How do I add leading zero to a 7 digit date using When clause in SQL - amazon-athena

I have a column of dates, which have some 7 digits and 8 digit characters as type string.
dates_stamp
9022023
10012023
14012023
4122022
How do I add a leading zero to dates?
select CASE
WHEN last_contact_date BETWEEN 0 AND 9999999
THEN ('0' , CAST(last_contact_date as VARCHAR(8)))
ELSE "FAIL"
END
FROM table1
This is my concept but not sure how to make it work

String functions in Amazon Athena are based on Trino. It contains an lpad function that should suit your needs.
lpad(Cast(last_contact_date as VarChar(8), last_contact_date, 8, '0')
I would strongly suggest complying with the ISO 8601 international standard. "YYYY-MM-DD" retains a way to sort after conversion to string.
substring(to_iso8601(last_contact_date),1,10)

Related

Convert Dates with Regular Expression 1/1/18 -> 2018-01-01

I have dates that are in m/d/yy but need to convert them to yyyy-dd-mm format. I have the following regex
(\d{1,2})\/(\d{1,2})\/(\d{1,2}) with this replace 20$3-$1-$2. I'm assuming year 2000 which is okay but this breaks for single digit days and months.
Is there any way to prefix the days and months with 0 if the match is on a single digit?
It might be usefull for you.
Add zero padding to single digit dates:
Regex to add leading zero in date record

substr at Amazon Athena

At Amazon Athena, I want to extract only the character string "2017-07-27" from the character string "2017-07-27 12:10:08".
SELECT SUBSTRING (event_datetime.s, 0, 10) FROM production limit 10
I tried it like this which only returns numbers 0 to 10.
At Athena, is it possible to cut character strings? If so, how can I do it?
Or, if you know how to cast "2017-07-27 12:10:08" to date type, that's fine.
Thank you.
You can use SUBSTR to substring a column value.
Here is the string function reference page.
In your case, this would lead to the following statement:
SELECT SUBSTR(event_datetime.s, 1, 10) FROM production limit 10
NOTE that the index position of the first character is 1 (not zero), the same as in standard SQL.

Regular expression numeric format with length size of 22 and decimal size of 3 regex

I am using a profiling tool to validate the data inside the tables. I want to check to see if the data in the tables matches the requirement of checking to see if the values input for current market value amount usd are length size of 22 and decimal size of 3. I am using ataccamba profiling tool which picks up the variables.
iif(
matches(#"^\d{22}.\d{3}$", Current_Market_Value_Amount__USD_),
true,false
I am looking to make this validation rule satisfy the requirement of:
Current Market Value Amount (USD) attribute should be in numeric format with length size of 22 and decimal size of 3
You need to escape the ., by doing \. because it is a meta-character that has the meaning 'match any'. To mean the literal . you prefix it with the \ escape character.
On a side note: your [0-9] repetitions can be replaced easily:
^[0-9]{18}\.[0-9]{3}$
This assumes the total string length must be 22.
You don't need more than:
^\d{22}\.\d{3}$
Since \d matches a digit, number inside bracket ensure that the preceding item is repeated x times (or from m to n in case of {m,n}).
By the way, Regex are not supported inside formula, I suggest you to take a look at How do I get regex support in excel via a function, or custom function?

How to convert numbers in a character variable to Numeric in sas

Can anyone help me to resolve this?
I have a very large raw dataset with a character variable that contains text strings along with numbers & dates defined in character format. Now I want to process the dataset and create a new numeric variable and populate values only when the text in the actual variable is either a number or a date value. Otherwise missing
RAWDATA:
ACTUAL_VARIABLE NEW_NUM_VARIABLE(Expected Values)
------------------ ---------------------------------
ODed on pills threw them all up - 2006
Y
1 1
5 5
ODed on pills
6 6
Less than once a week
N
N
2006-11-12 2006-11-12
Many Thanks in Advance
The easy way to do it (if you know the specific date format) is to use the input function.
09:27
If put(input(var,??yymmdd10.),yymmdd10.)=var then its a date!
else if input(var,best.) ne . then its a number.
Otherwiseits a character string.
This isn't as straightforward as it first looks, so I understand why it would be difficult to search for an answer. Just extracting a number is pretty easy, but when dates are included it becomes a bit more complicated (particularly when the format entered could change, e.g. yyyy-mm-dd, dd-mm-yyyy, dd/mm/yy etc).
One thing to note first. If you want to store the new values as a numeric field then you can't show a mix of numbers and dates. Dates are stored as numbers and formatted to show the date, but you can't apply a format at row level. Therefore I would suggest creating 2 new columns, 1 for numbers and 1 for dates.
My preferred approach is to use the anyalpha function to exclude any records with an alphabetic character, followed by the anypunct function to identify if a punctuation character exists (this should identify dates rather than just numbers). The anydtdte informat is then used to extract the date, this is a very useful informat as it reads dates stored in different ways (as per my note above).
There are clearly some caveats with this method.
If any numbers contain decimals then my method would incorrectly treat these as dates, therefore only integers will be assigned correctly.
It won't pick up dates that contain the month as words, e.g. 15-May-2015, as the anyalpha function would exclude them. They will need to contain numbers only, separated by any punctuation character.
Here's my code.
/* create initial dataset */
data have;
input actual_variable $ 50.;
datalines;
ODed on pills threw them all up - 2006
Y
1
5
ODed on pills
6
Less than once a week
N
N
2006-11-12
;
run;
/* extract dates and numbers */
data want;
set have;
if not anyalpha(actual_variable) then do; /* exclude records with an alphabetic character */
if anypunct(actual_variable) then new_date_variable = input(actual_variable,anydtdte10.); /* if a punctuation character exists then read in as a date */
else new_num_variable = input(actual_variable,best12.); /* else read in as a number */
end;
format new_date_variable yymmdd10.; /* show date field in required format */
run;

