多个日期行的按日期排序

Question

我正在尝试编写一个 sql 查询来合并行，因此结果将如下所示：

post_id	date_1	time_1	date_2	time_2
500	20220303	20:00:00
500			20220202	20:00:00

这是我到目前为止尝试过的方法：

SELECT 
    CASE meta_key WHEN"performance_date_1"THEN meta_value ELSE NULL END  AS date_1,
    CASE meta_key WHEN"performance_date_2"THEN meta_value ELSE NULL END  AS date_2
FROM test;

架构SQL:

CREATE TABLE test (
  meta_id bigint(20) AUTO_INCREMENT,
  post_id INT,
  meta_key varchar(255) NULL,
  meta_value longtext NULL,
  PRIMARY KEY (`meta_id`)
);
INSERT INTO test (post_id, meta_key, meta_value) VALUES (500, 'performance_date_1', '20220303');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (500, 'performance_time_1', '20:00:00');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (500, 'performance_date_2', '20220202');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (500, 'performance_time_2', '20:00:00');

INSERT INTO test (post_id, meta_key, meta_value) VALUES (501, 'performance_date_1', '20220403');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (501, 'performance_time_1', '20:00:00');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (501, 'performance_date_2', '20220407');
INSERT INTO test (post_id, meta_key, meta_value) VALUES (501, 'performance_time_2', '19:00:00');

SQL Fiddle https://www.db-fiddle.com/f/9Zxq4HJpuAiSPvabQzL4ux/1

如有任何帮助，我们将不胜感激。

Answer 1

您想使用条件聚合：

SELECT
    post_id,
    MAX(CASE meta_key WHEN 'performance_date_1' THEN meta_value END) AS date_1,
    MAX(CASE meta_key WHEN 'performance_time_1' THEN meta_value END) AS time_1,
    MAX(CASE meta_key WHEN 'performance_date_2' THEN meta_value END) AS date_2,
    MAX(CASE meta_key WHEN 'performance_time_2' THEN meta_value END) AS time_2
FROM test
GROUP BY
    post_id;

Answer 2

感谢 Tim Biegeleisen 让我走上正轨。我最终使用了您的查询作为子查询，然后 UNION ALL 我得到了我需要的结果。结果 table 看起来与我原来的问题略有不同，但对我有用。

查询SQL:

SELECT
    *
FROM 
  (SELECT
      post_id,
      MAX( CASE meta_key WHEN"performance_date_1"THEN meta_value ELSE NULL END ) AS new_date_1,
      MAX( CASE meta_key WHEN"performance_time_1"THEN meta_value ELSE NULL END ) AS new_time_1
  FROM test
  GROUP BY post_id
  UNION ALL
  SELECT
      post_id,
      MAX( CASE meta_key WHEN"performance_date_2"THEN meta_value ELSE NULL END ) AS new_date_1,
       MAX( CASE meta_key WHEN"performance_time_2"THEN meta_value ELSE NULL END ) AS new_time_1
  FROM test
  GROUP BY post_id) as T1
ORDER BY post_id

结果：

post_id	new_date	new_time
500	20220303	20:00:00
500	20220202	20:00:00
501	20220403	20:00:00
501	20220407	19:00:00

Answer 3

对于此示例数据，您可以将代码简化为：

SELECT post_id, 
       MAX(CASE WHEN SUBSTRING_INDEX(meta_key, '_', 2) = 'performance_date' THEN meta_value END) new_date_1,
       MAX(CASE WHEN SUBSTRING_INDEX(meta_key, '_', 2) = 'performance_time' THEN meta_value END) new_time_1
FROM test
GROUP BY post_id, SUBSTRING_INDEX(meta_key, '_', -1);

参见demo。

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