Django - Ajax分页
目前总结了2种方法：
1. Ajax 分页
尼玛各种google,stackoverflow,搞了好久才总结出这个，之前使用Pagination tag loading的方式不好用，并且不能进行ajax提交请求的页面无刷新的方式去分页
1.view.py
1 from django.core.paginator import Paginator, EmptyPage, PageNotAnInteger 2 from django.shortcuts import render 3 def xxx(request): 4 rows = cursor.fetchall() 5 paginator = Paginator(rows, 15) //how many items per page 6 page = request.POST.get('page') 7 try: 8 rows = paginator.page(page) 9 except PageNotAnInteger: 10 # If page is not an integer, deliver first page. 11 rows = paginator.page(1) 12 except EmptyPage: 13 # If page is out of range (e.g. 9999), deliver last page of results. 14 rows = paginator.page(paginator.num_pages) 15 return render(request,'ajax_page.html', {'rows': rows})
2. ajax_page.html
<div class="pagination"> <span class="step-links"> {% if rows.has_previous %} <a id='pre' href="#">previous</a> {% endif %} <span class="current"> Page {{ rows.number }} of {{ rows.paginator.num_pages }}. </span> {% if rows.has_next %} <a id="next" href="#" >next</a> {% endif %} </span> </div>
3. JS -Ajax 部分
1 {%block js%} 2 <script type="text/javascript"> 3 $('.step-links #next').click(function(){ 4 {% if rows.has_next %} 5 page={{ rows.next_page_number }}; 6 {% endif %} 7 $.ajax({type:"POST",url:"/submitjobs/",data:{"page":page},10 success:function(data){ 11 $("#jobs_table").html(data); 12 } 13 }) 15 }); 16 $('.step-links #pre').click(function(){ 17 {% if rows.has_previous %} 18 page={{ rows.previous_page_number }}; 19 {% endif %} 20 $.ajax({type:"POST",url:"/submitjobs/",data:{page":page},23 success:function(data){ 24 $("#jobs_table").html(data); 25 } 26 }) 27 }); 28 </script> 29 {%endblock%}
2. Datatable - plugin for Jquery (http://datatables.net/)
这种方法quick and dirty 唯一缺点就是不适用海量数据一般几百来页也是可以的，只需在js中调用dataTable方法就好。
数据填充已经在方法1中使用render.request将数据载入table.
{%block js%}
<script type="text/javascript">
$('#job_table').dataTable();
</script>
{%endblock%}
