Sql server 确定SQL查询中的剩余分钟数

我对此有点困惑。我可以得到小时数或总分钟数，但我同时需要小时和分钟（如果工作需要60分钟以上）。开始时间和结束时间（datetime数据类型）存储在数据库元组中的字段中

我需要几个小时（如果超过60分钟）和剩余的几分钟。我的sql查询中有以下计算。我知道总的时间是几分钟，我只需要在几个小时后再做剩余的东西

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

下面是我的输出示例

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

如果您有总分钟数，DIV（/）和MOD/rements（%）函数将完成此任务

DECLARE @TotalMinutes int
SET @TotalMinutes = 70

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

这将返回以下结果：

Total Minutes    Hours    Minutes
     70            1         10
    121            2          1

---

您的代码应该如下所示。请注意括号中小时和分钟的顺序变化。首先将所有分钟相加，然后除以60（整数）或将其模化，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

您可以将

datediff（）

与

hour

配合使用数小时（如果您没有求和），将

minute

配合使用数分钟，并根据需要进行除法或模运算：

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:

• 使用参数，不要将变量连接到直接执行的sql；这就是打开代码进行sql注入的方式
• 不要使用字符串文字作为别名（它已经被弃用了一段时间）

• format（）

  性能非常差：
• 为什么要用SQL格式化日期，而不是将日期带回应用程序并在表示层格式化

---

我认为简单除以60就足够了，如下所示：

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

---

格式不是高性能的。对于您的场景，您可以避免
谢谢，@SqlZim，这是我在应用程序中的建议。我已经添加了你作为书签的链接，以备将来阅读。我按照建议删除了format（），并清理了我的XAML以进行格式化。@lseting真棒！快乐编码！哦对小鲍比桌子…：-）