Netezza In-Built AGE在Redshift中用作UDF

问题描述 投票：0回答：2

我正在尝试在Redshift中实现Netezza AGE功能作为UDF。我能够在Python（Spyder IDE - Py 3.6）中得到正确的答案但是当我在Redshift中以UDF的形式执行它时，它会给我不正确的输出。

我试图在Redshift中执行select AGE_UDF('1994-04-04 20:10:52','2018-09-24 11:31:05');。这是RS UDF中使用的代码。

CREATE OR REPLACE FUNCTION AGE_UDF (START_DATE TIMESTAMP, END_DATE TIMESTAMP)
    RETURNS varchar(100)
stable
AS $$
    from datetime import datetime
    from dateutil import relativedelta

    START_DATE = datetime.strptime(START_DATE, '%Y-%m-%d %H:%M:%S')

    END_DATE = datetime.strptime(END_DATE, '%Y-%m-%d %H:%M:%S')

    difference = relativedelta.relativedelta(END_DATE, START_DATE)

    years = difference.years
    months = difference.months
    days = difference.days
    hours = difference.hours
    minutes = difference.minutes
    seconds = difference.seconds
    age=''
    if years == 0: 
        age='' 
    elif years == 1:
        age+=str(years)+' year '
    else:
        age+=str(years)+' years '

    if months == 0: 
        age+='' 
    elif months == 1:
        age+=str(months)+' mon '
    else:
        age+=str(months)+' mons '

    if days == 0: 
        age+='' 
    elif days == 1:
        age+=str(days)+' day '
    else:
        age+=str(days)+' days '

    age+=str(hours)+':'+str(minutes)+':'+str(seconds)
    return age
$$ language plpythonu;

RS输出：-8809.15：20：13

这是Python（3.6）中使用的代码。

from datetime import datetime
from dateutil import relativedelta

START_DATE = '1994-04-04 20:10:52'
START_DATE = datetime.strptime(START_DATE, '%Y-%m-%d %H:%M:%S')

END_DATE = '2018-09-24 11:31:05'
END_DATE = datetime.strptime(END_DATE, '%Y-%m-%d %H:%M:%S')

difference = relativedelta.relativedelta(END_DATE, START_DATE)

years = difference.years
months = difference.months
days = difference.days
hours = difference.hours
minutes = difference.minutes
seconds = difference.seconds
age=''
if years == 0: 
    age='' 
elif years == 1:
    age+=str(years)+' year '
else:
    age+=str(years)+' years '


if months == 0: 
    age+='' 
elif months == 1:
    age+=str(months)+' mon '
else:
    age+=str(months)+' mons '


if days == 0: 
    age+='' 
elif days == 1:
    age+=str(days)+' day '
else:
    age+=str(days)+' days '

age+=str(hours)+':'+str(minutes)+':'+str(seconds)
print(age)

Python输出：24年5月19日15:20:13

编辑：

我找到了实现Netezza功能的方法，我已经粘贴在这里了。我还在期待另一种有效的方式！干杯!!!

感谢您的支持和建议！

python amazon-web-services amazon-redshift user-defined-functions netezza
2个回答
1
投票

不需要Python。这是一个封装逻辑的SQL UDF。如果单位复数对你很重要，你需要扩展它（mons vs mon）。

/*
    Postgres AGE() Function
*/
CREATE OR REPLACE FUNCTION f_postgres_age(TIMESTAMP, TIMESTAMP) 
RETURNS VARCHAR(64) 
STABLE AS $$
-- Input: '1994-04-04 20:10:52', '2018-09-24 11:31:05' 
-- Output: 24 years 5 mons 19 days 15:20:13
-- Input: '1994-10-04 20:10:52', '2019-06-12 11:31:05' 
-- Output: 24 years 8 mons 7 days 15:20:13
-- Check: SELECT '1994-10-04 20:10:52'::TIMESTAMP 
--               + INTERVAL '24 years' + INTERVAL '8 months' + INTERVAL '7 days'
--               + INTERVAL '15 hours' + INTERVAL '20 minutes' + INTERVAL '13 seconds';
-- Result: 2019-06-12 11:31:05
SELECT CASE WHEN DATEDIFF(year, DATE_TRUNC('year', $1)
                              , DATE_TRUNC('year', CASE WHEN DATEPART(month, $1) > DATEPART(month, $2)
                                                        THEN $2 - INTERVAL '1 Year' ELSE $2 END)) > 0
            THEN DATEDIFF(year, DATE_TRUNC('year', $1)
                              , DATE_TRUNC('year', CASE WHEN DATEPART(month, $1) > DATEPART(month, $2)
                                                        THEN $2 - INTERVAL '1 Year' ELSE $2 END)) || ' years '
       ELSE '' END
    || CASE WHEN ABS(  DATEDIFF(month, DATE_TRUNC('month', $1), DATE_TRUNC('month', $2))
                     - DATEDIFF(month, DATE_TRUNC('year', $1)
                                     , DATE_TRUNC('year', CASE WHEN DATEPART(month, $1) > DATEPART(month, $2)
                                                               THEN $2 - INTERVAL '1 Year' ELSE $2 END))) > 0
            THEN DATEDIFF(month, DATE_TRUNC('month', $1), DATE_TRUNC('month', $2))
               - DATEDIFF(month, DATE_TRUNC('year', $1)
                               , DATE_TRUNC('year', CASE WHEN DATEPART(month, $1) > DATEPART(month, $2)
                                                         THEN $2 - INTERVAL '1 Year' ELSE $2 END)) || ' mons '
       ELSE '' END
    || CASE WHEN ABS( DATEDIFF(day, DATE_TRUNC('day', $1)+1, DATE_TRUNC('day', $2)) 
                    - DATEDIFF(day, DATE_TRUNC('month', $1), DATE_TRUNC('month', $2))) > 0
            THEN DATEDIFF(day, DATE_TRUNC('day', $1)+1, DATE_TRUNC('day', $2))
               - DATEDIFF(day, DATE_TRUNC('month', $1), DATE_TRUNC('month', $2)) || ' days '
       ELSE '' END
    || TO_CHAR((TIMESTAMP 'epoch' 
                + ( DATEDIFF(second, $1, DATE_TRUNC('day', $1)+1 )
                  + DATEDIFF(second, DATE_TRUNC('day', $2), $2) )
                * INTERVAL '1 Second '),'HH24:MI:SS') age
$$ LANGUAGE SQL
;
-1
投票

我找到了让输出与Netezza一样的方法！我们需要使用不同的输入创建4个不同的UDF！这里我添加了UDF（TIMESTAMP，TIMESTAMP）

create or replace function AGE_UDF_V2 (START_DATE TIMESTAMP, END_DATE TIMESTAMP)
  returns VARCHAR
stable
as $$
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 26 12:59:24 2018

@author: pnataraj
"""

from dateutil import relativedelta
from dateutil.parser import parse

if (START_DATE is None or END_DATE is None):
    return None

else:
    START_DATE = str(START_DATE).strip()
    END_DATE = str(END_DATE).strip()

    START_DATE = parse(START_DATE)
    END_DATE = parse(END_DATE)

    difference = relativedelta.relativedelta(START_DATE, END_DATE)
    years = difference.years
    months = difference.months
    days = difference.days
    hours = difference.hours
    minutes = difference.minutes
    seconds = difference.seconds
    age=''
    if years != 0:
        if years == 1 or years == -1:
            age+=str(years)+' year '
        else:
            age+=str(years)+' years '

    if months != 0:
        if months == 1 or months == -1:
            age+=str(months)+' mon '
        else:
            age+=str(months)+' mons '

    if days != 0: 
        if days == 1 or days == -1:
            age+=str(days)+' day '
        else:
            age+=str(days)+' days '

    if (hours !=0 or minutes !=0 or seconds != 0):
        if (hours < 0 or minutes < 0 or seconds < 0):
            age+=str("-"+format(abs(hours),"02")+":"+format(abs(minutes),"02")+":"+format(abs(seconds),"02"))
        else:
            age+=str(format(hours,"02")+":"+format(minutes,"02")+":"+format(seconds,"02"))
    elif(hours == 0 and minutes ==0 and seconds == 0):
        if len(age)>0:
            age = age
        else:
            age = "00:00:00"

    return age.strip()
$$ language plpythonu;

感谢所有的建议和帮助！希望对那些正在使用Nz进行AWS RS迁移的人有所帮助！

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