【原创】SQL栏目树的代码
本文由网友whl供稿,特此感谢!
/**
* Desc: 取栏目树 ,过滤用户权限和无效栏目
* Author: WHL
* Date: 2009-05-31 15:17
*/
/** 1. 取某用户有权限(np_cms_column_security表有记录且t.action_1 = ‘1’)的栏目的树 **/
create or replace view V_NP_CTREE_BS as select B.* from ( select A.*, lag(A.column_id) over(partition by A.column_id order by 0 ) RK from (select /*+choose */ t.* from np_cms_column t where t.is_active = '1' connect by prior t.column_id = t.parent_id start with t.column_id in (select t.column_id from np_cms_column_security t where t.subject_id = 'mazj' /*这里添加角色过滤*/ and t.action_1 = '1'))A) B where not exists (select 0 from (select distinct d.column_id from np_cms_column d connect by prior d.column_id = d.parent_id start with d.column_id in (select t.column_id from np_cms_column_security t where t.subject_id = 'mazj' /* 这里添加角色过滤*/ and t.action_1 = '0' /* 排除有权限树下的非授权ID,既 Action_1=0的*/ and exists (select 0 from (select distinct d.column_id from np_cms_column d connect by prior d.column_id = d.parent_id start with d.column_id in (select t.column_id from np_cms_column_security t where t.subject_id = 'mazj' /*这里添加角色过滤*/ and t.action_1 = '1')) C1 where C1.column_id = t.column_id)) and d.is_active = '1') C where C.column_id = B.column_id and B.RK is null) and B.RK is null union all select c.*, 0 RK from np_cms_column c where c.parent_id = 0;
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/** 2.得到栏目的虚拟父亲ID(考虑到把断层的节点接起来)**/
create or replace view V_NP_CTREE_PA as select B.*, (case B.column_id when 1 then 0 else nvl(B.father, 1) end) VFA from (select v.*, (select vv.column_id from V_NP_CTREE_BS vv where vv.column_id = v.parent_id) FATHER from V_NP_CTREE_BS v) B;
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/** 3. 取出门户需要的栏目树 **/
--create or replace view V_NP_CTREE_RS as select D.*, LPAD(' ', 2 * level - 1) || SYS_CONNECT_BY_PATH(D.COLUMN_NAME, '/') "Path" from (select c.* from V_NP_CTREE_PA c order by c.VFA, c.disorder desc, c.column_id desc) D connect by prior D.column_id = D.VFA start with D.column_id = 1;
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《【原创】SQL栏目树的代码》 的相关评论
看到一个这么长的sql,起初就会想起数据表设计不合理了,还有谁想直接在这上面维护。