从枢转的MySQL的concat结果值

Question

我想旋转后得到一个连接字符串值。

目前，仅是单值和工作正常。

这里的查询：

DROP PROCEDURE IF EXISTS getAllUserLunchReport;
DELIMITER $$
CREATE DEFINER=`root`@`localhost` PROCEDURE `getAllUserLunchReport`(IN `start_date` DATETIME, IN `end_date` DATETIME)
BEGIN

SET GLOBAL group_concat_max_len=4294967295;
SET @SQL = NULL;
SET @start_date = DATE(start_date);
SET @end_date = DATE(end_date);

    SELECT
    COALESCE(GROUP_CONCAT(DISTINCT
    CONCAT(
    'SUM(CASE WHEN date = "',DATE(issuedDateTime),'" AND lunchStatus = 1 THEN (SELECT CONCAT((SELECT rate FROM lunch_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1), "," ,(SELECT rate FROM lunch_gbd_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1))) ELSE 0 END) AS `',DATE(issuedDateTime),'`'
    ) ORDER BY issuedDateTime
), '0 as `NoMatchingRows`') INTO @SQL

        FROM `lunch_status` 
        WHERE DATE(issuedDateTime) BETWEEN @start_date AND @end_date;

    SET @SQL 
    = CONCAT
    (
    '
        SELECT concat(u.firstname," ",u.lastname) as Employee, w.*
        FROM users u
        INNER JOIN
        (SELECT userId, ', @SQL, ' 
        FROM
        (
            SELECT userId, lunchStatus, DATE(issuedDateTime) as date 
            FROM `lunch_status` 
            WHERE DATE(issuedDateTime) BETWEEN "',@start_date,'" AND "',@end_date,'" 
        ) as a
        GROUP BY userId) w
        ON u.id = w.userId
        ORDER BY Employee;
    '
    );

PREPARE stmt FROM @SQL;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;


END$$
DELIMITER ;

我只获得第一个值，不，我想，从而将其他值。

以下是我所期望的结果：

employee    |   2019-1-15  |  2019-1-16
----------------------------------------
Jack        |   30,140     |  30,140

单独执行拼接查询工作虽然。

注：在上面的存储过程，值如果分离器被移除串联。 CONCAT((select ...), (select...))给30140。

难道是与quotes的问题，但我已经想尽办法。

仅供参考，这里就是@SQL第一个查询后得到：

SUM(CASE WHEN date = "2019-01-15" AND lunchStatus = 1 THEN (SELECT CONCAT((SELECT rate FROM lunch_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1), ",", (SELECT rate FROM lunch_gbd_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1))) ELSE 0 END) AS `2019-01-15`,
SUM(CASE WHEN date = "2019-01-16" AND lunchStatus = 1 THEN (SELECT CONCAT((SELECT rate FROM lunch_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1), ",", (SELECT rate FROM lunch_gbd_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1))) ELSE 0 END) AS `2019-01-16`

看起来有效。

编辑：现在看来似乎是工作的罚款与“整”的价值观。例如

CONCAT((SELECT rate FROM lunch_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1), "111" ,(SELECT rate FROM lunch_gbd_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1))

给301140。

但与其他字符,/./-/_，只显示第一个参数：30。

Answer 1

当用户在使用,的CONCAT您的代码不工作的原因是因为CONCAT是SUM内，它将不能够解释30,140为一个数字，它可以为30140这样做的（因此有没有分隔，你得到的结果）。不知道你的表中的数据和预期的效果更多的细节，很难说究竟是如何解决这个问题，但也许你想是这样的：

CASE WHEN date = "2019-01-15" AND lunchStatus = 1 THEN CONCAT((SELECT SUM(rate) FROM lunch_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1), ",", (SELECT SUM(rate) FROM lunch_gbd_rate WHERE DATE(created_on) <= date ORDER BY created_on DESC LIMIT 1)) ELSE 0 END) AS `2019-01-15`