SELECT age,DATE_FORMAT(NOW(),'%Y/%m/%d'),(YEAR(NOW())-YEAR(age)-1) + ( DATE_FORMAT(age, '%m%d') <= DATE_FORMAT(NOW(), '%m%d') ) AS 年龄 FROM users AS u
mysql根据出生年月日计算年龄大小
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转载自blog.csdn.net/lihua5419/article/details/80967231
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