'Choose Account',content:str,buttons:function(close_modal)
{
$(this).find('.choose-login-popup').on('click','a',function()
{
group = $(this).attr('rel');
var mydata = $.extend(data,{group:$(this).attr('rel')});
$.ajax(site_url+url+'/fetch_salt',{type:'POST',data:mydata,success:check_salt})
close_modal();
});
}});
}
}
else
{
var salt = d;
var pw = $(elem).find('input[name="password"]').val();
data.password = hex_md5(challenge+hex_md5(salt+pw));
data.group = group;
$.ajax(site_url+url+'/attempt_login',{type:'POST',data:data,success:function(d)
{
if (d=='no_email')
{
show_error('Invalid username or password');//Invalid Email address
}
else if (d=='account_disabled')
{
show_error('This account has been disabled, please contact your administrator');
}
else if (d=='invalid_login')
{
show_error('Invalid username or password');//Email or Password did not match
}
else
{
window.location.href = d;
}
}});
}
}
}
$(elem).on('keyup','input',function(e)
{
if (e.keyCode=='13')
{
submit();
}
});
$(elem).find('.login-submit').on('click',function()
{
submit();
});
}
最佳答案
对于所有麻烦的家伙,我深表歉意,我最近在我的电脑上安装了附加软件,并努力摆脱它。我认为它损坏/劫持了我的 Firefox。重新安装 Firefox 后问题就消失了,回调现在执行。
关于javascript - $.ajax 成功回调在 Firefox 中未触发,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18725437/