Mysql group_concat 的重复键和 1 个查询中多列的重复计数（查询优化）
Mysql group_concat of repeated keys and count of repetition of multiple columns in 1 query ( Query Optimization )
此问题与查询优化有关，以避免通过 PHP.
多次调用数据库
这是一个场景，我有两个 table，一个包含您可以称之为参考的信息 table，另一个是数据 table，字段 key1和key2在table中是通用的，根据这些字段，我们可以加入他们。
我不知道是否可以使查询比我现在做的更简单，我想实现的是：
I would like to find distinct key1,key2,info1,info2 from main_info
table, whenever serial value is less than 10 and key1,key2 of both
table matches, and then group them by info1,info2, while grouping
count the repeated key1,key2 for duplicates of info1,info2 fields
and group_concat those keys
tablemain_info的内容
MariaDB [demos]> select * from main_info;
+------+------+-------+-------+----------+
| key1 | key2 | info1 | info2 | date |
+------+------+-------+-------+----------+
| 1 | 1 | 15 | 90 | 20120501 |
| 1 | 2 | 14 | 92 | 20120601 |
| 1 | 3 | 15 | 82 | 20120801 |
| 1 | 4 | 15 | 82 | 20120801 |
| 1 | 5 | 15 | 82 | 20120802 |
| 2 | 1 | 17 | 90 | 20130302 |
| 2 | 2 | 17 | 90 | 20130302 |
| 2 | 3 | 17 | 90 | 20130302 |
| 2 | 4 | 16 | 88 | 20130601 |
+------+------+-------+-------+----------+
9 rows in set (0.00 sec)
tableproduct1
的内容
MariaDB [demos]> select * from product1;
+------+------+--------+--------------+
| key1 | key2 | serial | product_data |
+------+------+--------+--------------+
| 1 | 1 | 0 | NaN |
| 1 | 1 | 1 | NaN |
| 1 | 1 | 2 | NaN |
| 1 | 1 | 3 | NaN |
| 1 | 2 | 0 | 12.556 |
| 1 | 2 | 1 | 13.335 |
| 1 | 3 | 1 | NaN |
| 1 | 3 | 2 | 13.556 |
| 1 | 3 | 3 | 14.556 |
| 1 | 4 | 3 | NaN |
| 1 | 5 | 3 | NaN |
| 2 | 1 | 0 | 12.556 |
| 2 | 1 | 1 | 13.553 |
| 2 | 1 | 2 | NaN |
| 2 | 2 | 12 | 129 |
| 2 | 3 | 22 | NaN |
+------+------+--------+--------------+
16 rows in set (0.00 sec)
通过 PHP 我在当前上下文 serial 中将 table main_info 的 info1 和 info2 字段分组,table的product_dataproduct1,一个接一个的多次（这里我是运行查询了两次大家可以看到）
对于字段 serial - 第一个查询
MariaDB [demos]> select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
-> (
-> SELECT distinct
-> if(b.serial < 10,a.key1,null) AS `key1`,
-> if(b.serial < 10,a.key2,null) AS `key2`,
-> if(b.serial < 10,a.info1,null) AS `info1`,
-> if(b.serial < 10,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ;
+------+------+-------+-------+--------------+-------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids |
+------+------+-------+-------+--------------+-------------+
| NULL | NULL | NULL | NULL | 1 | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 |
+------+------+-------+-------+--------------+-------------+
5 rows in set (0.00 sec)
对于字段 product_data - 第二个查询
MariaDB [demos]> select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
-> (
-> SELECT distinct
-> if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
-> if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
-> if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
-> if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ;
+------+------+-------+-------+--------------------+------------------+
| key1 | key2 | info1 | info2 | product_data_count | product_data_ids |
+------+------+-------+-------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 |
| 2 | 2 | 17 | 90 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------------+------------------+
4 rows in set (0.01 sec)
我想使用一个查询得到这样的输出，Group by info1, info2
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
下面是tables
的结构
DROP TABLE IF EXISTS `main_info`;
CREATE TABLE `main_info` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`info1` int(11) NOT NULL,
`info2` int(11) NOT NULL,
`date` int(11) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `main_info` WRITE;
