结果列名称中的日期答案

【问题标题】：dates in result column name结果列名称中的日期
【发布时间】：2022-01-20 07:00:21
【问题描述】：

亲爱的，你们好。请帮我查询数据库。

架构

架构 (MySQL v8.0)

create table TEST_TABLE (
    DT date,
    OBJECT_ID NUMERIC,
    PARAM_VALUE NUMERIC
);

insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-01',1, 21);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-02',1, 34);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-03',1, 78);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-04',1, 11);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-05',1, 26);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-06',1, 11);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-07',1, 12);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-08',1, 74);

insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-01',2, 18);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-02',2, 96);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-03',2, 33);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-04',2, 77);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-05',2, 55);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-06',2, 28);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-07',2, 54);
insert TEST_TABLE (DT, OBJECT_ID, PARAM_VALUE) values ('2022-01-08',2, 16);

查询 #1

select * from TEST_TABLE;

DT	OBJECT_ID	PARAM_VALUE
2022-01-01	1	21
2022-01-02	1	34
2022-01-03	1	78
2022-01-04	1	11
2022-01-05	1	26
2022-01-06	1	11
2022-01-07	1	12
2022-01-08	1	74
2022-01-01	2	18
2022-01-02	2	96
2022-01-03	2	33
2022-01-04	2	77
2022-01-05	2	55
2022-01-06	2	28
2022-01-07	2	54
2022-01-08	2	16

查询 #2

set @TARGET_DATE = '2022-01-08';

select
OBJECT_ID,
group_concat(if(DT = DATE_ADD(@TARGET_DATE, interval -1 day) , PARAM_VALUE, null)) as `dt-1`,
group_concat(if(DT = DATE_ADD(@TARGET_DATE, interval -2 day) , PARAM_VALUE, null)) as `dt-2`
from TEST_TABLE group by OBJECT_ID;

OBJECT_ID	dt-1	dt-2
1	12	11
2	54	28

我想收到列名中包含日期的结果

OBJECT_ID	2022-01-07	2022-01-06
1	12	11
2	54	28

对不起我的英语

【问题讨论】：

我想收到列名中包含日期的结果研究Prepared Statements.

标签： mysql sql

【解决方案1】：

正如Akina所说，准备好的声明：

SET @TARGET_DATE = '2022-01-08';

SELECT GROUP_CONCAT(DISTINCT CONCAT(
        'GROUP_CONCAT(IF(DT="',DT,'", PARAM_VALUE, NULL)) AS "',DT,'"')
        ORDER BY DT DESC
        SEPARATOR ', \r\n' )
        INTO @columns
FROM TEST_TABLE
WHERE DT >= DATE_ADD(@TARGET_DATE, interval -2 day) 
AND DT <= DATE_ADD(@TARGET_DATE, interval -1 day);

SELECT CONCAT('SELECT OBJECT_ID,',@columns,' 
                 FROM TEST_TABLE 
                 GROUP BY OBJECT_ID;') INTO @sql;

PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;

对此的最终查询将与您最初的尝试不同，但它是基于它的：

SELECT OBJECT_ID,
       GROUP_CONCAT(IF(DT="2022-01-07", PARAM_VALUE, NULL)) AS "2022-01-07", 
       GROUP_CONCAT(IF(DT="2022-01-06", PARAM_VALUE, NULL)) AS "2022-01-06" 
FROM TEST_TABLE 
GROUP BY OBJECT_ID;

我特意将@columns 的创建和@sql 最终查询分开，希望它更容易理解，但是一旦你掌握了它的窍门，就可以对准备好的语句使用单个查询。

Demo fiddle

【讨论】：