• 分页器插件


    分页器的主体封装进一个类中,调用时将该类实例化后传入前端即可

    page.py  写有分页器主体的类的文件

    """
    自定义分页组件的使用方法:

    ######################### django用法 ######################## pager_obj = Pagination(request.GET.get('page',1),len(HOST_LIST),request.path_info,request.GET,per_page_zount=5,max_pager_count=11) 当前页(默认为1) 总数据量(从数据库取) 当前url(不带问号) ?后面的内容 每页显示的数据数 最多显示的页码数量 host_list = HOST_LIST[pager_obj.start:pager_obj.end] html = pager_obj.page_html() return render(request,'hosts.html',{'host_list':host_list,"page_html":html})


    ####################### Flask用法 ###########################
      pager_obj=Pagination(request.args.get('page',1),len(HOST_LIST),request.path,request.args,per_page_count=10)
      print(request.args)
      host_list=HOST_LIST[pager_obj.start:pager_obj.end]
      html=pager_obj.page_html()
      return=render_template('pager.html',host_list=host_list,page_html=html,condition=path)
    """
    
    class Pagination(object):
        """
        自定义分页
        """
        def __init__(self,current_page,total_count,base_url,params,per_page_count=10,max_pager_count=11):
            try:
                current_page = int(current_page)
            except Exception as e:
                current_page = 1
            if current_page <=0:
                current_page = 1
            self.current_page = current_page
            # 数据总条数
            self.total_count = total_count
    
            # 每页显示10条数据
            self.per_page_count = per_page_count
    
            # 页面上应该显示的最大页码
            max_page_num, div = divmod(total_count, per_page_count)
            if div:
                max_page_num += 1
            self.max_page_num = max_page_num
    
            # 页面上默认显示11个页面(当前页在中间)
            self.max_pager_count = max_pager_count
            self.half_max_pager_count = int((max_pager_count - 1) / 2)
    
            # URL前缀
            self.base_url = base_url
    
            # request.GET
            import copy
            params = copy.deepcopy(params)
            params._mutable = True
            # 包含当前列表页面所有的搜索条件
            # {source:[2,], status:[2], gender:[2],consultant:[1],page:[1]}
            # self.params[page] = 8
            # self.params.urlencode()
            # source=2&status=2&gender=2&consultant=1&page=8
            # href="/hosts/?source=2&status=2&gender=2&consultant=1&page=8"
            # href="%s?%s" %(self.base_url,self.params.urlencode())
            self.params = params
    
        @property
        def start(self):
            return (self.current_page - 1) * self.per_page_count
    
        @property
        def end(self):
            return self.current_page * self.per_page_count
    
        def page_html(self):
            # 如果总页数 <= 11
            if self.max_page_num <= self.max_pager_count:
                pager_start = 1
                pager_end = self.max_page_num
            # 如果总页数 > 11
            else:
                # 如果当前页 <= 5
                if self.current_page <= self.half_max_pager_count:
                    pager_start = 1
                    pager_end = self.max_pager_count
                else:
                    # 当前页 + 5 > 总页码
                    if (self.current_page + self.half_max_pager_count) > self.max_page_num:
                        pager_end = self.max_page_num
                        pager_start = self.max_page_num - self.max_pager_count + 1
                    else:
                        pager_start = self.current_page - self.half_max_pager_count
                        pager_end = self.current_page + self.half_max_pager_count
    
            page_html_list = []
            # {source:[2,], status:[2], gender:[2],consultant:[1],page:[1]}
            self.params['page'] = 1
            first_page = '<li><a href="%s?%s">首页</a></li>' % (self.base_url,self.params.urlencode(),)
            page_html_list.append(first_page)
            # 上一页
    
            if self.current_page==1:
                prev_page = '<li class="disabled"><a >上一页</a></li>'
            else:
                self.params['page'] = self.current_page - 1
                prev_page='<li><a href="%s?%s">上一页</a></li>' % (self.base_url,self.params.urlencode(),)
            page_html_list.append(prev_page)
    
    
    
            #当前页
            for i in range(pager_start, pager_end + 1):
                self.params['page'] = i
                if i == self.current_page:
                    temp = '<li class="active"><a  href="%s?%s">%s</a></li>' % (self.base_url,self.params.urlencode(), i,)
                else:
                    temp = '<li><a href="%s?%s">%s</a></li>' % (self.base_url,self.params.urlencode(), i,)
                page_html_list.append(temp)
    
            # 下一页
    
            if self.current_page==self.max_page_num:
                next_page = '<li class="disabled"><a >下一页</a></li>'
            else:
                self.params['page'] = self.current_page + 1
                next_page = '<li><a href="%s?%s">下一页</a></li>' % (self.base_url, self.params.urlencode(),)
            page_html_list.append(next_page)
    
    
            self.params['page'] = self.max_page_num
            last_page = '<li><a href="%s?%s">尾页</a></li>' % (self.base_url, self.params.urlencode(),)
            page_html_list.append(last_page)
    
            return ''.join(page_html_list)

    在前端页面中只需将传入的两个参数展现出来即可

    {#    可导入bootstrap进行优化#}
    
               <div>{{ host_list }}</div>
    
    
                <div class="pager">
                    <nav aria-label="Page navigation">
                        <ul class="pagination">
    
                            {{ page_html|safe }}
                        </ul>
                    </nav>
                </div>        
  • 相关阅读:
    LeetCode中等题(三)
    java之使用poi对excel的.xls和.xlsx访问
    JAVA编程-----------40、字符串排序
    JAVA编程------------38、写一个函数,求一个字符串的长度,在main函数中输入字符串,并输出其长度
    JAVA编程---------37、n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡是报到3的人退出圈子,问最后留下的是原来第几号的那位。
    JAVA编程--------------36、有n个整数,使其前面各数顺序向后移m个位置最后m个数变成前面m个数
    JAVA编程-------------24、将一组数的最大数放在第一位,最小的数放在最后一位
    JAVA编程------------33、打印杨辉三角
    JAVA编程--------------32、取一个整数的4-7位
    JAVA编程---------------31、将一个数组逆序输出
  • 原文地址:https://www.cnblogs.com/zhuminghui/p/8494643.html
Copyright © 2020-2023  润新知