DATEDIFF() 只返回带有 2 个小数点的年龄

本文介绍了DATEDIFF() 只返回带有 2 个小数点的年龄的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

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

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.

这篇关于DATEDIFF() 只返回带有 2 个小数点的年龄的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！

本站部分内容来源互联网,如果有图片或者内容侵犯您的权益请联系我们删除！

相关文档推荐