时间:2021-07-01 10:21:17 帮助过:31人阅读
需求 字段 描述 备注 ID 主键,32位UUID TYPE_CODE 编码 如:1-01-003 PARENT_ID 父节点ID,32
需求
字段
描述
备注
ID
主键,32位UUID
TYPE_CODE
编码
如:1-01-003
PARENT_ID
父节点ID,,32位UUID
SORT_NUM
排序编号
正整数
假设顶级节点的TYPE_CODE为字符1,写存储过程把表中所有的节点TYPE_CODE生成好;
二级节点前面补一个龄,三级补两个零,依次类推;
实现关键点
不知道系统有多少层级,需要递归调用
通过递归调用自身;
如何动态在TYPE_CODE前面填充‘0’;通过计算‘-’的个数来确定层级,从而确定前缀的个数
tree_level:= (length(p_code)-length(replace(p_code,'-',''))) + 1;
前面填充前缀‘0’字符
lpad(to_char(cnt),tree_level,'0')
存储过程代码
CREATEOR REPLACE PROCEDURE INI_TREE_CODE
(
V_PARENT_ID IN VARCHAR2
)AS
p_id varchar2(32);
p_code varchar2(256);
sub_num number(4,0);
tree_level number(4,0);
cnt number(4,0) default 0;
cursor treeCur(oid varchar2) is
select id,TYPE_CODE from eval_index_type
where parent_id = oid
order by sort_num;
BEGIN
sub_num := 0;
select id,type_code into p_id,p_code
from eval_index_type
where id = V_PARENT_ID
order by sort_num;
for curRow in treeCur(p_id) loop
cnt := cnt +1;
tree_level :=(length(p_code)-length(replace(p_code,'-',''))) + 1;
update eval_index_type set type_code =p_code || '-' || lpad(to_char(cnt) ,tree_level,'0')
where id = curRow.id;
select COUNT(*) into sub_num fromeval_index_type where parent_id = p_id;
if sub_num > 0 then
INI_TREE_CODE (curRow.id);
end if;
end loop;
ENDINI_TREE_CODE;
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