如何从Oracle SQL Developer中的DateTime字段中提取时间？ [英] How to extract only Time from a DateTime field in Oracle SQL Developer?

问题描述

我尝试过 To_Timestamp 和SQL Developer的其他方法，但只有这一个对我有用。

 选择To_Number（To_Char（DateTime_FieldName，'HH24'））
 || '：'|| to_number（to_char（DateTime_FieldName，'MI'））
 || '：'|| to_number（to_char（DateTime_FieldName，'SS'））
 from TABLE_NAME 
  

是否有更好的解决方案？

解决方案

假设你的目标是生成一个代表时间的字符串（这是什么查询您发布的回报，尽管无关紧要的 to_number 调用）

  SELECT to_char << column_name>>'HH24：MI：SS'）
 FROM table_name 
  

如果要返回不同的数据类型，您需要告诉我们您要返回的数据类型。例如，如果您真的要返回一个 INTERVAL DAY TO SECOND

  SELECT numtodsinterval（< column name>>  -  trunc（< column name>>），'day'）
 FROM table_name 
  / pre> 
I tried To_Timestamp and other methods for SQL Developer but only this one worked fine for me.
Select To_Number(To_Char(DateTime_FieldName, 'HH24'))
    || ':' || to_number(to_char(DateTime_FieldName, 'MI'))
    || ':' ||to_number(to_char(DateTime_FieldName, 'SS'))
from TABLE_NAME
Is there a better solution to this?
解决方案
Assuming your goal is to generate a string representing the time (which is what the query you posted returns despite the extraneous to_number calls)
SELECT to_char( <<column_name>>, 'HH24:MI:SS' )
  FROM table_name
If you want to return a different data type, you'd need to tell us what data type you want to return.  If, for example, you really want to return an INTERVAL DAY TO SECOND
SELECT numtodsinterval( <<column name>> - trunc(<<column name>>), 'day' )
  FROM table_name


                        
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