INSERT INTO `main_info` VALUES (1,1,15,90,20120501),(1,2,14,92,20120601),(1,3,15,82,20120801),(1,4,15,82,20120801),(1,5,15,82,20120802),(2,1,17,90,20130302),(2,2,17,90,20130302),(2,3,17,90,20130302),(2,4,16,88,20130601);
UNLOCK TABLES;
DROP TABLE IF EXISTS `product1`;
CREATE TABLE `product1` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`serial` int(11) NOT NULL,
`product_data` varchar(1000) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `product1` WRITE;
INSERT INTO `product1` VALUES (1,1,0,'NaN'),(1,1,1,'NaN'),(1,1,2,'NaN'),(1,1,3,'NaN'),(1,2,0,'12.556'),(1,2,1,'13.335'),(1,3,1,'NaN'),(1,3,2,'13.556'),(1,3,3,'14.556'),(1,4,3,'NaN'),(1,5,3,'NaN'),(2,1,0,'12.556'),(2,1,1,'13.553'),(2,1,2,'NaN'),(2,2,12,'129'),(2,3,22,'NaN');
UNLOCK TABLES;
有人请帮我在一个查询中得到结果。
如何将您的两个查询与 JOIN 结合起来？
SQL:
SELECT
tbl1.key1, tbl1.key2, tbl1.info1, tbl1.info2, tbl1.serial_count, tbl1.serial_ids,
tbl2.product_data_count, tbl2.product_data_ids
FROM
(
select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
(
SELECT distinct
if(b.serial < 10,a.key1,null) AS `key1`,
if(b.serial < 10,a.key2,null) AS `key2`,
if(b.serial < 10,a.info1,null) AS `info1`,
if(b.serial < 10,a.info2,null) AS `info2`
FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
) as sub group by info1,info2
) tbl1
LEFT OUTER JOIN
(
select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
(
SELECT distinct
if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
) as sub group by info1,info2
) tbl2
ON tbl1.info1 = tbl2.info1 AND tbl1.info2 = tbl2.info2
ORDER BY 3,4
;
输出：
mysql> SELECT
-> tbl1.key1, tbl1.key2, tbl1.info1, tbl1.info2, tbl1.serial_count, tbl1.serial_ids,
-> tbl2.product_data_count, tbl2.product_data_ids
-> FROM
-> (
-> select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
-> (
-> SELECT distinct
-> if(b.serial < 10,a.key1,null) AS `key1`,
-> if(b.serial < 10,a.key2,null) AS `key2`,
-> if(b.serial < 10,a.info1,null) AS `info1`,
-> if(b.serial < 10,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ) tbl1
-> LEFT OUTER JOIN
-> (
-> select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
-> (
-> SELECT distinct
-> if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
-> if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
-> if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
-> if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ) tbl2
-> ON tbl1.info1 = tbl2.info1 AND tbl1.info2 = tbl2.info2
-> ORDER BY 3,4
-> ;
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
5 rows in set (0.01 sec)
mysql> select version();
+-----------------+
| version() |
+-----------------+
| 10.1.10-MariaDB |
+-----------------+
1 row in set (0.00 sec)
试试这个
SELECT
key1, key2, info1, info2,
SUM(Scount) AS serial_count, GROUP_CONCAT(Skey1, ' ', Skey2) AS serial_ids,
SUM(Pcount) AS product_data_count, GROUP_CONCAT(Pkey1, ' ', Pkey2) AS product_data_ids
FROM
(
SELECT DISTINCT
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.key1, NULL) AS `key1`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.key2, NULL) AS `key2`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.info1, NULL) AS `info1`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.info2, NULL) AS `info2`,
IF(b.serial < 10,a.key1, NULL) AS `Skey1`,
IF(b.serial < 10,a.key2, NULL) AS `Skey2`,
IF(b.product_data IS NOT NULL,a.key1, NULL) AS `Pkey1`,
IF(b.product_data IS NOT NULL,a.key2, NULL) AS `Pkey2`,
IF(b.serial < 10, 1, NULL) AS `Scount`,
IF(b.product_data IS NOT NULL, 1, NULL) AS `Pcount`
FROM main_info a INNER JOIN product1 b ON a.key1 = b.key1 AND a.key2= b.key2
UNION ALL
SELECT DISTINCT
NULL AS `key1`,
NULL AS `key2`,
NULL AS `info1`,
