INSERT INTO test.income (name, income) VALUES ("麻子", 20000); INSERT INTO test.income (name, income) VALUES ("李四", 12000); INSERT INTO test.income (name, income) VALUES ("张三", 10000); INSERT INTO test.income (name, income) VALUES ("王二", 16000); INSERT INTO test.income (name, income) VALUES ("土豪", 40000);
小任务1的查询语句:SELECT a1.name, a1.income, count(*) AS rank FROM income AS a1, income AS a2 WHERE a1.income < a2.income OR (a1.income = a2.income AND a1.name <= a2.name) GROUP BY a1.name, a1.income ORDER BY rank;小任务2的查询语句:SELECT (COUNT(*) + 1) DIV 2 FROM income;小任务3的查询语句: SELECT income AS median FROM (SELECT a1.name, a1.income, count(*) AS rank FROM income AS a1, income AS a2 WHERE a1.income < a2.income OR (a1.income = a2.income AND a1.name <= a2.name) GROUP BY a1.name, a1.income ORDER BY rank) a3
WHERE rank = (SELECT (COUNT(*) + 1) DIV 2 FROM income)至此,我们就找到了如何从一组数据中获得中位数的方法。下面,来介绍另外一种优化排名语句的方法。我们都知道如何给一组数据做排序操作,在本例中,实现方法如下: SELECT name, income FROM income ORDER BY income DESC那我们可不可以更进一步,对查询出的结果加一列,这一列的数据为排名呢?我们可以通过3个自定义变量的方法来实现这一目标:第一个变量用来记录当前行数据的收入 第二个变量用来记录上一行数据的收入 第三个变量用来记录当前行数据的排名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 income ORDER BY income DESC查询结果如下:然后再找出中位数的排名数字,进一步找出收入的中位数: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 income ORDER BY income DESC) AS a1 WHERE a1.rank = (SELECT (COUNT(*) + 1) DIV 2 FROM income)至此,我们找了两种方法来解决中位数的问题。撒花。本文永久更新链接地址