SQL min函数

Question

我正在尝试获取两列frequency和frequency - min(frequency)但我看到的第二列为零。什么可能是错的？

SELECT
    frequency, frequency - min(frequency)
    FROM
    words
GROUP BY id
ORDER BY frequency;

Answer 1

您的查询组按频率的唯一值。在每个这样的组中，最小频率只是频率本身，因此在减去两者时总是得到0。相反，您可以使用min的窗口版本：

SELECT   frequency, frequency - MIN(frequency) OVER() AS diff
FROM     words
ORDER BY frequency

Answer 2

添加一个返回最小频率值的子查询：

SELECT
    frequency, frequency - (select min(frequency) from words)
    FROM
    words
ORDER BY frequency;

修改

将其包装在派生表中：

SELECT frequency, frequency - minfreq, frequency + minfreq
FROM words
    CROSS JOIN (select min(frequency) minfreq from words) dt
ORDER BY frequency

Answer 3

试一试：

SELECT frequency, frequency-min_frequency
FROM ( 
SELECT frequency,  MIN(frequency) AS min_frequency 
FROM words
GROUP BY frequency
) as A 
ORDER BY frequency;

Answer 4

这肯定是一个奇怪的查询。您按ID分组，因此每个ID可以获得一个结果记录。但ID表明这是唯一标识记录的表格ID。因此GROUP BY id不会改变任何内容，您仍会获得结果中的所有记录。除了一个例外：min(frequency)现在表示每组的最低频率。由于“组”是一个记录，最小值当然是值本身。非聚合frequency也是由ID唯一标识的记录频率。因此，您的查询可以重写为：

SELECT
  frequency, frequency - frequency
FROM words
ORDER BY frequency;

我想您想要将每个记录的频率与表中的最低频率进行比较？您将在子查询中获得此值：

SELECT
  frequency, frequency - (select min(frequency) from words)
FROM words
ORDER BY frequency;

或者：

SELECT
  w.frequency, w.frequency - m.min_frequncy
FROM words w
CROSS JOIN (select min(frequency) as min_frequncy from words) m
ORDER BY frequency;