SQL樹形結構數據的查詢難點（持續更新）

--1、不是末級節點的記錄設置狀態爲【關閉狀態】
select
	a.id,a.parent_id,a.project_code,a.project_name,a.tree_level,a.sort,a.remark,a.creator_id,a.creator,a.create_date,a.creator_dept_id,a.creator_dept,a.company_id,a.company_name,a.bu_id,a.bu_name,
	case (select count(1) from table_name b where b.parent_id = a.id)
		when 0 then 'open'
		else 'closed' end tree_state
from table_name a
where 1=1
<if test="expand != null and expand == 'level'">
	<if test="treeLevels != null">
		and a.tree_level between 0 and #{treeLevels}
	</if>
</if>
order by a.sort, a.project_code

/** 關鍵sql
case (select count(1) from table_name b where b.parent_id = a.id)
		when 0 then 'open'
		else 'closed' end tree_state
*/

--2、查找滿足條件的末級節點及其所有父級節點
select distinct *  from table_name l
	connect by prior l.parent_id = l.id
	start with 1 = 1
	<if test="productionCostId != null  and productionCostId != '' ">
		and l.production_cost_id=#{productionCostId}
	and l.total_price=0
	and not exists(select 1 from table_name c where c.production_cost_id=#{productionCostId}
	and l.id=c.parent_id ) 
	</if>
	<if test="exendsSql != null  and exendsSql != '' ">
		${exendsSql}
	</if> 
	ORDER BY l.TREE_LEVEL,l.SORT

/** 關鍵sql
connect by prior l.parent_id = l.id
	start with 1 = 1
	and not exists(select 1 from table_name c where l.id=c.parent_id )
*/