python遍历CTE中的双循环？

Question

我在彼此之间有2个for循环.对于loop1中的每一行'A','B','C',我需要访问分层树以找到loop2中组'X'的所有父项.这使我使用CTE,我需要分别找到每一行的路径.在循环中使用CTE不是确定我可以匹配每个组ID的解决方案.提到这个链接,但无法做出很多循环层次CTE

使用flask框架的cron作业的代码片段:

    s = select([rt_issues]).\
    where(
        and_(
            rt_issues.c.status !='Closed',
            rt_issues.c.assigned_to != None
        ))
rs = conn.execute(s)
if rs.rowcount > 0:
    s4 = text('with recursive rec_grp as(select id, parent_id, name, head, 1 as level, array[id] as path_info from groups union all select grp1.id, grp1.parent_id, grp1.name, grp1.head, rc.level + 1, rc.path_info||grp1.id from groups grp1 join rec_grp rc on grp1.id = rc.parent_id) select distinct id, parent_id, name, head, path_info from rec_grp order by id')

    rs4 = conn.execute(s4)

    for r in rs:
        head_list = []
        hierarchical_grps = []
        for rr in rs4:
            if ((rr['path_info'][0] == r[rt_issues.c.assignee_group])):
                for g in rr['path_info']:
                    hierarchical_grps.append(g)
        hierarchical_grps = list(set(hierarchical_grps))            
        send_pending_mail(hierarchical_grps, r['id'])
        print hierarchical_grps, 'hierarchical_grps'

    exit(0)

我需要向问题层次结构中的assignee_group发送邮件给所有组头.如何实现这一目标.如何正确使用循环？我只使用sqlalchemy核心,postgresql,python与烧瓶.我需要相同的确切代码.

有效的是下面的代码:

 mgroup = None
 s = select([rt_issues]).\
     where(
         and_(
             rt_issues.c.status !='Closed',
             rt_issues.c.assigned_to != None
         ))
 rs = conn.execute(s)
 if rs.rowcount > 0:
     for r in rs:
         head_list = []
         hierarchical_grps = []
         mgroup = r[rt_issues.c.assignee_group]
         s4 = text('with recursive rec_grp as(select id, parent_id, name, head, 1 as level, array[id] as path_info from groups where id=' +str(mgroup) + 'union all select grp1.id, grp1.parent_id, grp1.name, grp1.head, rc.level + 1, rc.path_info||grp1.id from groupsgrp1 join rec_grp rc on grp1.id = rc.parent_id) select distinct id,parent_id, name, head, path_info from rec_grp order by id')

     rs4 = conn.execute(s4)
     for rr in rs4:
         if ((rr['path_info'][0] == r[rt_issues.c.assignee_group])):
             for g in rr['path_info']:
                 hierarchical_grps.append(g)
     hierarchical_grps = list(set(hierarchical_grps))
     print hierarchical_grps, 'hierarchical_grps'
     send_pending_mail(hierarchical_grps, r['id'])
 exit(0)

Answer 1

假设该head列是布尔值，这将收集具有head：

rs4 = con.execute(s4)
for rr in rs4:
    if rr['head']:
        head_list.append(rr['id'])

print 'group heads:', head_list

这是假设使用第二个示例中的查询（请注意对 from 子句的更正，“from groupsgrp1”应为“from groups grp1”）：

WITH RECURSIVE rec_grp AS (
  SELECT
    id,
    parent_id,
    name,
    head,
    1          AS level,
    ARRAY [id] AS path_info
  FROM groups
  WHERE id = 4
  UNION ALL
  SELECT
    grp1.id,
    grp1.parent_id,
    grp1.name,
    grp1.head,
    rc.level + 1,
    rc.path_info || grp1.id
  FROM groups grp1
    JOIN rec_grp rc ON grp1.id = rc.parent_id
)
SELECT DISTINCT
  id,
  parent_id,
  name,
  head,
  path_info
FROM rec_grp
ORDER BY id;