中央値計算理論を含むグループ化された列を含む

Question

this postのおかげで、私は請求書テーブルの対応するベンダーの中央値を計算することができました。

これは、使用したクエリた：VENDOR_ID、median_invoice：

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; 

代わりresultboxでこれだけ出力する1列の、私はそれが2つの列を表示したいと思います。

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; 

INSERT文：次の操作を行って

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'); 

は良いんでした：あなたは（それを選択）、それを返す必要が親クエリにVENDOR_ID使用するためには

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; 

Answer 1

で各ネストされたサブクエリ：

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