SQL 我如何在postgres sql的列中找到最大值？
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How do i find the largest value in a column in postgres sql?
提问by raveren
For example:
例如：
name | weight
jon 100
jane 120
joe 130
How do I only return the name of the person with the largest weight?
如何只返回权重最大的人的名字？
采纳答案by Hao
Use this:
用这个：
select name
from tbl
where weight = (select max(weight) from tbl)
回答by raveren
SELECT name FROM tbl ORDER BY weight DESC LIMIT 1
Much more performant than the other answer and results in one row only.
比另一个答案的性能要高得多，并且只产生一行。
回答by Matt Kleinsmith
ORDER BY DESC puts rows with null values at the top.
ORDER BY DESC 将具有空值的行放在顶部。
To avoid returning results corresponding to null values:
为避免返回对应于空值的结果：
SELECT name FROM tbl WHERE weight = (SELECT MAX(weight) FROM tbl);
SELECT name FROM tbl WHERE weight = (SELECT MAX(weight) FROM tbl);
Note: This query will return multiple results if multiple people have a weight equal to the maximum weight. To grab just one, add LIMIT 1to the end of the query.
注意：如果多人的权重等于最大权重，则此查询将返回多个结果。要仅获取一个，请添加LIMIT 1到查询的末尾。
Acknowledgements and more information:
致谢和更多信息：
Why do NULL values come first when ordering DESC in a PostgreSQL query?
回答by alexkovelsky
If you need to find multiple rows, e.g. date on which each person had maximum weight:
如果您需要查找多行，例如每个人体重最大的日期：
name | weight | day
don 110 1
don 120 20
don 110 30
joe 90 1
joe 80 15
joe 85 30
i.e. for "don" you want to get "don | 120 | 20"and for joe you want "joe | 90 | 1", then you can write:
即对于你想要的“don”和你想要"don | 120 | 20"的 joe "joe | 90 | 1"，那么你可以写：
SELECT name, max(weight), (array_agg(day ORDER BY weight DESC))[1] FROM tbl GROUP BY name
