Question

我正在尝试为我的django应用程序创建通知，我必须在模板中的div中显示数据库中的更改。我希望ajax定期点击我的数据库，如果有任何更改，那么它们应该显示在该div中。我希望我的数据来自3个不同的模型，其中任何一个的任何更改都应该在我的模板的div中通知。

我该怎么做？

模板：

 <div class="notification">
        {% for like in likes_notify %}
        {% if like.item.reward != None %}
            <div class="like_newsfeed">
                <a href="/employee/{{ emp_code }}/social-info/single-news-feed/{{ like.item.reward.pk }}/show/"> &nbsp;{{ like.user.get_full_name }} liked your post : {{ like.item.reward }}&nbsp;&nbsp;&nbsp; </a>
            </div>
           {% else %}
            <div class="like_news_post">{{ like.item.comment.pk }}
                <a href="/employee/{{ emp_code }}/social-info/single-news-feed/{{ like.item.comment.pk }}/show/">&nbsp;{{ like.user.get_full_name }} liked your comment "{{ like.item.comment }}".&nbsp;&nbsp;&nbsp; </a>
            </div>
        {% endif %}
    {% endfor %}
<br><hr>
    {% for comment in comments_notify %}
        {% if comment.item.reward != None %}
            <div class="comment_newsfeed">
                <a href="/employee/{{ emp_code }}/social-info/single-news-feed/{{ comment.item.reward.pk }}/show/"> &nbsp;{{ comment.user.get_full_name }} commented on your post : {{ comment.item.reward }} with "{{ comment }}".&nbsp;&nbsp;&nbsp; </a>
            </div>
        {% else %}
            <div class="comment_news_post">{{ comment }}
                <a href="/employee/{{ emp_code }}/social-info/single-news-feed/{{ comment.item.reward.pk }}/show/"> &nbsp;{{ comment.user.get_full_name }} commented on your post "{{ comment }}".&nbsp;&nbsp; </a>
            </div>
        {% endif %}
    {% endfor %}
<br><hr>
    {% for reward in rewards_notify %}
        <div class="reward_newsfeed">
            <a href="/employee/{{ emp_code }}/social-info/single-news-feed/{{ reward.pk }}/show/">&nbsp;{{reward.role_history.organisation_role.user.get_full_name }} got a reward :  {{ reward }}&nbsp;&nbsp;&nbsp;</a>
        </div>
    {% endfor %}
    </div>

JS：

  function worker(){
    var url =  '/employee/'+get_emp_code()+'/home/dashboard/';
//    $(".notification").load(url + ' .notification',data, function () { setTimeout(worker, 5000);  });
    $.ajax({
        type:'get',
    url: url,
    success: function(data) {
//      $('.notification').load(url + " .notification", data);
//      $('.notification').load(url + " .notification").html(data);
      $('.notification').html(data);
    },
    complete: function() {
      // Schedule the next request when the current one's complete
      setTimeout(worker, 3000);
    }
  });
}
$(document).ready(function(){
setTimeout(worker, 5000);
    });

我试过写这个js但是这会加载我的整个页面两次而不是div。我实际上已将一个基本模板扩展到此模板中，因此它也会在我的div中加载整个基本模板的数据。

查看：

def get_context(request,*args,**kwargs):
    context = {}
    likes = Like.objects.exclude(user = request.user).order_by('-date_edited')[:2]
    comments = Comment.objects.filter(user = request.user).order_by('-last_updated')[:2]
    rewards = Reward.objects.filter(role_history__organisation_role__organisation=request.user.organisation_user.organisation, approved=True).order_by('-last_updated')[:2]
    #notification = chain(likes,comments,rewards)

    context.update({'likes_notify':likes,'comments_notify':comments,
                    'rewards_notify':rewards})

这是我想要为其发送通知的视图中的部分。

Answer 1

终于找到了我的问题的答案。我刚用这段代码修改了我的js：

JS：

success: function(data) {
var dtr = $('.notification',data);
$('.notification').html(dtr);
}

使用jquery，ajax和django定期刷新页面

1 个答案: