在我的 Teradata 表中,该列下有纪元时间戳dhTimestamp
dhTimestamp
1435308067705
1434965874565
1434763800794
1434775876034
1434765207057
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如何在 Teradata 上将纪元时间戳转换为人类日期/时间格式?
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)
;
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但你的值似乎包括毫秒,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)
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夏令时存在问题(请参阅DevEx 上 Teradata 上的此线程),这应该可以修复它:
-- 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));
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