Question

在我的Teradata表格中，我在dhTimestamp列

下有时间戳记

dhTimestamp  
1435308067705  
1434965874565  
1434763800794  
1434775876034  
1434765207057

如何在Teradata上将纪元时间戳转换为人类日期/时间格式？

Answer 1

这是标准unixtime的SQL UDF：

/**********
Converting Unix/POSIX time to a Timestamp 

Unix time: Number of seconds since 1970-01-01 00:00:00 UTC not counting leap seconds (currently 24 in 2011)

Also working for negative numbers.
The maximum range of Timestamps is based on the range of INTEGERs:
1901-12-13 20:45:52 (-2147483648) to 2038-01-19 03:14:07 (2147483647)

Can be changed to use BIGINT instead of INTEGER

20101211 initial version - Dieter Noeth
**********/

REPLACE FUNCTION Epoch2Timestamp (UnixTime INT)
RETURNS TimeStamp(0)
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
RETURNS NULL ON NULL INPUT
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN 
CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0))
+ ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND)
;

SELECT
   Epoch2Timestamp(-2147483648)
  ,Epoch2Timestamp(0)
  ,Epoch2Timestamp(2147483647)
;

但是您的值似乎包含毫秒，这需要修改计算：

CAST(DATE '1970-01-01' + (UnixTime / 86400000) AS TIMESTAMP(3))
+ ((UnixTime / 1000.000 MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND)

编辑2016-07-01：

日常节省时间存在问题（请参阅有关DevEx的Teradata上的this thread），这应该解决它：

-- Unix time to Timestamp WITH TIME ZONE (+00:00)
REPLACE FUNCTION UnixTime_to_TimeStamp_TZ (UnixTime INT)
RETURNS TIMESTAMP(0) WITH TIME ZONE
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
  ((CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0) AT 0)) AT 0)
+ ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND);

-- Unixtime to Timestamp, implicit TIME ZONE of the local session
REPLACE FUNCTION UnixTime_to_TimeStamp (UnixTime INT)
RETURNS TIMESTAMP(0)
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
  CAST(((CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0) AT 0)) AT 0)
     + ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND) AS TIMESTAMP(0));

Answer 2

我可以使用以下查询将纪元列转换为时间戳。

if (Friends != null) { Friends.setOnCheckedChangeListener(new CompoundButton.OnCheckedChangeListener() { @Override public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) { if (isChecked) { friend_source = "Friend"; Newspaper.setChecked(false); Facebook.setChecked(false); Zomato.setChecked(false); RadioSource.setChecked(false); } } }); }

只需将 epochtimecolumn 替换为上述查询中的列即可在teradata中运行它。

希望有帮助！

Answer 3

这是将EPOCH转换为TERADATA LOCAL的最简单方法。

SELECT
   dhTimestamp as unix_epoc_time   , 

   to_timestamp(unix_epoc_time) utc,   

   cast(cast(utc as char(19))||'+00:00' as timestamp(0) with time zone) AT LOCAL

如果纪元的位数超过10位，则将数字切成10位后的数字，就可以了。
记住，Unix时间是UTC。
您的系统会将此utc视为本地。因此，让我们通过添加'+00:00'来理解它是UTC，然后使用AT LOCAL或使用以下两个“ America Central”，“ America Eastern”，“ America Mountain”将其转换为本地等。

如何在Teradata中将纪元时间转换为人类可读

3 个答案:

编辑2016-07-01：