Python Django Paginator
我试图分页一页,以便每页显示五种状态。输入这些代码后,无法分页。下面是我的django应用程序中分页和状态更新的代码 视图(Views.py文件): 模板(html文件):Python Django Paginator,python,django,Python,Django,我试图分页一页,以便每页显示五种状态。输入这些代码后,无法分页。下面是我的django应用程序中分页和状态更新的代码 视图(Views.py文件): 模板(html文件): {%block content%} {%用于请求中的请求%} {{Ask.user} {{Ask.status} {{Ask.pub|u date | timesince}}之前 {%endfor%} {%if Asks.has_previous%} {%endif%} 第{{Asks.paginator.num_pages
{%block content%}
{%用于请求中的请求%}
{{Ask.user}
{{Ask.status}
{{Ask.pub|u date | timesince}}之前
{%endfor%}
{%if Asks.has_previous%}
{%endif%}
第{{Asks.paginator.num_pages}页中的第{{Asks.number}页。
{%if Asks.has_next%}
{%endif%}
{%endblock%}
您的页面数据位于fp
变量中,但它从未提交到模板中。相反,您试图从任务中获取页面数据,但这与代码中的分页器无关。我不确定您试图分页的内容,但如果您想通过筛选数据进行分页,应该是这样的:
plan = Ask.objects.filter(user=request.user)
paginator=Paginator(plan, 5)
###...get you page number
try:
asks = paginator.page(page)
except (EmptyPage, InvalidPage):
asks = paginator.page(paginator.num_pages)
extra_data_context.update({'Asks': asks})
同时修复python代码的缩进。是的,我想对过滤后的数据进行分页。这样每页只显示五种状态。我尝试了上面的代码,但得到的结果是:TemplateSyntaxError at/qaskp/catch TypeError在呈现时:“Page”对象不可编辑
{% block content %}
{% for Ask in Asks %}
<tr>
<p> {{Ask.user}} </p> </strong>
<p>{{Ask.status}}</p>
<p> {{Ask.pub_date|timesince }} ago </p>
</tr>
{% endfor %}
<div class="pagination">
<span class="step-links">
{% if Asks.has_previous %}
<a href="?page={{ Asks.previous_page_number }}">previous</a>
{% endif %}
<span class="current">
Page {{ Asks.number }} of {{ Asks.paginator.num_pages }}.
</span>
{% if Asks.has_next %}
<a href="?page={{ Asks.next_page_number }}">next</a>
{% endif %}
</span>
</div>
{% endblock %}
plan = Ask.objects.filter(user=request.user)
paginator=Paginator(plan, 5)
###...get you page number
try:
asks = paginator.page(page)
except (EmptyPage, InvalidPage):
asks = paginator.page(paginator.num_pages)
extra_data_context.update({'Asks': asks})