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我想根据历史状态和原因计算订单数量和收入总和。

以下是我的表结构。

订单表:-

CREATE TABLE `order_item` (
  `id_order_item` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `unit_price` decimal(17,2) DEFAULT NULL,
  `fk_reason` int(11) DEFAULT NULL,
  PRIMARY KEY (`id_order_item`),
  KEY `fk_reason` (`fk_reason`)
) ENGINE=InnoDB;

历史表:-

CREATE TABLE `order_item_status_history` (
  `id_order_item_status_history` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `fk_order_item` int(10) unsigned NOT NULL,
  `fk_order_item_status` int(10) unsigned NOT NULL COMMENT ''New status'',
  `updated_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `created_at` datetime DEFAULT NULL,
  PRIMARY KEY (`id_order_item_status_history`),
  KEY `fk_order_item` (`fk_order_item`),
  CONSTRAINT `order_item_status_history_ibfk_1` FOREIGN KEY (`fk_order_item`) REFERENCES `order_item` (`id_order_item`) ON DELETE CASCADE ON UPDATE NO ACTION,
  CONSTRAINT `order_item_status_history_ibfk_3` FOREIGN KEY (`fk_order_item_status`) REFERENCES `order_item_status` (`id_order_item_status`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB;

状态表:-

CREATE TABLE `order_item_status` (
  `id_order_item_status` int(10) unsigned NOT NULL,
  `name` varchar(50) NOT NULL,
  `desc` varchar(255) NOT NULL,
  `deprecated` tinyint(1) DEFAULT ''0'',
  PRIMARY KEY (`id_order_item_status`),
) ENGINE=InnoDB;

原因表:-

CREATE TABLE `reason` (
  `id_reason` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(100) NOT NULL,
  `desc` varchar(255) NOT NULL,
  PRIMARY KEY (`id_cancel_reason`),
) ENGINE=InnoDB ;

我需要将订单分组到以下存储桶中,

  1. 订单状态为“已关闭”,如果订单之前已发货。(即先前的订单状态为“已发货”)
  2. 订单状态为“已关闭”,如果订单之前未发货(即之前的订单状态不是“已发货”) (在这种情况下,需要检查当前状态以及之前的订单状态。)
  3. 订单的状态为“欺诈” (在这种情况下,只需检查当前状态。) ......

如何根据上面定义的存储桶获得计数或订单以及收入。 我在计算第 3 点和第 4 点的订单并在单个查询中获取所有计数时遇到问题。

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1 回答 1

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要将所有这些都放入一个查询中,您可以CASE WHEN ..像这样使用

SELECT
whateverYouAreGroupingByIfNeeded,
SUM(CASE WHEN status = 'canceled' AND reason = 1 THEN 1 ELSE 0 END) AS count_whatever
SUM(CASE WHEN whatever = true THEN whateverYouWantToSummarize ELSE NULL END) AS sum_whatever
FROM yourTable
GROUP BY whatever

当您需要具体帮助时,最好展示您的尝试。

PS:如果您在加入时遇到问题,请阅读此内容。

于 2013-05-08T12:46:58.040 回答