[英]MySQL find averages based on multiple factors
我有桌子做這樣的事情
+--------------------------+--------+------+---------+
| | City | Year | Density |
+--------------------------+--------+------+---------+
| Project 1 | City A | 2008 | 500 |
+--------------------------+--------+------+---------+
| Project 2 | City B | 2012 | 800 |
+--------------------------+--------+------+---------+
| Project 3 | City C | 2012 | 400 |
+--------------------------+--------+------+---------+
| Project 4 | City A | 2008 | 600 |
+--------------------------+--------+------+---------+
| Project 5 | City C | 2013 | 700 |
+--------------------------+--------+------+---------+
| etc (c. 30,000 projects spread across 30 cities) |
+--------------------------+--------+------+---------+
(大約30,000個項目分布在30個城市。)
我可以這樣寫查詢:
SELECT Year, AVG(`Density`) as Density FROM table where City=’A’ GROUP BY Year
對於一個城市來說效果很好。 關於我如何編寫一個查詢來按年計算每個城市的平均值的問題,有人能為我指明正確的方向嗎? 我期望結果表看起來像這樣:
+------+--------+--------+--------+-------------+
| | City A | City B | City C | City D, etc |
+------+--------+--------+--------+-------------+
| 2005 | | | | |
+------+--------+--------+--------+-------------+
| 2006 | | | | |
+------+--------+--------+--------+-------------+
| 2008 | | | | |
+------+--------+--------+--------+-------------+
| 2009 | | | | |
+------+--------+--------+--------+-------------+
| 2010 | | | | |
+------+--------+--------+--------+-------------+
| etc | | | | |
+------+--------+--------+--------+-------------+
我試圖在where子句(where in (select distinct City))
中的where中)中使用子查詢,但是它的行為不符合我的預期。
還是我只需要為30個城市中的每個城市單獨划一條線?
我不是MySQL的專家,在概念上看不到我需要做什么。 如果有人可以給我任何指示,我將不勝感激。 謝謝。
您可以按多列分組:
SELECT city, year, AVG(density) AS density
FROM table
GROUP BY city, year
這將為每個城市/年份組合返回單獨的一行。 要將城市作為列,您需要對其進行透視。 見MySQL數據透視表
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.