如何在Stackoverflow中创建一个投票上下按钮？

Question

问题

1. 如何制作一个Ajax按钮(向上和向下箭头),使得数字可以增加或减少
2. 如何将用户的动作保存到变量NumberOfVotesOfQuestionID

我不确定是否应该使用数据库或不使用数据库.但是,我知道有一种更简单的方法可以节省投票数.

你怎么解决这些问题？

[编辑]

服务器端编程语言是Python.

Answer 1

这是一个使用jQuery/Django的脏/未经测试的理论实现.

我们将假设上下投票是针对本网站上的问题/答案,但显然可以根据您的实际用例进行调整.

模板

<div id="answer_595" class="answer">
  <img src="vote_up.png" class="vote up">
  <div class="score">0</div>
  <img src="vote_down.png" class="vote down">
  Blah blah blah this is my answer.
</div>

<div id="answer_596" class="answer">
  <img src="vote_up.png" class="vote up">
  <div class="score">0</div>
  <img src="vote_down.png" class="vote down">
  Blah blah blah this is my other answer.
</div>

使用Javascript

$(function() {
    $('div.answer img.vote').click(function() {
        var id = $(this).parents('div.answer').attr('id').split('_')[1];
        var vote_type = $(this).hasClass('up') ? 'up' : 'down';
        if($(this).hasClass('selected')) {
            $.post('/vote/', {id: id, type: vote_type}, function(json) {
                if(json.success == 'success') {
                    $('#answer_' + id)
                     .find('img.' + vote_type);
                     .attr('src', 'vote_' + vote_type + '_selected.png')
                     .addClass('selected');
                    $('div.score', '#answer_' + id).html(json.score);
                }
            });
        } else {
            $.post('/remove_vote/', {id: id, type: vote_type}, function(json) {
                if(json.success == 'success') {
                    $('#answer_' + id)
                     .find('img.' + vote_type);
                     .attr('src', 'vote_' + vote_type + '.png')
                     .removeClass('selected');
                    $('div.score', '#answer_' + id).html(json.score);
                }
            });                
        }
    });
});

Django的观点

def vote(request):
    if request.method == 'POST':
        try:
            answer = Answer.objects.get(pk=request.POST['id'])
        except Answer.DoesNotExist:
            return HttpResponse("{'success': 'false'}")

        try:
            vote = Vote.objects.get(answer=answer, user=request.user)
        except Vote.DoesNotExist:
            pass
        else:
            return HttpResponse("{'success': 'false'}")

        if request.POST['type'] == 'up':
            answer.score = answer.score + 1
        else:
            answer.score = answer.score - 1

        answer.save()

        Vote.objects.create(answer=answer,
                            user=request.user,
                            type=request.POST['type'])

        return HttpResponse("{'success':'true', 'score':" + answer.score + "}")
    else:
        raise Http404('What are you doing here?')

def remove_vote(request):
    if request.method == 'POST':
        try:
            answer = Answer.objects.get(pk=request.POST['id'])
        except Answer.DoesNotExist:
            return HttpResponse("{'success': 'false'}")

        try:
            vote = Vote.objects.get(answer=answer, user=request.user)
        except Vote.DoesNotExist:
            return HttpResponse("{'success': 'false'}")
        else:
            vote.delete()

        if request.POST['type'] == 'up':
            answer.score = answer.score - 1
        else:
            answer.score = answer.score + 1

        answer.save()

        return HttpResponse("{'success':'true', 'score':" + answer.score + "}")
    else:
        raise Http404('What are you doing here?')

让人惊讶.当我开始回答这个问题时,我并不打算写这么多,但我得到了一点点.您仍然缺少初次请求以在首次加载页面时获得所有投票,但我会将其作为练习留给读者.总之,如果你是实际上是使用Django和感兴趣的投票#1的测试更加/实FPGA实现,我建议你检查出的源代码为cnprog.com,用Python编写/ Django的#1的中国克隆.他们发布了他们的代码并且相当不错.

Answer 2

没有人提到过几点:

• 更改数据库的状态时,您不希望使用GET.否则我可以在我的网站上放一张图片src="http://stackoverflow.com/question_555/vote/up/answer_3/".
• 您还需要csrf(跨站点请求伪造)保护
• 您必须记录每次投票的人,以避免人们针对特定问题多次投票.无论是IP地址还是用户ID.