Mysql 计数不同的+；通过具有（值）和#x2B对组进行计数；按月分组

我有一个包含以下列的表：cid，date

示例表数据：注意：cid包含字符串值，例如：“otsytb8o7sbs50w9doghwzvfy0vb8f9h”许多值是重复的

 cid. date  
--------------------------------------------------------
1   2015-10-10 04:57:57
2   2015-10-10 05:03:58
3   2015-10-10 05:24:49
4   2015-10-10 05:28:24
5   2015-10-10 05:28:26
6   2015-10-10 05:28:40
7   2015-10-10 05:30:39
8   2015-10-10 05:33:04
9   2015-10-10 05:35:42
9   2015-10-10 05:36:03

我想得到以下信息：

独立

cid

作为UNIQQ访问的计数

---

具有的

cid的计数（计数），因此我已解决此问题：

SELECT uniqVisitors, COUNT(*) AS bounced, T1.month
FROM (
    SELECT cid,
    SUBSTR(DATE(date),1,7) AS month
    FROM table
    GROUP BY cid
    HAVING COUNT(*) <= 1
) T1
LEFT JOIN
( SELECT count( DISTINCT(cid) ) AS uniqVisitors,
    SUBSTR(DATE(date),1,7) AS month
    FROM table
    GROUP By month ) T2
ON T1.month = T2.month
GROUP BY month

只是澄清一下：请添加准确的输入数据，以生成您要查找的准确的输出数据。例如，无法从该输入中获取7345。@SamiKuhmonen这只是一个计数，表中有100000个值，您的请求是不可行的。此外，有人编辑了我的实际值，但这与我们的要求无关e仅计算包含重复数据的
cid
的数量，即相同的
cid
字符串value@Grant这是非常可行的。您已经提供了一个输入。现在从该数据中提供您想要的输出，而不是从其他数据中。不需要成百上千万行的真实数据。只需要一些输入和相应的输出这表明了您的需要。此请求是为了验证我们的尝试。如果我们无法验证，我们无法为您提供准确的答案。这就是为什么您需要提供问题中提到的示例数据的所需输出。输出必须等于“所需结果”：在我的问题表中，这是我想要的输出：uniqVisits=2345，Bounched=325，month=2015-10/行。
SELECT count(*) AS bounced,
( SELECT count( DISTINCT(cid) ) AS uniqVisits,
    SUBSTR(DATE(date),1,7) AS month
    FROM table ) AS uniqVisits
FROM (
   SELECT COUNT(cid) AS bounced,
   SUBSTR(DATE(date),1,7) AS month
   FROM table
   GROUP BY cid
   HAVING (count <= 1)
) AS x
GROUP BY month

SELECT c1.uniqVisits,
        c1.month,
        c2.bounced
    FROM ( SELECT COUNT(DISTINCT t1.cid)    AS `uniqVisits`,
SUBSTR(DATE(t1.date),1,7) AS `month`
FROM table t1
GROUP BY month
        ) c1
    JOIN ( SELECT COUNT(*)         AS `bounced`,
SUBSTR(DATE(t2.date),1,7) AS `month`
FROM table t2
GROUP BY month, cid HAVING (count <= 1)
        ) c2
    ON c2.month = c1.month
    ORDER BY c1.month

SELECT uniqVisitors, COUNT(*) AS bounced, T1.month
FROM (
    SELECT cid,
    SUBSTR(DATE(date),1,7) AS month
    FROM table
    GROUP BY cid
    HAVING COUNT(*) <= 1
) T1
LEFT JOIN
( SELECT count( DISTINCT(cid) ) AS uniqVisitors,
    SUBSTR(DATE(date),1,7) AS month
    FROM table
    GROUP By month ) T2
ON T1.month = T2.month
GROUP BY month

uniqVisitors | bounced | month  
---------------------------------
7237    6822    2015-10
12597   12136   2015-11
12980   12573   2015-12
12091   11695   2016-01
5396    5134    2016-02