Regular Expression to match valid dates

I'm trying to write a regular expression that validates a date. The regex needs to match the following
M/D/YYYY
MM/DD/YYYY
Single digit months can start with a leading zero (eg: 03/12/2008)
Single digit days can start with a leading zero (eg: 3/02/2008)
CANNOT include February 30 or February 31 (eg: 2/31/2008)
So far I have
^(([1-9]|1[012])[-/.]([1-9]|[12][0-9]|3[01])[-/.](19|20)\d\d)|((1[012]|0[1-9])(3[01]|2\d|1\d|0[1-9])(19|20)\d\d)|((1[012]|0[1-9])[-/.](3[01]|2\d|1\d|0[1-9])[-/.](19|20)\d\d)$
This matches properly EXCEPT it still includes 2/30/2008 & 2/31/2008.
Does anyone have a better suggestion?
Edit: I found the answer on RegExLib
^((((0[13578])|([13578])|(1[02]))[\/](([1-9])|([0-2][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/](([1-9])|([0-2][0-9])|(30)))|((2|02)[\/](([1-9])|([0-2][0-9]))))[\/]\d{4}$|^\d{4}$
It matches all valid months that follow the MM/DD/YYYY format.
Thanks everyone for the help.
This is not an appropriate use of regular expressions. You'd be better off using
[0-9]{2}/[0-9]{2}/[0-9]{4}
and then checking ranges in a higher-level language.
Here is the Reg ex that matches all valid dates including leap years. Formats accepted mm/dd/yyyy or mm-dd-yyyy or mm.dd.yyyy format
^(?:(?:(?:0?[13578]|1[02])(\/|-|\.)31)\1|(?:(?:0?[1,3-9]|1[0-2])(\/|-|\.)(?:29|30)\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:0?2(\/|-|\.)29\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:(?:0?[1-9])|(?:1[0-2]))(\/|-|\.)(?:0?[1-9]|1\d|2[0-8])\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$
courtesy Asiq Ahamed
I landed here because the title of this question is broad and I was looking for a regex that I could use to match on a specific date format (like the OP). But I then discovered, as many of the answers and comments have comprehensively highlighted, there are many pitfalls that make constructing an effective pattern very tricky when extracting dates that are mixed-in with poor quality or non-structured source data.
In my exploration of the issues, I have come up with a system that enables you to build a regular expression by arranging together four simpler sub-expressions that match on the delimiter, and valid ranges for the year, month and day fields in the order you require.
These are :-
Delimeters
[^\w\d\r\n:]
This will match anything that is not a word character, digit character, carriage return, new line or colon. The colon has to be there to prevent matching on times that look like dates (see my test Data)
You can optimise this part of the pattern to speed up matching, but this is a good foundation that detects most valid delimiters.
Note however; It will match a string with mixed delimiters like this 2/12-73 that may not actually be a valid date.
Year Values
(\d{4}|\d{2})
This matches a group of two or 4 digits, in most cases this is acceptable, but if you're dealing with data from the years 0-999 or beyond 9999 you need to decide how to handle that because in most cases a 1, 3 or >4 digit year is garbage.
Month Values
(0?[1-9]|1[0-2])
Matches any number between 1 and 12 with or without a leading zero - note: 0 and 00 is not matched.
Date Values
(0?[1-9]|[12]\d|30|31)
Matches any number between 1 and 31 with or without a leading zero - note: 0 and 00 is not matched.
This expression matches Date, Month, Year formatted dates
(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})
But it will also match some of the Year, Month Date ones. It should also be bookended with the boundary operators to ensure the whole date string is selected and prevent valid sub-dates being extracted from data that is not well-formed i.e. without boundary tags 20/12/194 matches as 20/12/19 and 101/12/1974 matches as 01/12/1974
Compare the results of the next expression to the one above with the test data in the nonsense section (below)
\b(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})\b
There's no validation in this regex so a well-formed but invalid date such as 31/02/2001 would be matched. That is a data quality issue, and as others have said, your regex shouldn't need to validate the data.
Because you (as a developer) can't guarantee the quality of the source data you do need to perform and handle additional validation in your code, if you try to match and validate the data in the RegEx it gets very messy and becomes difficult to support without very concise documentation.
Garbage in, garbage out.
Having said that, if you do have mixed formats where the date values vary, and you have to extract as much as you can; You can combine a couple of expressions together like so;
This (disastrous) expression matches DMY and YMD dates
(\b(0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](0?[1-9]|1[0-2])[^\w\d\r\n:](\d{4}|\d{2})\b)|(\b(0?[1-9]|1[0-2])[^\w\d\r\n:](0?[1-9]|[12]\d|30|31)[^\w\d\r\n:](\d{4}|\d{2})\b)
BUT you won't be able to tell if dates like 6/9/1973 are the 6th of September or the 9th of June. I'm struggling to think of a scenario where that is not going to cause a problem somewhere down the line, it's bad practice and you shouldn't have to deal with it like that - find the data owner and hit them with the governance hammer.
