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django通过ajax发起请求返回JSON格式数据的方法

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Release： 2016-06-10 15:10:48
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本文实例讲述了django通过ajax发起请求返回JSON格式数据的方法。分享给大家供大家参考。具体实现方法如下：

这是后台处理的：

def checkemail(request):
  user = None
  if request.POST.has_key('email'):
    useremail = request.POST['email']
    result = {}
    user = User.objects.filter(useremail__iexact = useremail)
  if user:
    result = "1"
    result = simplejson.dumps(result)
  else:
    result = "0"
    result = simplejson.dumps(result)
  return HttpResponse(result, mimetype='application/javascript')
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这是AJAX部分：

if(valid_email($('#reg-for-email').val())){
  var email = $('#reg-for-email').val();
  //这里把用户输入的EMAIL地址提交到后台数据库中去验证是否已存在。
  $.ajax({
    type:"POST" ,
    url:"/reg/checkemail",
    data:"email=" + email ,
    cache: false,
    success: function(result){
  if (result==1)
   {
    $("#reg-for-email-msg").removeClass("g-hide");
    $('#reg-for-email-msg').removeClass("msg-isok").addClass("msg-error").html("该邮箱已存在！");
    eok = true;
   }
   else
   {
    $("#reg-for-email-msg").addClass("g-hide");
    eok = false;
   }
    }
   })
}
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URL的配置是:

复制代码 代码如下:
url(r'^reg/checkemail/', 'reg.views.checkemail', name='ce'),

希望本文所述对大家的Python程序设计有所帮助。

Related labels：
ajax django json
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