如何点击筛选按钮再出结果
迅睿框架 更新 2024-01-31 14:37:50
dizanna
dizanna
如何点击下方确定再筛选，而不是点击菜单直接跳转，看了搜索函数写法没看明白
需要能看懂js语句才行参考文档：《搜索函数写法 dr_search_url》
{linkage code=pinpai pid=$params.pinpai} <a class="{if $t.id==$params.pinpai}on{/if}" href="{dr_search_url($params, 'pinpai', $t.id, 'news')}">{$t.name}</a> {/linkage} <a href="" class="btn">确定</a>{linkage code=pinpai pid=$params.pinpai} <a class="{if $t.id==$params.pinpai}on{/if}" href="javascript:$('#aa').attr("href", '{dr_search_url($params, 'pinpai', $t.id, 'news')}');">{$t.name}</a> {/linkage} <a href="" id="aa" class="btn">确定</a>这是很基础的js啊<div id="screencon" class="con"> <div id="subNavBox"> <ul class="subNav"> <li class="icon titie">品牌</li> <li id="screenAttribute9"> <a href="javascript:void(0)" class="on">不限</a> {linkage code=pinpai pid=$params.pinpai} <a class="{if $t.id==$params.pinpai}on{/if}" href="javascript:$('#aa').attr("href", '{dr_search_url($params, 'pinpai', $t.id, 'news')}');">{$t.name}</a> {/linkage} </li> <li class="icon titie">价格区间</li> <li id="screenAttribute10"> <a href="javascript:void(0)" class="on">不限</a> {linkage code=jiagequjian pid=$params.jiagequjian} <a class="{if $t.id==$params.jiagequjian}on{/if}" href="{dr_search_url($params, 'jiagequjian', $t.id, 'news')}">{$t.name}</a> {/linkage} </li> </ul> </div> <div class="screenbtn"> <a href="" class="btn">重置</a> <a href="" id="aa" class="btn">确定</a> </div> </div>