您如何像在Stackoverflow中一样制作一个上-下投票按钮？

Question

问题

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

我不确定是否应该对变量使用数据库。但是，我知道还有一种更简单的方法可以节省投票数。

您如何解决这些问题？

[编辑]

服务器端编程语言是Python。

Answer 1

这是使用jQuery / Django的肮脏/未经尝试的理论实现。

我们将假设上下投票是针对本网站上的问题/答案，但是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>

Java脚本

$(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视图

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?')

kes 当我开始回答这个问题时，我并不想写那么多，但是我被带走了。首次加载页面时，您仍然缺少最初获得所有投票的初始请求，但我将其留给读者练习。总之，如果你
是
在使用Django事实，有兴趣的投票＃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。他们发布了代码，这相当不错。