NULL AS `info2`,
NULL AS `Skey1`,
NULL AS `Skey2`,
NULL AS `Pkey1`,
NULL AS `Pkey2`,
IF(serial > 9, 1, NULL) AS `Scount`,
IF(product_data IS NULL, 1, NULL) AS `Pcount`
FROM product1 WHERE serial > 9 xor product_data IS NULL
) AS sub GROUP BY info1,info2
结果（来自问题的数据）
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
结果（评论中的数据）
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | 1 | NULL |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 2 | 4 | 16 | 88 | 1 | 2 4 | 1 | 2 4 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 2 | 1 | 17 | 90 | NULL | NULL | 3 | 2 1,2 2,2 3 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
注意：
问题背后的基本逻辑我确实能理解，所以主要根据预期结果回答。比如组字段（info1和info2）为null，除了serial_count和product_data_count可以为1或null之外，其他结果将始终为null，你真的吗意思是得到那个？请注意，此答案使用另一个带有 UNION ALL 的子查询来满足这一点。
从你的引述来看，在我看来你想做这样的事情 (SQLfiddle):
SELECT
m.info1,
m.info2,
COUNT(DISTINCT CONCAT(m.key1, ' ', m.key2)) key_count,
GROUP_CONCAT(DISTINCT CONCAT(m.key1, ' ', m.key2) ORDER BY m.key1, m.key2) key_pairs,
COUNT(DISTINCT p.serial) serial_count,
GROUP_CONCAT(DISTINCT p.serial ORDER BY p.serial) serials,
COUNT(DISTINCT p.product_data) data_count,
GROUP_CONCAT(DISTINCT p.product_data ORDER BY p.product_data) product_data
FROM
main_info m INNER JOIN
product1 p ON p.key1 = m.key1 AND p.key2 = m.key2
WHERE
p.serial < 10
GROUP BY
m.info1,
m.info2
计算不同的值并列出它们，这样正确吗？您不能只按 info1、info2 分组，结果中还包含 key1 或 key2 的列（例如 min(key1) 或 max(key2) 会起作用）。我在上面的查询中对此进行了调整，虽然它与您的结果有很大不同，但它可能是您实际需要的，可能有一些变化。
此问题与查询优化有关，以避免通过 PHP.
多次调用数据库这是一个场景，我有两个 table，一个包含您可以称之为参考的信息 table，另一个是数据 table，字段 key1和key2在table中是通用的，根据这些字段，我们可以加入他们。
我不知道是否可以使查询比我现在做的更简单，我想实现的是：
I would like to find distinct
key1,key2,info1,info2frommain_infotable, whenever serial value is less than 10 andkey1,key2of both table matches, and then group them byinfo1,info2, while grouping count the repeatedkey1,key2for duplicates ofinfo1,info2fields andgroup_concatthose keys
tablemain_info的内容
MariaDB [demos]> select * from main_info;
+------+------+-------+-------+----------+
| key1 | key2 | info1 | info2 | date |
+------+------+-------+-------+----------+
| 1 | 1 | 15 | 90 | 20120501 |
| 1 | 2 | 14 | 92 | 20120601 |
| 1 | 3 | 15 | 82 | 20120801 |
| 1 | 4 | 15 | 82 | 20120801 |
| 1 | 5 | 15 | 82 | 20120802 |
| 2 | 1 | 17 | 90 | 20130302 |
| 2 | 2 | 17 | 90 | 20130302 |
| 2 | 3 | 17 | 90 | 20130302 |
| 2 | 4 | 16 | 88 | 20130601 |
+------+------+-------+-------+----------+
9 rows in set (0.00 sec)
tableproduct1
MariaDB [demos]> select * from product1;
+------+------+--------+--------------+
| key1 | key2 | serial | product_data |
+------+------+--------+--------------+
| 1 | 1 | 0 | NaN |
| 1 | 1 | 1 | NaN |
| 1 | 1 | 2 | NaN |
| 1 | 1 | 3 | NaN |
| 1 | 2 | 0 | 12.556 |
| 1 | 2 | 1 | 13.335 |
| 1 | 3 | 1 | NaN |
| 1 | 3 | 2 | 13.556 |
| 1 | 3 | 3 | 14.556 |
| 1 | 4 | 3 | NaN |
| 1 | 5 | 3 | NaN |
| 2 | 1 | 0 | 12.556 |
| 2 | 1 | 1 | 13.553 |
| 2 | 1 | 2 | NaN |
| 2 | 2 | 12 | 129 |
| 2 | 3 | 22 | NaN |
+------+------+--------+--------------+
16 rows in set (0.00 sec)
通过 PHP 我在当前上下文 serial 中将 table main_info 的 info1 和 info2 字段分组,table的product_dataproduct1,一个接一个的多次（这里我是运行查询了两次大家可以看到）
对于字段 serial - 第一个查询
MariaDB [demos]> select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
-> (
-> SELECT distinct
-> if(b.serial < 10,a.key1,null) AS `key1`,
-> if(b.serial < 10,a.key2,null) AS `key2`,
-> if(b.serial < 10,a.info1,null) AS `info1`,
-> if(b.serial < 10,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ;
+------+------+-------+-------+--------------+-------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids |
