Question

如何计算2天之间的小时数？

我试过这个：

ABS(TO_DATE(t11.JAM_AKHIR,'HH24:MI:SS') - TO_DATE(t11.JAM_AWAL,'HH24:MI:SS')) * 24 AS SELISIH_JAM,

结果：

| NO| JAM_AWAL    | JAM AKHIR    | SELISIH |
| 1 | 16.30    | 18.30        | 2      |
| 2 | 18.30    | 06.30        | 12      |
| 3 | 18.30    | 08.00        | 10.5      |    (WRONG)

但是，NO 3是错误的。然后我试了一下：

CASE WHEN TO_DATE(t11.JAM_AWAL,'HH24:MI:SS') < TO_DATE(t11.JAM_AKHIR,'HH24:MI:SS')
     THEN                           
          ABS(TO_DATE(t11.JAM_AKHIR,'HH24:MI:SS') - TO_DATE(t11.JAM_AWAL,'HH24:MI:SS')) *24

     WHEN TO_DATE(t11.JAM_AWAL,'HH24:MI:SS') > TO_DATE(t11.JAM_AKHIR,'HH24:MI:SS')
     THEN                           
          ABS(TO_DATE(t11.JAM_AWAL,'HH24:MI:SS') - TO_DATE(t11.JAM_AKHIR,'HH24:MI:SS')) *24                               
END SELISIH_JAM,

但仍然失败。请帮帮我，谢谢：）

Answer 1

我认为你需要在jam_akhir的to_date中添加一天，如果它大于jam_awal，例如：

public class OnSetPropertyListener<DllClass>
{
    public void OnSetProperty(PropertyInfo propertyInfo, DllClass instance) 
    {
        // set another value, for example: "new value"
    }
}

Answer 2

只是较短的选择; - ）

<强> ^demo

select no, jam_akhir, jam_awal,
       mod (24 *(to_date(jam_akhir, 'hh24:mi') - to_date(jam_awal, 'hh24:mi') ) + 24, 24)
  from t11

2天之间的总和

2 个答案: