级联查找成员资格的根referer-PHP/MySQL查询

我运行的会员网站和mySQL数据库字段有ID、refereid和rootid字段要知道 谁邀请谁，谁推荐谁

当新成员注册时，在新成员记录中，我将她的邀请者成员id写在参考id中

反过来，A邀请B，B邀请C，C邀请D，D邀请E

E有正确的字段引用id作为数据库中D的id

我想要E的rootid字段中的A的id，它是原始的invitor 这就像是下线建设，但我不能正确地填写我的3个查询，其中需要invitor的refereid，并找到他们的rootid和设置在自己的rootid字段rootid字段

如果我能以某种方式保存邀请树和下线，那会更好

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困惑于如何从这里开始

如果我没有弄错的话，这就是你所描述的

CREATE TABLE members
(
  id INT UNSIGNED AUTO_INCREMENT NOT NULL,
  refer_id INT UNSIGNED,
  rootrefer_id INT UNSIGNED,
  name VARCHAR(100) NOT NULL,
  PRIMARY KEY(id)
);


/*Adam is an original member and was not reffered by anyone.*/
  INSERT INTO members (name) VALUES ('adam');

/*
Barbara is referred by Adam her id = 2
her refer_id, and rootrefer_id = 1
*/
  INSERT INTO members (name, refer_id, rootrefer_id) 
  VALUES('barbara', 1, 1);

/*
Cindy is referred by Barbara her id = 3, 
her refer_id = 2, her rootrefer_id = 1, just like Barbara
*/
  INSERT INTO members (name, refer_id, rootrefer_id)
  VALUES('cindy', 2, 1);

/*
Cindy in turn refers Dan, Dans id = 4
his refer_id = 3, and his rootrefer_id = 1
*/
  INSERT INTO members (name, refer_id, rootrefer_id)
  VALUES ('dan', 3, 1);

/*
Dan invites his pal Eddie, Eds id = 5
his refer_id = 4, and his rootrefer_id also = 1
*/
  INSERT INTO members (name, refer_id, rootrefer_id)
  VALUES ('eddie', 4, 1);

您可以这样查询：

#select original, non-refered members
#SELECT * FROM members WHERE refer_id IS NULL;

#select referred members
#SELECT * FROM members WHERE refer_id IS NOT NULL;

#select all referers and refered
SELECT 
  m1.id AS ref_id, 
  m1.name AS ref_name,
  m2.id AS id,
  m2.name AS name  
  FROM members m1
 JOIN members m2 ON m1.id=m2.refer_id;

#selct all root referers and refered
SELECT 
  m1.id AS root_id, 
  m1.name AS root_name,
  m2.id AS id,
  m2.name AS name  
  FROM members m1
 JOIN members m2 ON m1.id=m2.rootrefer_id;

#select all referers who are also referred, their referers and referred
SELECT
  mrr.id AS referrer_id,
  mrr.name AS referrer_name,
  m.id AS id,
  m.name AS name,
  mrd.id AS refered_id,
  mrd.name AS refered_name
  FROM members m
 JOIN members mrr ON mrr.id=m.refer_id
 JOIN members mrd ON m.id=mrd.refer_id;

---

试试看

你费了不少劲，谢谢你，但是根id不能随新成员一起插入，它要在以后处理时“找到”——让我再解释一下。亚当邀请了芭芭拉，芭芭拉邀请了辛迪，所以1，亚当，是第一张唱片，第二张是2，芭芭拉，1，第三张是3，辛迪，2，现在我们希望最后一张唱片是3，辛迪，2，1，当我们找到芭芭拉的父母时，从芭芭拉的唱片中取1。cindy的孩子也应该把1作为根id，因为这个链中的第一个邀请者是adam（id 1）。我将实现这一行：-它在数学上保持对父母和孩子的跟踪，直到第n个级别