包括具有中位数计算理论的分组列答案

【问题标题】：Include grouped column with median calculating theory包括具有中位数计算理论的分组列
【发布时间】：2017-06-05 00:34:34
【问题描述】：

感谢this post，我能够计算出发票表中相应供应商的中位数。

这是使用的查询：

SELECT AVG(middle_values) AS 'median'
FROM (
    SELECT t1.invoice_total AS 'middle_values'
    FROM
    (
        SELECT @row:=@row+1 as `row`, iv.invoice_total
        FROM invoices AS iv, (SELECT @row:=0) AS r
        WHERE iv.vendor_id = 97
        ORDER BY iv.invoice_total
    ) AS t1,
    (
        SELECT COUNT(*) as 'count'
        FROM invoices iv
        WHERE iv.vendor_id = 97
    ) AS t2
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;

我希望它显示两列，而不是只在结果框中输出一列：vendor_id、median_invoice。

CREATE TABLE IF NOT EXISTS `invoices` (
  `invoice_id` int(11) NOT NULL AUTO_INCREMENT,
  `vendor_id` int(11) NOT NULL,
  `invoice_number` varchar(50) NOT NULL,
  `invoice_date` date NOT NULL,
  `invoice_total` decimal(9,2) NOT NULL,
  `payment_total` decimal(9,2) NOT NULL DEFAULT '0.00',
  `credit_total` decimal(9,2) NOT NULL DEFAULT '0.00',
  `terms_id` int(11) NOT NULL,
  `invoice_due_date` date NOT NULL,
  `payment_date` date DEFAULT NULL,
  PRIMARY KEY (`invoice_id`),
  KEY `invoices_fk_vendors` (`vendor_id`),
  KEY `invoices_fk_terms` (`terms_id`),
  KEY `invoices_invoice_date_ix` (`invoice_date`),
  CONSTRAINT `invoices_fk_terms` FOREIGN KEY (`terms_id`) REFERENCES `terms` (`terms_id`),
  CONSTRAINT `invoices_fk_vendors` FOREIGN KEY (`vendor_id`) REFERENCES `vendors` (`vendor_id`)
) ENGINE=InnoDB AUTO_INCREMENT=119 DEFAULT CHARSET=latin1;

插入语句：

INSERT INTO `invoices` VALUES (118, 97, '456792', '2011-08-03', 565.60, 0.00, 0.00, 2, '2011-09-02', NULL);
INSERT INTO `invoices` VALUES (117, 97, '456791', '2011-08-03', 4390.00, 0.00, 0.00, 2, '2011-09-02', NULL);
INSERT INTO `invoices` VALUES (116, 97, '456701', '2011-08-02', 270.50, 0.00, 0.00, 2, '2011-09-01', NULL);
INSERT INTO `invoices` VALUES (115, 97, '456789', '2011-08-01', 8344.50, 0.00, 0.00, 2, '2011-08-31', NULL);
INSERT INTO `invoices` VALUES (114, 123, '963253249', '2011-08-02', 127.75, 127.75, 0.00, 3, '2011-09-01', '2011-09-04');
INSERT INTO `invoices` VALUES (113, 37, '547480102', '2011-08-01', 224.00, 0.00, 0.00, 3, '2011-08-31', NULL);
INSERT INTO `invoices` VALUES (112, 110, '0-2436', '2011-07-31', 10976.06, 0.00, 0.00, 3, '2011-08-30', NULL);
INSERT INTO `invoices` VALUES (111, 123, '263253257', '2011-07-30', 22.57, 22.57, 0.00, 3, '2011-08-29', '2011-09-03');

执行以下操作并不好：

SELECT t1.vendor_id, AVG(middle_values) AS 'median'
FROM (
    SELECT vendor_id, t1.invoice_total AS 'middle_values'
    FROM
    (
        SELECT @row:=@row+1 as `row`, iv.invoice_total
        FROM invoices AS iv, (SELECT @row:=0) AS r
        WHERE iv.vendor_id = 97
        ORDER BY iv.invoice_total
    ) AS t1,
    (
        SELECT COUNT(*) as 'count'
        FROM invoices iv
        WHERE iv.vendor_id = 97
    ) AS t2, invoices
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;

【问题讨论】：

标签： mysql group-by median

【解决方案1】：

为了在父查询中使用 vendor_id，您需要在每个嵌套子查询中返回（选择它）：

SELECT t3.vendor_id, AVG(middle_values) AS 'median'
FROM (
    SELECT t1.invoice_total AS 'middle_values', t1.vendor_id
    FROM
    (
        SELECT @row:=@row+1 as `row`, iv.invoice_total, iv.vendor_id
        FROM invoices AS iv, (SELECT @row:=0) AS r
        WHERE iv.vendor_id = 97
        ORDER BY iv.invoice_total
    ) AS t1,
    (
        SELECT COUNT(*) as 'count'
        FROM invoices iv
        WHERE iv.vendor_id = 97
    ) AS t2
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3

【讨论】：