Php Mysql组通过在列中显示总计值，然后再计算值，如何停止此操作？

我正在处理一个查询，以在多个列中找到重复的值，因此我将首先关注查询的单个部分，以获得更好的解释

在一天结束时，我需要知道的是，这4列中是否有一个重复，以及复制的列在哪个列中

以下是单个查询：

select  count(*) as cnt, 'CUST_REF' as what_column
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_ref having count(cust_ref) > 1;

所以这很有效，除了输出是2行。看起来第一行是列中点击数>1的总数，然后下一行是实际重复数，如下所示：

cnt what_column
9440    CUST_REF
2   CUST_REF

我的问题是，在没有列总数的情况下，如何才能得到第二行？（此列的值2是正确的）即，我只想要：

cnt what_column    
2   CUST_REF

综合起来：

我将所有这些与一个

联合体

放在一起，因此对于4列，它是这样的：

select  count(*) as cnt, 'CUST_REF' as what_column
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_ref having count(cust_ref) > 1
 union
 select  count(*) as cnt, 'CUST_PO' as what_column
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_po having count(cust_po) > 1
  union
 select count(*) as cnt, 'SHIP_BL' as what_column
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by ship_bl having count(ship_bl) > 1
  union
 select count(*) as cnt, 'CUST_SHIPID' as what_column
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_shipid having count(cust_shipid) > 1;

它的输出呈现了下面的内容，我想把所有显示重复项的字段分组在一起，同时去掉总计数

cnt what_column
9440    CUST_REF
2   CUST_REF
332 CUST_PO
3   CUST_PO
2   CUST_PO
8   CUST_PO
4   CUST_PO
9   CUST_PO
37  CUST_PO
6   CUST_PO
5   CUST_PO
7   CUST_PO
11  CUST_PO
6609    SHIP_BL
2   SHIP_BL
5   SHIP_BL
8   SHIP_BL
3   SHIP_BL
4   SHIP_BL
6   SHIP_BL
7   SHIP_BL
9183    CUST_SHIPID
2   CUST_SHIPID
3   CUST_SHIPID
6   CUST_SHIPID

同样，在一天结束时，我需要知道的是，这4列中的任何一列都有一个副本，以及该副本所在的列

对于下面的注释，我无法共享表数据。但是让我们这样看，在将列添加回

HAVING

中的select之后：

select cust_ref as val, count(*) as cnt, 'CUST_REF' as what_column
     from sometable 
      where status != 'whateverStatus' 
        and custm_id = 1234
     group by cust_ref having count(cust_ref) > 1;

HAVING

中的所有列名都是此表中的实际列名，

what_列

只是一个别名，它显示了在哪个列/查询中找到了dupe

假设数据是这样的，我已经用*标记了前两列中的重复项。我希望它能让他们大胆：

id | cust_ref | cust_po | ship_bl |cust_shipid
997| **1234** | 9656    | 5656    | 9876
998| **1234** | **6353**| 2436    | 9394
999| 4327     | **6353**| 4388    | 4353

我很确定我最终会：

val cnt what_column
      3 CUST_REF
1234  2 CUST_REF

---

希望有帮助

您已经找到了重复项。因此，如果只需要不包含cnt列的列，则执行子查询：

select distinct what_column 
 from (
select  count(*) as cnt, 'CUST_REF' as what_column
from sometable 
 where status != 'whateverStatus' 
 and custm_id = 1234
group by cust_ref having count(cust_ref) > 1
union
 select  count(*) as cnt, 'CUST_PO' as what_column
 from sometable 
 where status != 'whateverStatus' 
  and custm_id = 1234
 group by cust_po having count(cust_po) > 1
union
 select count(*) as cnt, 'SHIP_BL' as what_column
from sometable 
 where status != 'whateverStatus' 
and custm_id = 1234
 group by ship_bl having count(ship_bl) > 1
union
select count(*) as cnt, 'CUST_SHIPID' as what_column
  from sometable 
where status != 'whateverStatus' 
and custm_id = 1234
group by cust_shipid having count(cust_shipid) > 1);

