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需求：

分类树是无限级别的，用户有部门的岗位两个属性。现需要将树跟据部门和岗位两个属性进行划分。

代码：

$catepv = "dw1(dep,mer),dw2(*),-dw2(dp,pw)";

  $userpv = "dw2(pw1)";

  function trans($pv){

  $cate = split('\)\,',$pv);

  foreach($cate as $c){

     $p = split('\(',$c);

     $v['dep'] = $p[0];

      $_p = $p[1];

     if(substr($_p,strlen($_p)-1)==")"){

       $_p = substr($_p,0,strlen($_p)-1);

      }

     $v['job'] = split(',',$_p);

     $_v[] = $v;

    }

   return $_v;

  }

  

  function pv_dep($d,$_u){

    $all = false;

    $f = false;

   foreach($d as $_d){

          if(trim($_d['dep']) == '*'){

            $all = true;

          }elseif(substr($_d['dep'],0,1)=='-' && substr($_d['dep'],1) == $_u['dep']){

            if(count($_d['job'])==1 && $_d['job'][0] == '*'){

               $f = false;

             }

           }elseif($_d['dep'] == $_u['dep']){

             $f = true;

           }

     }

   if($all||$f){

      return true;

    }else{

     return false;

     }

   }

  

  function pv_job($d,$_u){

     $f = false;

     $all = false;

     $user=array();

     foreach($_u['job'] as $uj){

        echo '
check: '.$uj;

        foreach($d as $_d){

         if(substr($_d['dep'],0,1) == '-' && substr($_d['dep'],1) == $_u['dep']){

               foreach($_d['job'] as $dj){

                  if($uj == $dj){

               //     echo '->lost by '.$dj;

                    $user[$uj] = false;

                }

             }

           }elseif($_d['dep'] == $_u['dep']){

              if(count($_d['job'])==1 && $_d['job'][0] == '*'){

                 $user[$uj] = true;

                // echo '->get by '.$uj;

              }else{

                 foreach($_d['job'] as $dj){

                   if($uj == $dj){

                      $user[$uj] = true;

                    }

                  }

               }

            }

         }

      }

    if(count($user)==0){

       return false;

     }else{

      foreach($user as $_ut){

         if($_ut){

          return true;

          }

       }

      }

   }

  function pv($d,$u){

     

     $d = trans($d);

     $u = trans($u);

     $_u = $u[0];

     if (pv_dep($d,$_u)){

         echo 'got dep.';

        if (pv_job($d,$_u)){

         echo '
Get';

        }else{

         echo '
Lost by Job!';

         }

      }else{

       echo '
Lost by Department!';

       }

   }

  pv($catepv,$userpv);

测试数据：（()以外表示部门()内表示岗位，*是通配符。下同）

    $catepv = "dw1(dep,mer),dw2(*),-dw2(dp,pw)";
    $userpv = "dw2(pw1)";

结果：Get

测试数据：

    $catepv = "dw1(dep,mer),dw2(p),-dw2(dp,pw)";
    $userpv = "dw2(pw1)";

结果：Lost by Job!

测试数据：

    $catepv = "dw1(dep,mer),dw2(p),-dw2(dp,pw)";
    $userpv = "dw3(pw1)";

结果：Lost by Job!

 

测试数据：

    $catepv = "dw1(dep,mer),dw2(*),-dw2(dp,pw)";
    $userpv = "dw2(pw1)";

结果：Get