Question

我正在尝试构建一个简单的排名系统，我按“分数”然后按“ID”命令主题。我最初构建这是PHP，在SQL中设置'rownum'变量并像这样调用'Rank'：

public function rank() {
    global $database;
    $sql = "SET @rownum :=0";
    $database->query($sql);
    $sql = "SELECT rank FROM ( 
        SELECT @rownum:=@rownum+1 AS rank, id, score 
        FROM subjects
        ORDER BY score DESC, id ASC) AS derived_table 
        WHERE id = {$this->id}";

    $result_set = $database->query($sql);
    $row = $database->fetch_array($result_set);
    return array_shift($row);
    }

我可以从Queryset索引它，但我还没弄明白如何做到这一点。关于如何在Django中完成此任务的任何想法？

模特：

class Subject(models.Model):
    def __unicode__(self):
        return self.name
    def __str__(self):
        return self.name

    name = models.CharField(max_length=40)
    score = models.IntegerField(default=0)
    created = models.DateTimeField(default=datetime.datetime.now)

提前感谢！

Answer 1

尝试

order_by('score', 'pk') #pk is the ID or primary key

查看Django Book: Chapter 5 Models

Answer 2

使用extra方法可以完成此操作。

Blog.objects.extra(
    select={
        'entry_count': 'SELECT COUNT(*) FROM blog_entry WHERE blog_entry.blog_id = blog_blog.id'
    },
)

Answer 3

试试这个：

someSubject = Subject.objects.get(name='someSubject')
someSubject_rank = Subject.objects.filter(score__gt=someSubject.score).count()+1

Answer 4

如果你只想显示数据而不是以任何方式处理每一行的排名字段，那么使用Django的forloop.counter template variable是简单而优雅的方式：

# your views.py    
return (super(View, self).get_context_data(
        subjects=Subject.objects.order_by("-score", "pk"), 
        **kwargs)
)

# your template.html
{% for subject in subjects %}
    <tr>
        <td>{{ forloop.counter }}</td>
        <td>{{ subject.name }}</td>
        <td>{{ subject.score }}</td>
    </tr>
{% endfor %}

Django排名解决方案

4 个答案: