問題描述

誰能幫助我如何在 PL/SQL 中將 2016-07-01 01:12:22 PM 轉換為 2016-07-01 13:12:22?我使用了以下但沒有運氣.

Can anyone help me how to convert 2016-07-01 01:12:22 PM to 2016-07-01 13:12:22 in PL/SQL? I used the following but no luck.

SELECT TO_TIMESTAMP ('08-FEB-19 06.41.41.000000 PM', 'DD-Mon-RR HH24:MI:SS.FF') 
FROM dual

我預計:08-FEB-19 18.41.41.000000

I expect: 08-FEB-19 18.41.41.000000

我收到以下錯誤:

ORA-01830:日期格式圖片在轉換整個輸入字符串之前結束

ORA-01830: date format picture ends before converting the entire input string

推薦答案

您當前的時間戳沒有任何意義，因為它指定的時間為 18 小時，即下午 6 點，但隨后它也指定了AM 子午線指示器，表示早于中午.因此，您可以從 TO_TIMESTAMP 模式中刪除 AM:

Your current timestamp makes no sense, because it specifies the time as 18 hours, which is 6pm, but then it also specifies the AM meridian indicator, which means earlier than noon. So, you may remove the AM from your TO_TIMESTAMP pattern:

SELECT TO_TIMESTAMP ('08-FEB-19 18.41.41.000000', 'DD-Mon-RR HH24.MI.SS.FF')
FROM dual;

08-FEB-19 06.41.41.000000000 PM

請注意，Oracle 內部確實沒有 12 或 24 小時格式時間戳之類的東西.相反，如果您想以 24 小時格式查看您的 Oracle 時間戳，您可以使用適當的 24 小時格式掩碼調用 TO_CHAR 之類的操作:

Note that there is really no such thing as 12 or 24 hour format timestamp internally in Oracle. Rather, if you want to view your Oracle timestamp in 24 hour format, you may do something like call TO_CHAR with an appropriate 24 hour format mask:

SELECT
    TO_CHAR(TO_TIMESTAMP ('08-FEB-19 18.41.41.000000', 'DD-Mon-RR HH24.MI.SS.FF'),
        'DD-Mon-RR HH24.MI.SS.FF') AS ts
FROM dual;

08-Feb-19 18.41.41.000000000

演示

如果您想轉換帶有 12 小時時間和 AM/PM 組件的時間戳字符串，我們可以嘗試:

If you wanted to convert a timestamp string with 12 hour time and an AM/PM component, we can try:

SELECT
    TO_CHAR(
        TO_TIMESTAMP ('08-FEB-19 06.41.41.000000 PM', 'DD-Mon-RR HH.MI.SS.FF PM'),
        'DD-Mon-RR HH24.MI.SS.FF')
FROM dual;

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