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Sql server SQL Server:如何获取正确的百分比_Sql Server - Fatal编程技术网

Sql server SQL Server:如何获取正确的百分比

Sql server SQL Server:如何获取正确的百分比,sql-server,Sql Server,我有一个简单的SQL Server用户贷款申请表 CREATE TABLE [dbo].[Application] ( [Id] INT NOT NULL, [ApplicantId] INT NOT NULL, [LoanAmount] INT NOT NULL, [LoanTerm] SMALLINT NOT NULL, [Accepted]

我有一个简单的SQL Server用户贷款申请表

CREATE TABLE [dbo].[Application] 
(
    [Id]          INT            NOT NULL,
    [ApplicantId] INT            NOT NULL,
    [LoanAmount]  INT            NOT NULL,
    [LoanTerm]    SMALLINT       NOT NULL,
    [Accepted]    BIT            NOT NULL,
    [BuyerId]     SMALLINT       NOT NULL,
    [Commission]  DECIMAL (9, 2) NOT NULL,
    [PostTime]    SMALLINT       NOT NULL,
    [CreatedOn]   DATETIME       NOT NULL,
    PRIMARY KEY CLUSTERED ([Id] ASC),
    CONSTRAINT [FK_Application_ToTable] FOREIGN KEY ([ApplicantId]) REFERENCES [dbo].[Applicant] ([Id])
);
我想在表上创建一个视图,提供两条信息。首先是处理申请的平均时间,其次是接受申请的百分比

这是我不确定的第二个因素——我完全是SQL新手

以下是我目前掌握的情况:

CREATE VIEW [dbo].[PostPerformance]
AS 
    SELECT AVG(PostTime) AS 'AvgPostTime'
    FROM [Application]
正如您所看到的那样,
AvgPostTime
已经足够简单,但我不确定如何计算在特定时期内接受的贷款的百分比

每行将有一个“true”或“false”接受字段-我需要计算为true的条目数,并计算选择中的总行数,然后计算百分比


如有任何建议,将不胜感激。

您尚未提供样本数据。我猜你在找这个:

SELECT 
    AVG(PostTime) AS 'AvgPostTime'
    , 100.0 * count(case when [Accepted] = 1 then 1 end) / count(*) AS [Percentage]
FROM [Application]

你能给我们提供一些数据到表中,以及期望的结果吗?