mysql在同一个选项卡上进行计数和求和的复杂查询

我正在使用下面的查询，它非常慢（大约需要30-45分钟），有时会卡住，生成500个错误，这对应用程序不好

SELECT id, account_number, email, referred_by, dnt,
                             (SELECT count(*)
                              FROM tbl.registrations t2
                              WHERE t2.referred_by = t1.account_number
                             ) AS level1,
                             (SELECT count(*)
                              FROM  tbl.registrations t2 join
                                    tbl.registrations t3
                                    on t3.referred_by = t2.account_number
                              WHERE t2.referred_by = t1.account_number
                             ) AS level2,
                             (SELECT count(*)
                              FROM tbl.registrations t2 join
                                    tbl.registrations t3
                                    on t3.referred_by = t2.account_number join
                                    tbl.registrations t4
                                    on t4.referred_by = t3.account_number
                              WHERE t2.referred_by = t1.account_number
                             ) AS level3
                          FROM tbl.registrations t1 GROUP BY id;

这个表中大约有35000行，数据库结构如下

| id | account_number  | referred_by  |
+----+-----------------+--------------+
|  1 | ac01            | 5            |
+----+-----------------+--------------+
|  2 | ac02            | 5            |
+----+-----------------+--------------+
|  3 | ac03            | 4            |
+----+-----------------+--------------+

1级、2级和3级的计算如下

total_referred - total number of members referred by account_number
total_reffered2 - total number of members that THEY(all accounts referred by account_number) all referred
total_reffered3 - total number of members that referred by all members of total_reffered2

(hierarchy for 1)      (hierarchy for 2)   (hierarchy for 3)
    1                      2                   3
       4                      13                  21
          5                       14              22
          6                       15                  23
             31               16                         39
             32               17                         40
          7                       18                  24
       8                              35              25
          9                           36              26
          10                          37          27
          11                          38              28
             33                   19                  29
             34                   20                  30
          12

比如说

如果每个会员都有10次推荐，那么

level1 = 10
level2 = 100
level3 = 1000

请检查我做错了什么？请帮我重写这个查询或优化，顺便说一句，我已经尝试添加索引了。谢谢

这是一个via的示例。我输入了一个包含40个条目的列表，其中对应的ID引用了这些条目，显示了以下的层次结构

total_referred - total number of members referred by account_number
total_reffered2 - total number of members that THEY(all accounts referred by account_number) all referred
total_reffered3 - total number of members that referred by all members of total_reffered2

(hierarchy for 1)      (hierarchy for 2)   (hierarchy for 3)
    1                      2                   3
       4                      13                  21
          5                       14              22
          6                       15                  23
             31               16                         39
             32               17                         40
          7                       18                  24
       8                              35              25
          9                           36              26
          10                          37          27
          11                          38              28
             33                   19                  29
             34                   20                  30
          12

SQLFiddle查询基于引用引用的任何可能列创建3列。如果一个ID没有被其ID引用的帐户，那么这些帐户将没有值，需要使用COALESCE进行解析以防止为空

select r.*,
       coalesce( presum.lvl1Cnt, 0 ) as Lvl1Total,
       coalesce( presum.lvl1cnt + presum.lvl2cnt, 0 ) as Lvl2Total,
       coalesce( presum.lvl1cnt + presum.lvl2cnt + presum.lvl3cnt, 0 ) as Lvl3Total
   from
      registrations r
         LEFT JOIN 
            ( SELECT 
                    r1.referred_by, 
                    count(distinct r1.id) as Lvl1Cnt,
                    count(distinct r2.id) as Lvl2Cnt,
                    count(distinct r3.id) as Lvl3Cnt
                 from 
                    registrations r1 
                       LEFT JOIN registrations r2 
                          ON r1.id = r2.referred_by 
                          LEFT JOIN registrations r3 
                             ON r2.id = r3.referred_by 
                 group by 
                    r1.referred_by ) presum
      ON r.id = presum.referred_by

---

presum内部查询在整个系统中执行一次。我只想确保您在（refered_，id）上的表上有和索引，以帮助优化按id值refered_的分组

如果

id

在您的注册表中是唯一的，那么

groupby

是不必要的。尝试删除它，看看它对性能有何影响。@GordonLinoff没什么我已经尝试过了，什么都没发生！此查询非常慢，并引发超时错误您是否尝试对语句运行EXPLAIN？只是想知道发生了什么（查询优化器是如何处理问题的），以及在哪里搜索优化。@Zagor23对不起，我没有尝试过，你能帮我做吗？@Zagor23要解释的输出吗