联动菜单多选条件筛选，如何全选的时候选择带有一级条件的全部的信息,迅睿CMS,Codeigniter/Laravel/ThinkPHP技术文档,PHP开发文档,迅睿CMS框架官方教程

联系官方销售客服

1835022288

028-61286886

问题求助问题分享经验开发需求

开发框架版主：迅睿框架研发组

联动菜单多选条件筛选，如何全选的时候选择带有一级条件的全部的信息

万俟肸

类型：迅睿CMS 更新时间：2020-06-22 18:11:33

联动菜单多选条件筛选，红框部分是一级菜单，蓝框部分是二级菜单，在二级中有一个二五仔就是第一个全选。

如何判定哪个一级下面有二级被选中了，

如何全选的时候选择带有一级条件的全部的信息，并且不影响其他的一级菜单

{linkage code=gzddyx pid=0 return=c1}

                    <li class="enzpsxItem">
                      <div class="enzpsxItemBox">
                          <div class="enzpsxItemTop">
                              <div class="enzpsxItemLeft">
                                    <label></label>
                                    <span>{$c1.name}</span>
                                </div>
                                <div class="enzpsxItemRight">
                                    <span>View details</span>
                                    <img src="{HOME_THEME_PATH}/images/icon74.png" class="zpTag3">
                                    <img src="{HOME_THEME_PATH}/images/icon75.png" class="zpTag4">
                                </div>
                           </div>
                            <ul class="enzpsxItemBottom">
                                <li class="jobArea">
                                    <a href="{Router::search_url($params, 'gongzuoquyu', $c1.cname)}#zpFrom">
                                    <label class="{if dr_is_double_search($params.gongzuoquyu, $c1.cname)}on{/if}"></label>
                                    <span>ALL {$c1.name}</span>
                                    </a>
                                </li>
                                {linkage code=gzddyx pid=$c1.id return=c2}
                                <li class="jobArea" link="">
                                    <a href="{Router::search_url($params, 'gongzuoquyu', dr_get_double_search($params.gongzuoquyu, $c2.cname))}#zpFrom">
                                    <label class="{if dr_is_double_search($params.gongzuoquyu, $c2.cname)} on {/if}"></label>
                                    <span>{$c2.name}</span>
                                    </a>
                                </li>
                                {/linkage}
                                <!--{$debug_c1}-->
                           </ul>
                               
                      </div>
                  </li>   
                {/linkage}

回帖

#1楼迅睿框架创始人

2020-06-22 17:27:53
Chrome 0

这个恐怕我都很难写出来
#2楼每天吃包子

2020-06-22 17:32:18
Chrome 0

感觉难道很大
#3楼万俟肸

2020-06-22 17:33:04
Chrome 0

每天吃包子两天了啥也没干就伺候这个沙雕了
#4楼易学习

2020-06-22 17:37:05
Chrome 0
模板标签生成工具

必火！模板制作的必备插件，自动生成字段的模板标签，解决新手老手调用困难问题

我应该可以搞定，主要是没看明白是什么逻辑

#5楼万俟肸

2020-06-22 18:10:37

Chrome 0

回复迅睿框架创始人

               {linkage code=gzddyx pid=0 return=c1}

                    <li class="enzpsxItem">
                      <div class="enzpsxItemBox">
                          <div class="enzpsxItemTop">
                              <div class="enzpsxItemLeft">
                                    <label></label>
                                    <span>{$c1.name}</span>
                                </div>
                                <div class="enzpsxItemRight">
                                    <span>View details</span>
                                    <img src="{HOME_THEME_PATH}/images/icon74.png" class="zpTag3">
                                    <img src="{HOME_THEME_PATH}/images/icon75.png" class="zpTag4">
                                </div>
                           </div>
                            <ul class="enzpsxItemBottom">
                                <li class="jobArea" link="">
                                    <a href="{Router::search_url($params, 'gongzuoquyu', dr_get_double_search($params.gongzuoquyu, $c1.cname))}#zpFrom">
                                    <label class="{if dr_is_double_search($params.gongzuoquyu, $c1.cname)} on {/if}"></label>
                                    <span>ALL {$c1.name}</span>
                                    </a>
                                </li>
                                {linkage code=gzddyx pid=$c1.id return=c2}
                                <li class="jobArea" link="">
                                    <a href="{Router::search_url($params, 'gongzuoquyu', dr_get_double_search($params.gongzuoquyu, $c2.cname))}#zpFrom">
                                    <label class="{if dr_is_double_search($params.gongzuoquyu, $c2.cname)} on {/if}"></label>
                                    <span>{$c2.name}</span>
                                    </a>
                                </li>
                                {/linkage}
                                <!--{$debug_c1}-->
                           </ul>
                               
                      </div>
                  </li>   
                  <!--
                  id：{$c1.name}、site：{$c1.site}、pid：{$c1.pid}、pids：{$c1.pids}、name：{$c1.name}、cname：{$c1.cname}、child：{$c1.child}、hidden：{$c1.hidden}、childids：{$c1.childids}、displayorder：{$c1.displayorder}、ii：{$c1.ii}-->
                {/linkage}   {$debug_c1}

再配合一点JS，手起刀落人抬走

满意答案

#6楼万俟肸

2020-06-22 18:11:33
Chrome 0

@万俟肸：手起刀落人抬走

模板标签生成工具