要得到一組數(shù)據(jù)的中位數(shù)（例如某個地區(qū)或某家公司的收入中位數(shù)），我們一般要將這一任務細分為 3 個小任務：

1.將數(shù)據(jù)排序，并給每一行數(shù)據(jù)給出其在所有數(shù)據(jù)中的排名；

2.找出中位數(shù)的排名數(shù)字；

3.找出中間排名對應的值；

下面以某公司員工月收入為例，示例 MySQL 的一些復雜語句的使用。

方法一

創(chuàng)建測試表

首先創(chuàng)建一個收入表，建表語句為：

CREATE TABLE IF NOT EXISTS `employee` (   `id`     INT                  AUTO_INCREMENT PRIMARY KEY,   `name`   VARCHAR(10) NOT NULL DEFAULT '',   `income` INT         NOT NULL DEFAULT '0' )   ENGINE = InnoDB   DEFAULT CHARSET = utf8; INSERT INTO `employee` (`name`, `income`) VALUES ('麻子', 20000); INSERT INTO `employee` (`name`, `income`) VALUES ('李四', 12000); INSERT INTO `employee` (`name`, `income`) VALUES ('張三', 10000); INSERT INTO `employee` (`name`, `income`) VALUES ('王二', 16000); INSERT INTO `employee` (`name`, `income`) VALUES ('土豪', 40000);

完成任務 1

將數(shù)據(jù)排序，并給每一行數(shù)據(jù)給出其在所有數(shù)據(jù)中的排名：

SELECT t1.name, t1.income, COUNT(*) AS rank FROM employee AS t1,      employee AS t2 WHERE t1.income < t2.income    OR (t1.income = t2.income AND t1.name <= t2.name) GROUP BY t1.name, t1.income ORDER BY rank;

查詢結果為：

完成小任務 2

找出中位數(shù)的排名數(shù)字：

SELECT (COUNT(*) + 1) DIV 2 as rank FROM employee;

查詢結果為：

完成小任務 3

SELECT income AS median FROM (SELECT t1.name, t1.income, COUNT(*) AS rank       FROM employee AS t1,            employee AS t2       WHERE t1.income < t2.income          OR (t1.income = t2.income AND t1.name <= t2.name)       GROUP BY t1.name, t1.income       ORDER BY rank) t3 WHERE rank = (SELECT (COUNT(*) + 1) DIV 2 FROM employee)

查詢結果為：

至此，我們就找到了如何從一組數(shù)據(jù)中獲得中位數(shù)的方法。

方法二

下面，來介紹另外一種優(yōu)化排名語句的方法。

我們都知道如何給一組數(shù)據(jù)做排序操作，在本例中，實現(xiàn)方法如下：

SELECT name, income FROM employee ORDER BY income DESC

查詢結果為：

那我們可不可以更進一步，對查詢出的結果加一列，這一列的數(shù)據(jù)為排名呢？

我們可以通過 3 個自定義變量的方法來實現(xiàn)這一目標：

第一個變量用來記錄當前行數(shù)據(jù)的收入

第二個變量用來記錄上一行數(shù)據(jù)的收入

第三個變量用來記錄當前行數(shù)據(jù)的排名

SET @curr_income := 0; SET @prev_income := 0; SET @rank := 0; SELECT `name`,        @curr_income := income                                      AS income,        @rank := if(@prev_income != @curr_income, @rank + 1, @rank) AS rank,        @prev_income := @curr_income                                AS dummy FROM employee ORDER BY income DESC

查詢結果如下：

然后再找出中位數(shù)的排名數(shù)字，進一步找出收入的中位數(shù)：

SET @curr_income := 0; SET @prev_income := 0; SET @rank := 0; SELECT income AS median FROM (SELECT `name`,              @curr_income := income                                      AS income,              @rank := if(@prev_income != @curr_income, @rank + 1, @rank) AS rank,              @prev_income := @curr_income                                AS dummy       FROM employee       ORDER BY income DESC) AS t1 WHERE t1.rank = (SELECT (COUNT(*) + 1) DIV 2 FROM employee)

查詢結果為：

至此，我們找了兩種方法來解決中位數(shù)的問題。撒花。

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