---

您对似乎是一个非常简单的问题的解释非常复杂，并且您还没有清楚地解释您想要计算为“重复”的内容-您想要一个值出现多次的总记录计数，还是一个值出现多次的计数

您将重复值的计数与域的计数混淆，从而进一步混淆了问题-恰好查询输出中的第二行是2-这不是您要查找的值，只是碰巧是相同的基数

此列的值2是正确的

这表明你想要后者。在这种情况下，因为：

select  cust_ref, count(*) as cnt, 'CUST_REF' as what_column
from sometable 
where status != 'whateverStatus' 
   and custm_id = 1234
group by cust_ref having count(cust_ref) > 1;

将给出前者，您只需要计算该查询输出的行数。您可以通过两种方式执行此操作：

SELECT COUNT(*) AS number_of_values_in_more_than_row, what_column
FROM (
   select  count(*) as cnt, 'CUST_REF' as what_column, cust_ref
   from sometable 
   where status != 'whateverStatus' 
      and custm_id = 1234
   group by cust_ref 
   having count(cust_ref) > 1
)
GROUP BY what_column

……或

select  count(DISTINCT cust_ref) as cnt, 'CUST_REF' as what_column
from sometable 
where status != 'whateverStatus' 
    and custm_id = 1234
group by cust_ref 
having count(DISTINCT cust_ref) > 1;

---

最终有效的答案是在外部查询中使用having子句，这将返回所需的正确数字：

SELECT sum(cnt) as dupes, COUNT(*) AS number_of_values_in_more_than_row, what_column
  FROM (
select  count(*) as cnt, 'CUST_REF' as what_column,cust_ref
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_ref having count(cust_ref) > 1
 union
 select  count(*) as cnt, 'CUST_PO' as what_column,cust_po
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_po having count(cust_po) > 1
  union
 select count(*) as cnt, 'SHIP_BL' as what_column,ship_bl
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by ship_bl having count(ship_bl) > 1
  union
 select count(*) as cnt, 'CUST_SHIPID' as what_column,cust_shipid
 from sometable 
  where status != 'whateverStatus' 
    and custm_id = 1234
 group by cust_shipid having count(cust_shipid) > 1
 )x
 GROUP BY what_column having count(number_of_values_in_more_than_row) >0;

---

您应该为表架构添加一个清晰的数据样本和预期结果。您可以显示样本数据和表架构吗？我不明白你在这里想要实现什么。如果显示实际列值，

选择cust\u ref，count（*）作为cnt，

，您会得到什么？您是按未选择的字段分组的-这不是一件好事。请尝试在

选择sum（cnt），what\u column from（…）中将您的联合包装为按what\u column

分组的部分，你可能会发现每一行都有不同的cust_ref（例如），这将汇总所有特定类型的总数。伙计们，在阅读了你的评论之后，我用一些额外的信息更新了这个问题。希望对您有所帮助。因此，这将返回每个what_列。我怎么知道它是正确的？如果我在select子句中添加任何其他内容，我将返回额外的行，但是我如何确保它不会返回，因为值是9442，而不是2？您首先必须确定除了异常值之外，您可以获得的最多重复数。然后按计数过滤。因此，当count<11时（如果11是不包括异常值的最大重复数），您的上一个代码段没有产生任何结果，但上面的较大代码段给出的值比第94列CUST\U PO 2 CUST\U REF 10 CUST\U SHIPID 26 SHIP\U BL中的值多-这看起来正确！实际上，这并没有返回正确的结果，如果我从数据集中删除cust_ref的所有值（其中有1组重复项和1个唯一编号），因此将cust_ref字段中的所有值都置为空，则查询仍返回cust_ref 1，就像这个数字一样，列94 CUST\u PO 1 CUST\u REF 10 CUST\u SHIPID 26 SHIP\u BL中的值比行多，为什么它显示1，而列中根本没有值？