+------+------+-------+-------+--------------+-------------+
| NULL | NULL | NULL | NULL | 1 | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 |
+------+------+-------+-------+--------------+-------------+
5 rows in set (0.00 sec)
对于字段 product_data - 第二个查询
MariaDB [demos]> select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
-> (
-> SELECT distinct
-> if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
-> if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
-> if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
-> if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ;
+------+------+-------+-------+--------------------+------------------+
| key1 | key2 | info1 | info2 | product_data_count | product_data_ids |
+------+------+-------+-------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 |
| 2 | 2 | 17 | 90 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------------+------------------+
4 rows in set (0.01 sec)
我想使用一个查询得到这样的输出，Group by info1, info2
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
下面是tables
的结构DROP TABLE IF EXISTS `main_info`;
CREATE TABLE `main_info` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`info1` int(11) NOT NULL,
`info2` int(11) NOT NULL,
`date` int(11) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `main_info` WRITE;
INSERT INTO `main_info` VALUES (1,1,15,90,20120501),(1,2,14,92,20120601),(1,3,15,82,20120801),(1,4,15,82,20120801),(1,5,15,82,20120802),(2,1,17,90,20130302),(2,2,17,90,20130302),(2,3,17,90,20130302),(2,4,16,88,20130601);
UNLOCK TABLES;
DROP TABLE IF EXISTS `product1`;
CREATE TABLE `product1` (
`key1` int(11) NOT NULL,
`key2` int(11) NOT NULL,
`serial` int(11) NOT NULL,
`product_data` varchar(1000) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
LOCK TABLES `product1` WRITE;
INSERT INTO `product1` VALUES (1,1,0,'NaN'),(1,1,1,'NaN'),(1,1,2,'NaN'),(1,1,3,'NaN'),(1,2,0,'12.556'),(1,2,1,'13.335'),(1,3,1,'NaN'),(1,3,2,'13.556'),(1,3,3,'14.556'),(1,4,3,'NaN'),(1,5,3,'NaN'),(2,1,0,'12.556'),(2,1,1,'13.553'),(2,1,2,'NaN'),(2,2,12,'129'),(2,3,22,'NaN');
UNLOCK TABLES;
有人请帮我在一个查询中得到结果。
如何将您的两个查询与 JOIN 结合起来？
SQL:
SELECT
tbl1.key1, tbl1.key2, tbl1.info1, tbl1.info2, tbl1.serial_count, tbl1.serial_ids,
tbl2.product_data_count, tbl2.product_data_ids
FROM
(
select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
(
SELECT distinct
if(b.serial < 10,a.key1,null) AS `key1`,
if(b.serial < 10,a.key2,null) AS `key2`,
if(b.serial < 10,a.info1,null) AS `info1`,
if(b.serial < 10,a.info2,null) AS `info2`
FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
) as sub group by info1,info2
) tbl1
LEFT OUTER JOIN
(
select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
(
SELECT distinct
if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
) as sub group by info1,info2
) tbl2
ON tbl1.info1 = tbl2.info1 AND tbl1.info2 = tbl2.info2
ORDER BY 3,4
;
输出：
mysql> SELECT
-> tbl1.key1, tbl1.key2, tbl1.info1, tbl1.info2, tbl1.serial_count, tbl1.serial_ids,
-> tbl2.product_data_count, tbl2.product_data_ids
-> FROM
-> (
-> select * , count(*) as serial_count,GROUP_CONCAT(key1,' ',key2) as serial_ids from
-> (
-> SELECT distinct
-> if(b.serial < 10,a.key1,null) AS `key1`,
-> if(b.serial < 10,a.key2,null) AS `key2`,
-> if(b.serial < 10,a.info1,null) AS `info1`,
-> if(b.serial < 10,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ) tbl1
-> LEFT OUTER JOIN
-> (
-> select * , count(*) as product_data_count,GROUP_CONCAT(key1,' ',key2) as product_data_ids from
-> (
-> SELECT distinct
-> if(b.product_data IS NOT NULL,a.key1,null) AS `key1`,
-> if(b.product_data IS NOT NULL,a.key2,null) AS `key2`,
