MySQL 获取所有商店的最新条目

Question

我正在尝试从记录每 15 分钟累积销售总额的表中获取最新行。

表格的基本布局是

StoreID, Time, Sales, Tax, Date

我希望能够返回每个商店 ID 的最新记录。到目前为止我尝试过的查询是：

SELECT t.StoreID, MAX(t.Time) as 'Last Reported', t.Sales+t.Tax as Sales, t.Date FROM ( SELECT * FROM trickledata WHERE Date = '20180724' ) t GROUP BY t.StoreID;

这可行，但它不会暂时返回相应的销售额。例如。

+---------+---------------+-------+------------+ | StoreID | Last Reported | Sales | Date | +---------+---------------+-------+------------+ | 100 | 11:45:00 | 0.00 | 2018-07-24 | | 111 | 12:00:00 | 0.00 | 2018-07-24 | | 115 | 12:00:00 | 0.00 | 2018-07-24 | | 121 | 12:00:00 | 0.00 | 2018-07-24 | | 122 | 12:00:00 | 0.00 | 2018-07-24 | | 123 | 12:00:00 | 0.00 | 2018-07-24 | | 124 | 12:00:00 | 0.00 | 2018-07-24 | | 125 | 12:00:00 | 0.00 | 2018-07-24 | | 126 | 12:00:00 | 0.00 | 2018-07-24 | | 127 | 12:00:00 | 0.00 | 2018-07-24 | | 128 | 12:00:00 | 0.00 | 2018-07-24 | | 129 | 12:00:00 | 0.00 | 2018-07-24 | | 130 | 12:00:00 | 0.00 | 2018-07-24 | | 131 | 12:00:00 | 0.00 | 2018-07-24 | | 135 | 12:00:00 | 0.00 | 2018-07-24 | | 137 | 12:00:00 | 0.00 | 2018-07-24 | | 138 | 12:00:00 | 0.00 | 2018-07-24 | | 141 | 12:00:00 | 0.00 | 2018-07-24 | | 150 | 12:00:00 | 0.00 | 2018-07-24 | | 160 | 12:00:00 | 0.00 | 2018-07-24 | | 164 | 12:00:00 | 0.00 | 2018-07-24 |

但当时有销售，因为当我查询商店 121 的具体日期和时间时，我得到：

+---------+------------+----------+-------+------+ | StoreID | Date | Time | Sales | Tax | +---------+------------+----------+-------+------+ | 121 | 2018-07-24 | 12:00:00 | 65.79 | 4.55 | +---------+------------+----------+-------+------+

CREATE TABLE 语句：

``CREATE TABLE `trickledata` (
    `StoreID` INT(11) NULL DEFAULT NULL,
    `Date` DATE NULL DEFAULT NULL,
    `Time` TIME NULL DEFAULT NULL,
    `Sales` DECIMAL(10,2) NULL DEFAULT NULL,
    `Tax` DECIMAL(10,2) NULL DEFAULT NULL,
    UNIQUE INDEX `Date` (`Date`, `Time`, `StoreID`),
    INDEX `by_date` (`Date`),
    INDEX `by_store` (`StoreID`)
)
COLLATE='utf8mb4_general_ci'
ENGINE=InnoDB
;
``

样本数据： INSERT LOW_PRIORITY IGNORE INTO `trickle`.`trickledata_copy` (`StoreID`, `Date`, `Time`, `Sales`, `Tax`) VALUES ('0100', '20180724', '120000', '212.63', '15.37'), ('0100', '20180724', '114500', '212.63', '15.37'), ('0100', '20180724', '113000', '212.63', '15.37'), ('0100', '20180724', '111500', '212.63', '15.37'), ('0100', '20180724', '110000', '212.63', '15.37'), ('0100', '20180724', '104500', '212.63', '15.37'), ('0100', '20180724', '103000', '212.63', '15.37'), ('0100', '20180724', '101500', '.00', '.00'), ('0100', '20180724', '100000', '.00', '.00'), ('0100', '20180724', '94500', '.00', '.00'), ('0100', '20180724', '93000', '.00', '.00'), ('0100', '20180724', '91500', '.00', '.00'), ('0100', '20180724', '90000', '.00', '.00'), ('0100', '20180724', '84500', '.00', '.00'), ('0100', '20180724', '83000', '.00', '.00'), ('0100', '20180724', '81500', '.00', '.00'), ('0100', '20180724', '80000', '.00', '.00'), ('0100', '20180724', '74500', '.00', '.00'), ('0100', '20180723', '253000', '986.07', '71.44'), ('0100', '20180723', '251500', '986.07', '71.44'), ('0100', '20180723', '250000', '986.07', '71.44'), ('0100', '20180723', '244500', '986.07', '71.44'), ('0100', '20180723', '243000', '986.07', '71.44'), ('0100', '20180723', '241500', '986.07', '71.44'), ('0100', '20180723', '240000', '986.07', '71.44'), ('0100', '20180723', '234500', '986.07', '71.44'), ('0100', '20180723', '233000', '986.07', '71.44'), ('0100', '20180723', '231500', '986.07', '71.44'), ('0100', '20180723', '230000', '986.07', '71.44'), ('0100', '20180723', '224500', '986.07', '71.44'), ('0100', '20180723', '223000', '986.07', '71.44'), ('0100', '20180723', '221500', '986.07', '71.44'), ('0100', '20180723', '220000', '986.07', '71.44'), ('0100', '20180723', '214500', '986.07', '71.44'), ('0100', '20180723', '213000', '986.07', '71.44'), ('0100', '20180723', '120000', '986.07', '71.44'), ('0100', '20180723', '210000', '986.07', '71.44'), ('0100', '20180723', '204500', '986.07', '71.44');

编辑#2： 使用该示例数据并运行查询会产生 0.00 错误。但是跑步 SELECT * FROM trickledata WHERE Time = '120000 产生 $212.63 的结果

我的 MariaDB 版本也是 10.1.29

编辑#3： 我尝试了以下方法：

''SELECT t.* FROM trickledata t
    -> JOIN (
    -> SELECT StoreID, MAX(Time) AS latest, Sales, Date
    -> FROM trickledata
    -> WHERE Date = '20180724'
    -> GROUP BY StoreID
    -> ) m ON t.StoreID = m.StoreID AND t.Time = m.latest;''

但是，它会返回我中午的每个日期。我只想要今天的数据。

编辑#4： 我发现如果我删除唯一索引，我的初始查询会按预期工作。但我不明白为什么。我还需要该唯一索引来防止添加重复项，因为这是从 csv 文件中读取信息。

Answer 1

尝试使用相关子查询过滤时间，获取一天的最长时间并存储。

SELECT `t1`.`storeid` `StoreId`,
       `t1`.`time` `Last Reported`,
       `t1`.`sales` + `t1`.`tax` `Sales`,
       `t1`.`date` `Date`
       FROM `trickledata` `t1`
       WHERE `t1`.`date` = '2018-07-24'
             AND `t1`.`time` = (SELECT max(`t2`.`time`)
                                       FROM `trickledata` `t2`
                                       WHERE `t2`.`storeid` = `t1`.`storeid`
                                             AND `t2`.`date` = `t1`.`date`);

如果有重复（同一天同一商店的两个或多个报告），它们将被显示。如果您想排除它们，您必须定义另一个标准，该标准应该在报告中定义一个额外的顺序。