Django分页开销与排序 [英] Django pagination overhead with sorting

问题描述

我尝试使用某些排序方式在Django中实现分页结构，但我无法确定如何正确执行。

视图。 py

  def search（request）：
 eList = Employer.objects.filter（eminence__lt = 4 ）.order_by（' -  eminence'）
 
 paginator = Paginator（eList，3）＃显示每页3个联系人
 
 page = request.GET.get（'page' ）
 try：
 employerList = paginator.page（page）
除了PageNotAnInteger：
 employerList = paginator.page（1）
除了EmptyPage：
 employerList = paginator.page（paginator.num_pages）
 
返回render_response（请求，雇主/ search.html，{'雇主列表'：雇主列表}）
  / pre> 
 
 
  search.html  
 < div class =pagination> 
< span class =step-links> 
 {％如果employerList.has_previous％} 
< a href =？page = {{employerList.previous_page_number}}> previous< / a> 
 {％endif％} 
 
< span class =current> {{employerList.paginator.num_pages}}的页面{{employerList.number}}。< / span> 
 {％如果employerList.has_next％} 
< a href =？page = {{employerList.next_page_number}}> next< / a> 
 {％endif％} 
< / span> 
< / div> 
  
这个例子很好，但是你可以看到，对于每个导航，我需要得到所有的雇主对象。之后， Paginator 根据页码处理查询中的对象。但是，我认为分页应该在查询期间完成，只有根据页码才能得到我想要的X对象。当然，我可以修改代码，但是我不明白为什么人们使用 Paginator 虽然有这样的开销。我可能会忽略一个细节... 
 
 
 
我的第二个问题是如何应用排序我的列表？我应该修改我的 url 并传递排序方法和页码，并根据他们通过sort方法得到quesy，并将其赋予 Paginator ？
 
 
 
这似乎是合理和罚款，但我只是想知道有没有更好的做法在django。
 
 
 
谢谢
解决方案
 Django对于延迟加载数据非常聪明，这使得这些查询非常有效。我们来看看你的请求会发生什么... 
  eList = Employer.objects.filter（eminence__lt = 4）.order_by '-eminence'）
 ##没有数据库查询。 
 
 paginator = Paginator（eList，3）
 ##没有数据库查询。 
 
 employerList = paginator.page（2）
 ## SELECT COUNT（*）FROM`yourproject_employer` 
 ## WHERE`yourproject_employer`.`eminence`< 4 
 
＃强制迭代。与循环结果相同：
 foo = list（employerList.object_list）
 ## SELECT * FROM`yourproject_employer` 
 ## WHERE`yourproject_employer`.`eminence`< 4 
 ## ORDER BY`yourproject_employer`.`eminence` DESC 
 ## LIMIT 3 OFFSET 3 
  
至于你的排序问题，我会说你只是按照你的建议修改 GET 参数。只要将非常小心地传递给数据库。例如，我会列出可能的排序并对此进行验证。这也意味着您不必公开数据库的内部工作。
  VALID_SORTS = {
 pk：pk，
pkd：-pk，
em：eminence，
emd：-eminence，
} 
 DEFAULT_SORT ='pk'
 def search（request）：
 sort_key = request.GET.get（'sort'，DEFAULT_SORT）＃将pk替换为默认值。 
 sort = VALID_SORTS.get（sort_key，DEFAULT_SORT）
 
 eList = Employer.objects.filter（eminence__lt = 4）.order_by（sort）
  
 
I try to implement pagination structure in Django with some sort options however, I can't figure out how can I do that properly.

views.py
def search(request):
    eList = Employer.objects.filter(eminence__lt=4).order_by('-eminence')

    paginator = Paginator(eList, 3) # Show 3 contacts per page

    page = request.GET.get('page')
    try:
        employerList = paginator.page(page)
    except PageNotAnInteger:
        employerList = paginator.page(1)
    except EmptyPage:
        employerList = paginator.page(paginator.num_pages)

return render_response(request, 'employer/search.html', {'employerList':employerList})
search.html
<div class="pagination">
    <span class="step-links">
        {% if employerList.has_previous %}
            <a href="?page={{ employerList.previous_page_number }}">previous</a>
        {% endif %}

        <span class="current"> Page {{ employerList.number }} of {{ employerList.paginator.num_pages }}.</span>
        {% if employerList.has_next %}
            <a href="?page={{ employerList.next_page_number }}">next</a>
        {% endif %}
    </span>
</div>
This example works great however as you can see, for every navigation I need to get all Employer objects. After that Paginator handles the objects in query depending on page number. However, I think the pagination should be done during query and only get X objects that I want depending on page number. I can modify, of course, the code in that manner but I can't  figure out why people use Paginator although there is such overhead. I may overlook a detail...

My second question is how can I apply sort my list ? Should I modify my url and passing a sort method and page number, and depending on them get the quesy by sort method and give it to Paginator ?

This seems reasonable and fine by me but I just wonder is there any better approach to do in django.

Thanks
解决方案
Django is very smart about lazy-loading the data, which makes these queries very efficient.  Let's walk through what happens with your requests...
eList = Employer.objects.filter(eminence__lt=4).order_by('-eminence')
## No database query.

paginator = Paginator(eList, 3)
## No database query.

employerList = paginator.page(2)
## SELECT COUNT(*) FROM `yourproject_employer`
## WHERE `yourproject_employer`.`eminence` < 4

 # Force iteration.  Same as looping over results:
foo = list(employerList.object_list)
## SELECT * FROM `yourproject_employer`
## WHERE `yourproject_employer`.`eminence` < 4
## ORDER BY `yourproject_employer`.`eminence` DESC
## LIMIT 3 OFFSET 3
As for your sorting question, I would say to simply modify the GET arguments as you suggested.  Just be very careful passing this to the database.  I would, for example, make a list of possible sorts and verify against that.  This also means you don't have to expose the inner-workings of your database.
VALID_SORTS = {
    "pk": "pk",
    "pkd": "-pk",
    "em": "eminence",
    "emd": "-eminence",
}
DEFAULT_SORT = 'pk'
def search(request):
    sort_key = request.GET.get('sort', DEFAULT_SORT) # Replace pk with your default.
    sort = VALID_SORTS.get(sort_key, DEFAULT_SORT)

    eList = Employer.objects.filter(eminence__lt=4).order_by(sort)


                        
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