DATEDIFF() 只返回带有 2 个小数点的年龄 [英] DATEDIFF() to just return age with 2 decimal points

问题描述

我有一个 SELECT 语句，要求提供测试时个人的年龄:

I have a SELECT statement requesting the age of he individual when a test was made:

SELECT
   DATEDIFF(dd,BIRTH_DATE,TEST_DATE)/365.25 [Age at Result]
FROM TABLE
WHERE ID = '100'

结果类似于2.056125.

我在另一个帖子中看到了 转换为秒并除以 86400.0，但我仍然得到 6 个小数点.

I saw on another post to convert to seconds and divide by 86400.0, but I was still getting 6 decimal points.

我回顾的是将年龄设为 2.05 甚至四舍五入到 2.00.

What I was looking back was to get the age as 2.05 or even round to 2.00.

谢谢

推荐答案

您可以以所需的精度强制转换为小数:

You can cast to a decimal with the precision you want:

SELECT CAST(DATEDIFF(day, BIRTH_DATE, TEST_DATE)/365.25 as DECIMAL(10, 2)) as [Age at Result]
FROM TABLE
WHERE ID = 100;

注意:我删除了100"周围的单引号.如果 id 是字符串，则仅使用单引号.

Note: I removed the single quotes around "100". Only use single quotes if it the id is a string.

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