-> if(b.product_data IS NOT NULL,a.info1,null) AS `info1`,
-> if(b.product_data IS NOT NULL,a.info2,null) AS `info2`
-> FROM main_info a inner join product1 b on a.key1 = b.key1 AND a.key2= b.key2
-> ) as sub group by info1,info2
-> ) tbl2
-> ON tbl1.info1 = tbl2.info1 AND tbl1.info2 = tbl2.info2
-> ORDER BY 3,4
-> ;
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
| 2 | 1 | 17 | 90 | 1 | 2 1 | 3 | 2 2,2 3,2 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
5 rows in set (0.01 sec)
mysql> select version();
+-----------------+
| version() |
+-----------------+
| 10.1.10-MariaDB |
+-----------------+
1 row in set (0.00 sec)
试试这个
SELECT
key1, key2, info1, info2,
SUM(Scount) AS serial_count, GROUP_CONCAT(Skey1, ' ', Skey2) AS serial_ids,
SUM(Pcount) AS product_data_count, GROUP_CONCAT(Pkey1, ' ', Pkey2) AS product_data_ids
FROM
(
SELECT DISTINCT
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.key1, NULL) AS `key1`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.key2, NULL) AS `key2`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.info1, NULL) AS `info1`,
IF(b.serial < 10 OR b.product_data IS NOT NULL,a.info2, NULL) AS `info2`,
IF(b.serial < 10,a.key1, NULL) AS `Skey1`,
IF(b.serial < 10,a.key2, NULL) AS `Skey2`,
IF(b.product_data IS NOT NULL,a.key1, NULL) AS `Pkey1`,
IF(b.product_data IS NOT NULL,a.key2, NULL) AS `Pkey2`,
IF(b.serial < 10, 1, NULL) AS `Scount`,
IF(b.product_data IS NOT NULL, 1, NULL) AS `Pcount`
FROM main_info a INNER JOIN product1 b ON a.key1 = b.key1 AND a.key2= b.key2
UNION ALL
SELECT DISTINCT
NULL AS `key1`,
NULL AS `key2`,
NULL AS `info1`,
NULL AS `info2`,
NULL AS `Skey1`,
NULL AS `Skey2`,
NULL AS `Pkey1`,
NULL AS `Pkey2`,
IF(serial > 9, 1, NULL) AS `Scount`,
IF(product_data IS NULL, 1, NULL) AS `Pcount`
FROM product1 WHERE serial > 9 xor product_data IS NULL
) AS sub GROUP BY info1,info2
结果（来自问题的数据）
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | NULL | NULL |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
结果（评论中的数据）
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| key1 | key2 | info1 | info2 | serial_count | serial_ids | product_data_count | product_data_ids |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| NULL | NULL | NULL | NULL | 1 | NULL | 1 | NULL |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 2 | 14 | 92 | 1 | 1 2 | 1 | 1 2 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 3 | 15 | 82 | 3 | 1 3,1 4,1 5 | 3 | 1 3,1 4,1 5 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 1 | 1 | 15 | 90 | 1 | 1 1 | 1 | 1 1 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 2 | 4 | 16 | 88 | 1 | 2 4 | 1 | 2 4 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 2 | 1 | 17 | 90 | NULL | NULL | 3 | 2 1,2 2,2 3 |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+
注意：
问题背后的基本逻辑我确实能理解，所以主要根据预期结果回答。比如组字段（info1和info2）为null，除了serial_count和product_data_count可以为1或null之外，其他结果将始终为null，你真的吗意思是得到那个？请注意，此答案使用另一个带有 UNION ALL 的子查询来满足这一点。
从你的引述来看，在我看来你想做这样的事情 (SQLfiddle):
SELECT
m.info1,
m.info2,
COUNT(DISTINCT CONCAT(m.key1, ' ', m.key2)) key_count,
GROUP_CONCAT(DISTINCT CONCAT(m.key1, ' ', m.key2) ORDER BY m.key1, m.key2) key_pairs,
COUNT(DISTINCT p.serial) serial_count,
GROUP_CONCAT(DISTINCT p.serial ORDER BY p.serial) serials,
COUNT(DISTINCT p.product_data) data_count,
GROUP_CONCAT(DISTINCT p.product_data ORDER BY p.product_data) product_data
FROM
main_info m INNER JOIN
product1 p ON p.key1 = m.key1 AND p.key2 = m.key2
WHERE
p.serial < 10
GROUP BY
m.info1,
m.info2
计算不同的值并列出它们，这样正确吗？您不能只按 info1、info2 分组，结果中还包含 key1 或 key2 的列（例如 min(key1) 或 max(key2) 会起作用）。我在上面的查询中对此进行了调整，虽然它与您的结果有很大不同，但它可能是您实际需要的，可能有一些变化。