只显示一些页码由django分页 [英] Display only some of the page numbers by django pagination
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问题描述
views.py：
def blog（request）：
blogs_list = Blog.objects.all（）
paginator = Paginator（blogs_list，1）
尝试：
page = int（request.GET.get（'page'，'1'））
除了：
page = 1
try：
blogs = paginator.page（page）
except（EmptyPage，InvalidPage）：
blogs = paginator.page（page）
return render（request，'blogs.html'，{
'blogs'：blogs
}）
模板片段： p>
< div class =prev_next>
{％if blogs.has_previous％}
< a class =prev btn btn-infohref =？page = {{blogs.previous_page_number}}> Prev< / A>
{％endif％}
{％if blogs.has_next％}
< a class =next btn btn-infohref =？page = {{blogs.next_page_number}} >接着< / A>
{％endif％}
< div class =pages>
< ul>
{％for pg in blogs.paginator.page_range％}
{％if blogs.number == pg％}
< li>< a href =？page = {{ pg}}class =btn btn-default> {{pg}}< / a>< / li>
{％else％}
< li>< a href =？page = {{pg}}class =btn> {{pg}}< / a> ; /立GT;
{％endif％}
{％endfor％}
< / ul>
< / div>
< span class =clear_both>< / span>
< / div>
现在看起来像这样：
我该怎么办只显示7页数，而不是全部从当前页码开始，如下所示：
上一页1（2）3 4 5下一页
我希望我很清楚，如果不是，请问。您的帮助和指导将非常感谢。谢谢。
解决方案
首先我会更改以下内容：
try：
blogs = paginator.page（page）
except（EmptyPage，InvalidPage）：
blogs = paginator.page（page）＃提升相同的错误
但你可以在你的上下文中传递一个范围。
index = paginator.page_range.index（blogs.number）
max_index = len（paginator.page_range）
start_index =索引 - 3 if index> = 3 else 0
end_index = index + 3 if index< = max_index - 3 else max_index
page_range = paginator.page_range [start_index：end_index]
现在，您应该可以循环使用来构建正确的链接？page = 。
===编辑===
所以你的看法会是这样的：
paginator = Paginator（Blog.objects.all（），1）
try：
page = int（request.GET.get（'page'，'1'））
除了：
page = 1
try：
blogs = paginator。页面（页）
except（EmptyPage，InvalidPage）：
blogs = paginator.page（1）
＃获取当前页面的索引
index = blogs .number - 1＃编辑更容易没有索引
＃这个值是你的页面的最大索引，所以最后一页 - 1
max_index = len（paginator.page_range）
＃你想要3，如果index< = max_index - 3 else max_index $，那么我们可以计算在哪里切片列表
start_index = index - 3 if index> = 3 else 0
end_index = b $ b＃我的新页面范围
page_range = paginator.page_range [start_index：end_index]
return render（request，'blogs.html'，{
'blogs'：blogs，
'page_range'：page_range，
}）
所以现在我们必须编辑你的模板来接受我们的新的页码列表：
< div class =prev_next>
{％if blogs.has_previous％}
< a class =prev btn btn-infohref =？page = {{blogs.previous_page_number}}> Prev< / a>
{％endif％}
{％if blogs.has_next％}
< a class =next btn btn-infohref =？page = {{blogs.next_page_number}} >接着< / A>
{％endif％}
< div class =pages>
< ul>
{％for page_range％}
{％if blogs.number == pg％}
< li>< a href =？page = {{pg}} class =btn btn-default> {{pg}}< / a>< / li>
{％else％}
< li>< a href =？page = {{pg}}class =btn> {{pg}}< / a> ; /立GT;
{％endif％}
{％endfor％}
< / ul>
< / div>
< span class =clear_both>< / span>
< / div>
I am using the django paginator in the template. Its working ok, but not good when there's large numbers of pages.
views.py:
def blog(request):
blogs_list = Blog.objects.all()
paginator = Paginator(blogs_list, 1)
try:
page = int(request.GET.get('page', '1'))
except:
page = 1
try:
blogs = paginator.page(page)
except(EmptyPage, InvalidPage):
blogs = paginator.page(page)
return render(request, 'blogs.html', {
'blogs':blogs
})
snippet of the template:
<div class="prev_next">
{% if blogs.has_previous %}
<a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
{% endif %}
{% if blogs.has_next %}
<a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
{% endif %}
<div class="pages">
<ul>
{% for pg in blogs.paginator.page_range %}
{% if blogs.number == pg %}
<li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
{% else %}
<li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
{% endif %}
{% endfor %}
</ul>
</div>
<span class="clear_both"></span>
</div>
Now it looks like this:
What do I do to display only 7 page numbers and not all of it ranging from the current page number, like this:
Prev 1 (2) 3 4 5 Next
I hope I was clear, if not please ask. Your help and guidance will be very much appreciated. Thank you.
解决方案
First of all I would change the following:
try:
blogs = paginator.page(page)
except(EmptyPage, InvalidPage):
blogs = paginator.page(page) # Raises the same error
But you could pass a range within your context.
index = paginator.page_range.index(blogs.number)
max_index = len(paginator.page_range)
start_index = index - 3 if index >= 3 else 0
end_index = index + 3 if index <= max_index - 3 else max_index
page_range = paginator.page_range[start_index:end_index]
Now you should be able to loop over the range to construct the right links with ?page=.
=== Edit ===
So your view would be something like this:
def blog(request):
paginator = Paginator(Blog.objects.all(), 1)
try:
page = int(request.GET.get('page', '1'))
except:
page = 1
try:
blogs = paginator.page(page)
except(EmptyPage, InvalidPage):
blogs = paginator.page(1)
# Get the index of the current page
index = blogs.number - 1 # edited to something easier without index
# This value is maximum index of your pages, so the last page - 1
max_index = len(paginator.page_range)
# You want a range of 7, so lets calculate where to slice the list
start_index = index - 3 if index >= 3 else 0
end_index = index + 3 if index <= max_index - 3 else max_index
# My new page range
page_range = paginator.page_range[start_index:end_index]
return render(request, 'blogs.html', {
'blogs': blogs,
'page_range': page_range,
})
So now we have to edit your template to accept our new list of page numbers:
<div class="prev_next">
{% if blogs.has_previous %}
<a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
{% endif %}
{% if blogs.has_next %}
<a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
{% endif %}
<div class="pages">
<ul>
{% for pg in page_range %}
{% if blogs.number == pg %}
<li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
{% else %}
<li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
{% endif %}
{% endfor %}
</ul>
</div>
<span class="clear_both"></span>
</div>
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