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我需要记录每天针对每个项目请求和发出的项目数量。purchase_doc表格是：

purchase_doc 表

该requested_items表包含以下要求的项目：

请求的项目

该movement表包含以下要求的项目：

移动

需要的输出（要插入的数据）是：

data_to_insert

这样做的一种方法是从前 2 个查询中获取发布和请求的项目，然后针对每个项目 id 构建一个发布和请求的项目数组，然后将这些值插入 daily_movement 表中，如下所示：

SELECT n.item_id AS n__item_id, SUM(n.qty) AS qty
FROM requested_items n LEFT JOIN purchase_doc doc ON n.doc_id = doc.id 
WHERE (doc.type = 'Item Request' AND doc.created_at > DATE_SUB(NOW(), INTERVAL 24 HOUR)) 
GROUP BY n.item_id

SELECT n.item_id AS item_id, SUM(n.qty) AS qty
FROM movement n LEFT JOIN purchase_doc doc ON n.doc_id = doc.id 
WHERE (doc.type = 'Store Issue' AND doc.created_at > DATE_SUB(NOW(), INTERVAL 24 HOUR)) 
GROUP BY n.item_id

从这些和其他选择中，我需要每天为每个项目插入一行，其中包含该项目的请求、问题等数量，以这种方式：

INSERT INTO daily_movement date, item_id, requested_qty, issued_qty VALUES ( NOW(), 23, 4, 5), ( NOW(), 25, 5, 5), ( NOW(), 113, 6, 8);

但是会有太多的选择（因为我还需要每个项目执行其他活动），然后是插入。

我的问题是：是否可以通过单个 SELECT ... INSERT 语句来做到这一点。如果没有，有人可以建议一种更优雅的方式来做到这一点

4

3 回答 3

1

我在想这个，但它可能过于简单：

INSERT INTO `daily_movement`
  (`date`, `item_id`, `requested_qty`, `issued_qty`)
SELECT NOW(), `r`.`item_id`, SUM(`r`.`qty`), SUM(`m`.`qty`)
  FROM `purchase_doc` `d`
  JOIN `requested_items` `r`
    ON `r`.`doc_id` = `d`.`id`
  LEFT JOIN `movement` `m`
    ON `m`.`doc_id` = `d`.`id`
WHERE
    (`d`.`type` = 'Item Request' OR `d`.`type` = 'Store Issue')
  AND
    `d`.`created_at` > DATE_SUB(NOW(), INTERVAL 24 HOUR)
GROUP BY `r`.`item_id`

编辑

这是我的最终答案，UNION很难解决 MySQL 的不足FULL OUTER JOIN：

INSERT INTO `daily_movement`
  (`date`, `item_id`, `week_no`, `requested_qty`, `issued_qty`)
SELECT *
FROM (
  (
    SELECT COALESCE(`r`.`item_id`, `a`.`item_id`) AS `item_id`, CURDATE() AS `date`, NULL AS `week_no`, SUM(`r`.`qty`) AS `requests`, COALESCE(`a`.`issued`, 0) AS `issued`
    FROM `purchase_doc` `d`
    LEFT JOIN `requested_items` `r`
      ON `r`.`doc_id` = `d`.`id` 
    LEFT JOIN (
      SELECT `m`.`item_id`, SUM(`m`.`qty`) AS `issued`
      FROM `purchase_doc` `d`
      JOIN `movement` `m`
        ON `m`.`doc_id` = `d`.`id` 
      WHERE `d`.`type` = 'Store Issue'
        AND `d`.`created_at` > DATE_SUB(NOW(), INTERVAL 24 HOUR)
      GROUP BY `m`.`item_id`
    ) `a`
      ON `a`.`item_id` = `r`.`item_id`
    WHERE `d`.`type` = 'Material Requisition'
      AND `d`.`created_at` > DATE_SUB(NOW(), INTERVAL 24 HOUR)
    GROUP BY `r`.`item_id`
  ) UNION DISTINCT (
    SELECT COALESCE(`m`.`item_id`, `a`.`item_id`) AS `item_id`, CURDATE() AS `date`, NULL AS `week_no`, COALESCE(`a`.`requests`, 0) AS `requests`, SUM(`m`.`qty`) AS `issued`
    FROM `purchase_doc` `d`
    LEFT JOIN `movement` `m`
      ON `m`.`doc_id` = `d`.`id` 
    LEFT JOIN (
      SELECT `r`.`item_id`, SUM(`r`.`qty`) AS `requests`
      FROM `purchase_doc` `d`
      JOIN `requested_items` `r`
        ON `r`.`doc_id` = `d`.`id` 
      WHERE `d`.`type` = 'Material Requisition'
        AND `d`.`created_at` > DATE_SUB(NOW(), INTERVAL 24 HOUR)
      GROUP BY `r`.`item_id`
    ) `a`
      ON `a`.`item_id` = `m`.`item_id`
    WHERE `d`.`type` = 'Store Issue'
      AND `d`.`created_at` > DATE_SUB(NOW(), INTERVAL 24 HOUR)
    GROUP BY `m`.`item_id`
  )
  ORDER BY `item_id`
) `u`

http://sqlfiddle.com/#!2/3923d/13

于 2012-07-25T15:14:48.110 回答
0

您可以在此处或此处使用联合运算符在一次选择中获取所有结果

于 2012-07-25T14:49:34.160 回答
0

您可以使用这样的查询 -

编辑：

INSERT INTO daily_movement(date, item_id, requested_qty, issued_qty)
SELECT i.item_id, SUM(ri.qty) requested_qty, SUM(m.qty) issued_qty FROM
  (SELECT item_id FROM requested_items UNION SELECT item_id FROM movement) i
  LEFT JOIN (
    SELECT n.item_id, n.qty
    FROM requested_items n LEFT JOIN purchase_doc doc ON n.doc_id = doc.id 
    WHERE doc.type = 'Item Request' AND doc.created_at > DATE_SUB(NOW(), INTERVAL 24 HOUR)
  ) ri
  ON ri.item_id = i.item_id
  LEFT JOIN (
    SELECT n.item_id, n.qty
    FROM movement n LEFT JOIN purchase_doc doc ON n.doc_id = doc.id 
    WHERE doc.type = 'Store Issue' AND doc.created_at > DATE_SUB(NOW(), INTERVAL 24 HOUR)
  ) m
  ON m.item_id = i.item_id
GROUP BY
  i.item_id;
于 2012-07-25T14:58:02.303 回答