mysql - 如何在SQL视图中计算年龄
我想要达到什么目标?
我正在尝试将客户详细信息生成到一个表中。客户名称和地址位于2个不同的表中,因此我创建了以下VIEW
SELECT dbo.Customer.CustomerId, dbo.Customer.GivenName ' ' dbo.Customer.FamilyName AS [CustomerName], dbo.Customer.DOB, dbo.Customer.Gender,
dbo.Address.PostCode AS [Post Code], dbo.Address.Town AS Suburb, dbo.State.Code AS State
FROM dbo.Address INNER JOIN
dbo.Cusotmer ON dbo.Address.AddressId = dbo.Customer.AddressID INNER JOIN
dbo.State ON dbo.Address.StateId = dbo.State.StateId
问题
如何在同一视图表中将DOB变为Age?
DOB是此格式中的日期列:2015-03-25
最佳答案:
0 个答案:
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