Mysql group_concat의 반복 키 및 1개의 쿼리에서 여러 열의 반복 횟수(Query Optimization)

Mysql group_concat의 반복 키 및 1개의 쿼리에서 여러 열의 반복 횟수(Query Optimization)

이 질문은 PHP를 통한 데이터베이스 호출을 피하기 위한 쿼리 최적화에 관한 것입니다.

이 시나리오에는 2개의 테이블이 있습니다.하나는 참조 테이블이라고 부를 수 있는 정보가 포함되어 있고 다른 하나는 데이터 테이블, 필드입니다.key1그리고.key2두 표에서 공통으로 볼 수 있습니다.이러한 필드에 근거하여 이들 필드를 결합할 수 있습니다.

제가 지금 하고 있는 것보다 더 간단하게 질의를 할 수 있을지 모르겠습니다.제가 달성하고 싶은 것은 다음과 같습니다.

나는 다른 것을 찾고 싶다key1,key2,info1,info2부터main_info테이블, 시리얼 값이 10 미만이고key1,key2두 테이블이 일치하는지 확인한 다음, 다음으로 그룹화합니다.info1,info2그룹화는 반복 횟수를 카운트합니다.key1,key2을 복제하기 위해info1,info2필드 및group_concat그 열쇠들

표 내용main_info

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) 

표 내용product1

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)

PHP I 그룹 필드 경유info1그리고.info2테이블의main_info, 현재 컨텍스트에서serial,product_data테이블의product1, 여러 번 연속해서 (여기서는 보시는 바와 같이 두 번 쿼리를 실행하고 있습니다)

필드용serial- 첫 번째 쿼리

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)

필드용product_data- 두 번째 질문

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)

Group by info1, info2라는 하나의 쿼리를 사용하여 이와 같이 출력하고 싶습니다.

+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 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      |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+

아래는 테이블의 구조입니다.

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;

한 번의 질문으로 결과를 얻을 수 있도록 도와주세요.

두 개의 쿼리를 JOIN과 결합하는 것은 어떻습니까?

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
 ;

출력:

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)

이거 드셔보세요.

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

결과(질문의 데이터)

+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 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              |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+

RESULT(코멘트

+------+------+-------+-------+--------------+-------------+--------------------+------------------+
| 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      |
+------+------+-------+-------+--------------+-------------+--------------------+------------------+

주의:

질문의 이면에 있는 기본 논리에 대해 정말 이해할 수 있는 부분이 있기 때문에, 주로 예상한 결과에 근거해 대답하십시오.예를 들어 if group 필드(info1그리고.info2)는 null입니다.다른 결과는 항상 null입니다.serial_count그리고.product_data_count1이 될 수도 있고 0이 될 수도 있는데, 정말 그걸 받으려고 한 거야?이 답변에서는 다음 서브쿼리가 사용됩니다.UNION ALL그걸 만족시키기 위해서요.

당신의 견적을 보면 다음과 같은 일을 하고 싶은 것 같습니다(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

고유한 값을 세어 나열하는 것이 맞습니까?info1, info2로만 그룹화할 수 없으며, 결과에서 key1 또는 key2에 대한 열도 가질 수 있습니다(예: min(key1) 또는 max(key2)가 작동합니다).위의 문의를 통해 조정했습니다만, 당신의 결과와는 상당히 다르지만, 몇 가지 변경이 있을 수 있습니다.

언급URL : https://stackoverflow.com/questions/35538092/mysql-group-concat-of-repeated-keys-and-count-of-repetition-of-multiple-columns