Question

我对此感到有点难过。我可以获得数小时或总分钟数，但我需要小时和分钟（如果工作时间超过60分钟）。开始时间和结束时间（datetime数据类型）存储在数据库中元组的字段中。

我需要花费数小时（如果超过60分钟）和剩余时间。我的SQL查询中有以下计算。我知道给我一个完整的分钟，我只需要在删除几个小时之后留下任何遗留物。

如果开始时间是上午11:15，结束时间是下午12:25，我需要1小时10分钟。

                sqlQuery = "select distinct(job_#1) as 'Job Number', " +
                "client_name as 'Client Name', " +
                "Convert(varchar,min(date),101) as 'Start Date', " +
                "convert(varchar, max(date), 101) as 'End Date', " +
                "FORMAT(sum(datediff(mi, START_TIME,END_TIME)%(60*24)/60),'#','en-us' ) as 'Hours', " + // calculate the total time taken on the job - formatted numeric
                "FORMAT(sum(datediff(mi, START_TIME,END_TIME)%(60*24)%60),'#','en-us' ) as 'Minutes', " + // calculate the total time taken on the job - formatted numeric
                "FORMAT(sum((datediff(MINUTE,start_time,end_time))/60.00)*25.00, 'C', 'en-us') as 'Total Labor' " + // multiple the job time by a $$ per hour rate - formatted as currency
                "from " + ssqltable + " " +                                      // table from which data is pulled 
                "where job_#1 IS NOT NULL " +                        // find all jobs that have a job number leaving out blank job numbers
                "and date>='" + myDate + "' " +                      // filled in when selected or passes null and shows all data
                "group by job_#1, Client_Name " +                    // aggregate grouping 
                "order by job_#1;"; // column(s) to order the data

以下是我的输出示例..

237452  AADC        05/18/2017  05/18/2017  1   **208** $111.67
237353  Wolverine   05/18/2017  05/18/2017      **110** $45.83
237492  Beeman      05/11/2017  05/16/2017      **74**  $30.83

Answer 1

您可以datediff()使用hour几小时（如果您没有汇总），minute分钟，并根据需要划分或模数：

select 
    [Job Number] = [job_#1]
  , [Client Name]= client_name
  , [Start Date] = convert(char(10),min(date),101)
  , [End Date]   = convert(char(10),max(date),101)
  , Hours        = sum(datediff(minute,Start_Time,End_Time))/60
  , Minutes      = sum(datediff(minute,Start_Time,End_Time))%60
  , [Total Labor]= (sum(datediff(minute,Start_Time,End_Time))/60.00) * 25.00
from tbl
where job_#1 is not null
  and [date] >= @date_parameter
group by [job_#1], client_name
order by [job_#1]

注意：

使用参数，不要将变量连接到直接执行的sql;这就是你打开代码到SQL注入的方式。 Bobby Tables: A guide to preventing SQL injection
不要使用字符串文字作为别名（现在已经弃用了一段时间）。
Bad habits to kick : declaring varchar without (length) - Aaron Bertrand
Bad Habits to Kick : Using shorthand with date/time operations - Aaron Bertrand
format()效果非常糟糕：format() is nice and all, but… - Aaron Bertrand
为什么要在SQL中格式化日期，而不是仅仅将日期带回应用程序并在表示层进行格式化？

Answer 2

如果你有总分钟数，那么DIV（/）和MOD / Remainder（％）函数就可以了。

DECLARE @TotalMinutes int
SET @TotalMinutes = 70

SELECT @TotalMinutes / 60 AS Hours, 
       @TotalMinutes % 60 AS Minutes

返回以下结果：

Total Minutes    Hours    Minutes
     70            1         10
    121            2          1

Answer 3

您的代码应如下所示。请注意括号中的小时和分钟的变化。首先按SUM计算所有分钟，然后除以或取模60（整数）得到小时和分钟。

      sqlQuery = "select distinct(job_#1) as 'Job Number', " +
            "client_name as 'Client Name', " +
            "Convert(varchar,min(date),101) as 'Start Date', " +
            "convert(varchar, max(date), 101) as 'End Date', " +
            "FORMAT(sum(datediff(mi, START_TIME,END_TIME)/60),'#','en-us' ) as 'Hours', " + // calculate the total time taken on the job - formatted numeric
            "FORMAT(sum(datediff(mi, START_TIME,END_TIME))%60),'#','en-us' ) as 'Minutes', " + // calculate the total time taken on the job - formatted numeric
            "FORMAT(sum(datediff(MINUTE,start_time,end_time)/60.00)*25.00, 'C', 'en-us') as 'Total Labor' " + // multiple the job time by a $$ per hour rate - formatted as currency
            "from " + ssqltable + " " +                                      // table from which data is pulled 
            "where job_#1 IS NOT NULL " +                        // find all jobs that have a job number leaving out blank job numbers
            "and date>='" + myDate + "' " +                      // filled in when selected or passes null and shows all data
            "group by job_#1, Client_Name " +                    // aggregate grouping 
            "order by job_#1;"; // column(s) to order the data

Answer 4

我认为简单的除以60足够如下：

        sum( datediff(mi,START_TIME,END_TIME)/60 ) as [Hours]
        sum( datediff(mi,START_TIME,END_TIME)%60 ) as [Minutes]

格式效果不高。对于您的方案，您可以避免

确定SQL查询中的剩余分钟数

4 个答案: