CRM手记-5 | Django

Posted 2020-10-21 胡说八道

# 过滤补充;
# 点击userprofile表报错;
# 因为当没有自定义admin_class时,默认使用的是BaseKingAdmin；
# 所以,BaseKingAdmin需要做一些默认的配置;
    # admin_base.py
        class BaseKingAdmin(object):
            list_display = []
            list_filter = []
            search_fields = []
    # 此时列表为空,build_table_row()的将没有返回值,前端显示空;
    # 做一个判断：
    # kingadmin_tag.py
        def build_table_row(obj,admin_class):
            """生成一条记录的html element"""

            ele = ""
            if admin_class.list_display:
                for column_name in admin_class.list_display:

                    column_obj = admin_class.model._meta.get_field(column_name)
                    if column_obj.choices: #get_xxx_display
                        column_data = getattr(obj,‘get_%s_display‘% column_name)()
                    else:
                        column_data = getattr(obj,column_name)

                    td_ele = "<td>%s</td>"% column_data
                    ele += td_ele
            else:
                td_ele = "<td>%s</td>" % obj
                ele += td_ele
            return mark_safe(ele)
 
    # 问题:没有注册admin_class,但是还是显示过滤;
        # table_obj_list.html
            <div>
                <!-- 判断是否有list_filter -->
                {% if admin_class.list_filter %}
                    <form action="">
                        {% for filter_column in admin_class.list_filter %}
                            {% build_filter_ele filter_column admin_class %}
                        {% endfor %}
                        <input type="submit" value="过滤">
                    </form>
                {% endif %}
            </div>
    # django-admin中如果没有注册不显示字段,表头显示表名;
        # table_obj_list.html
            <tr>
            {% if admin_class.list_display %}
                {% for column in admin_class.list_display %}
                    <th>{{ column }}</th>
                {% endfor %}
                {% else %}
                <th>{{ admin_class.model._meta.model_name }}</th>
            {% endif %}
            </tr>
    # 报错,因为模板语言不允许前缀下划线_meta
    # 写一个simple_tag
        # kingadmin_tag.py
            @register.simple_tag
            def get_model_name(admin_class):
                return admin_class.model._meta.model_name.upper()
        # table_obj_list.html
            <th>{{ get_model_name admin_class }}</th>

##############################################################

# 分页功能开发；
# 处理一：django-pure-pagination
    # veiws.py
        @login_required
        def table_obj_list(request,app_name,model_name):
            ...
            # 分页
            try:
                page = request.GET.get(‘page‘, 1)
            # 出现异常也取第一页
            except PageNotAnInteger:
                page = 1
            # 实例化分页器类，5条数据一页
            p = Paginator(querysets, 2, request=request)
            # 通过Paginator内部的page()方法，获取Page分页对象
            # 传入当前页码(指定页面)
            current_page = p.page(page)

            return render(request,‘kingadmin/table_obj_list.html‘, {
                ‘querysets‘:querysets,
                ‘admin_class‘:admin_class,
                ‘current_page‘:current_page
            })
    # table_obj_list.html
        <nav aria-label="Page navigation">
            <!-- 分页-->
            <ul class="pagination">
                <!-- 上一页 -->
                <!-- 如果有前一页就显示，没有就不显示，或者其他操作-->
                {% if current_page.has_previous %}
                    <li class="long"><a href="?{{ current_page.previous_page_number.querystring}}">上一页</a></li>
                {% else %}
                {% endif %}

                <!-- 遍历显示所有页码 -->
                {% for page in current_page.pages %}
                    <!-- 如果页码存在，就执行 -->
                    {% if page %}
                        <!-- 如果遍历到的页码是当前页，就激活样式 -->
                        {% ifequal page current_page.number%}
                            <li class="active"><a href="?{{ page.querystring }}">{{ page }}</a></li>
                        {% else %}
                            <li><a href="?{{ page.querystring }}">{{ page }}</a></li>
                        {% endifequal %}
                    {% else %}
                        <li class="none"><a href="">...</a></li>
                    {% endif %}
                {% endfor%}
                <!-- 下一页 -->
                {% if current_page.has_next %}
                    <li class="long"><a href="?{{ current_page.next_page_number.querystring}}">下一页</a></li>
                {% else %}
                {% endif %}
            </ul>
        </nav>
# 问题:分页之后再点击过滤,分页不作为条件;
#       过滤之后分页,在原url基础之上添加?page=,将过滤后的数据传递给分页;
    # views.py
        def get_filter_result(request,querysets):
            filter_conditions = {}
            for key,val in request.GET.items():
                # 排除分页的影响
                if key != ‘page‘:
                    if val:
                        filter_conditions[key] =  val
                else:
                    pass
            return querysets.filter(**filter_conditions),filter_conditions

###################################################################