Finally, if you want to match a YYYYMMDD string with no delimiters you can take some of the uncertainty out and the expression looks like this
\b(\d{4})(0[1-9]|1[0-2])(0[1-9]|[12]\d|30|31)\b
But note again, it will match on well-formed but invalid values like 20010231 (31th Feb!) :)
Test data
In experimenting with the solutions in this thread I ended up with a test data set that includes a variety of valid and non-valid dates and some tricky situations where you may or may not want to match i.e. Times that could match as dates and dates on multiple lines.
I hope this is useful to someone.
Valid Dates in various formats
Day, month, year
2/11/73
02/11/1973
2/1/73
02/01/73
31/1/1973
02/1/1973
31.1.2011
31-1-2001
29/2/1973
29/02/1976
03/06/2010
12/6/90
month, day, year
02/24/1975
06/19/66
03.31.1991
2.29.2003
02-29-55
03-13-55
03-13-1955
12\24\1974
12\30\1974
1\31\1974
03/31/2001
01/21/2001
12/13/2001
Match both DMY and MDY
12/12/1978
6/6/78
06/6/1978
6/06/1978
using whitespace as a delimiter
13 11 2001
11 13 2001
11 13 01
13 11 01
1 1 01
1 1 2001
Year Month Day order
76/02/02
1976/02/29
1976/2/13
76/09/31
YYYYMMDD sortable format
19741213
19750101
Valid dates before Epoch
12/1/10
12/01/660
12/01/00
12/01/0000
Valid date after 2038
01/01/2039
01/01/39
Valid date beyond the year 9999
01/01/10000
Dates with leading or trailing characters
12/31/21/
31/12/1921AD
31/12/1921.10:55
12/10/2016 8:26:00.39
wfuwdf12/11/74iuhwf
fwefew13/11/1974
01/12/1974vdwdfwe
01/01/99werwer
12321301/01/99
Times that look like dates
12:13:56
13:12:01
1:12:01PM
1:12:01 AM
Dates that runs across two lines
1/12/19
74
01/12/19
74/13/1946
31/12/20
08:13
Invalid, corrupted or nonsense dates
0/1/2001
1/0/2001
00/01/2100
01/0/2001
0101/2001
01/131/2001
31/31/2001
101/12/1974
56/56/56
00/00/0000
0/0/1999
12/01/0
12/10/-100
74/2/29
12/32/45
20/12/194
2/12-73
Maintainable Perl 5.10 version
/
(?:
(?<month> (?&mon_29)) [\/] (?<day>(?&day_29))
| (?<month> (?&mon_30)) [\/] (?<day>(?&day_30))
| (?<month> (?&mon_31)) [\/] (?<day>(?&day_31))
)
[\/]
(?<year> [0-9]{4})
(?(DEFINE)
(?<mon_29> 0?2 )
(?<mon_30> 0?[469] | (11) )
(?<mon_31> 0?[13578] | 1[02] )
(?<day_29> 0?[1-9] | [1-2]?[0-9] )
(?<day_30> 0?[1-9] | [1-2]?[0-9] | 30 )
(?<day_31> 0?[1-9] | [1-2]?[0-9] | 3[01] )
)
/x
You can retrieve the elements by name in this version.
say "Month=$+{month} Day=$+{day} Year=$+{year}";
( No attempt has been made to restrict the values for the year. )
To control a date validity under the following format :
YYYY/MM/DD or YYYY-MM-DD
I would recommand you tu use the following regular expression :
(((19|20)([2468][048]|[13579][26]|0[48])|2000)[/-]02[/-]29|((19|20)[0-9]{2}[/-](0[4678]|1[02])[/-](0[1-9]|[12][0-9]|30)|(19|20)[0-9]{2}[/-](0[1359]|11)[/-](0[1-9]|[12][0-9]|3[01])|(19|20)[0-9]{2}[/-]02[/-](0[1-9]|1[0-9]|2[0-8])))
Matches
2016-02-29 | 2012-04-30 | 2019/09/31
Non-Matches
2016-02-30 | 2012-04-31 | 2019/09/35
You can customise it if you wants to allow only '/' or '-' separators.
This RegEx strictly controls the validity of the date and verify 28,30 and 31 days months, even leap years with 29/02 month.
Try it, it works very well and prevent your code from lot of bugs !
FYI : I made a variant for the SQL datetime. You'll find it there (look for my name) : Regular Expression to validate a timestamp
Feedback are welcomed :)
Sounds like you're overextending regex for this purpose. What I would do is use a regex to match a few date formats and then use a separate function to validate the values of the date fields so extracted.
Perl expanded version
Note use of /x modifier.
/^(
(
( # 31 day months
(0[13578])
| ([13578])
| (1[02])
)
[\/]
(
([1-9])
| ([0-2][0-9])
| (3[01])
)
)
| (
( # 30 day months
(0[469])
| ([469])
| (11)
)
[\/]
(
([1-9])
| ([0-2][0-9])
| (30)
)
)
| ( # 29 day month (Feb)
(2|02)
[\/]
(
([1-9])
| ([0-2][0-9])
)
)
)
[\/]
# year
\d{4}$
| ^\d{4}$ # year only
/x
Original
^((((0[13578])|([13578])|(1[02]))[\/](([1-9])|([0-2][0-9])|(3[01])))|(((0[469])|([469])|(11))[\/](([1-9])|([0-2][0-9])|(30)))|((2|02)[\/](([1-9])|([0-2][0-9]))))[\/]\d{4}$|^\d{4}$
if you didn't get those above suggestions working, I use this, as it gets any date I ran this expression through 50 links, and it got all the dates on each page.
^20\d\d-(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)-(0[1-9]|[1-2][0-9]|3[01])$
This regex validates dates between 01-01-2000 and 12-31-2099 with matching separators.
^(0[1-9]|1[012])([- /.])(0[1-9]|[12][0-9]|3[01])\2(19|20)\d\d$
var dtRegex = new RegExp(/[1-9\-]{4}[0-9\-]{2}[0-9\-]{2}/);
if(dtRegex.test(date) == true){
var evalDate = date.split('-');
if(evalDate[0] != '0000' && evalDate[1] != '00' && evalDate[2] != '00'){
return true;
}
}
Regex was not meant to validate number ranges(this number must be from 1 to 5 when the number preceding it happens to be a 2 and the number preceding that happens to be below 6).
Just look for the pattern of placement of numbers in regex. If you need to validate is qualities of a date, put it in a date object js/c#/vb, and interogate the numbers there.
I know this does not answer your question, but why don't you use a date handling routine to check if it's a valid date? Even if you modify the regexp with a negative lookahead assertion like (?!31/0?2) (ie, do not match 31/2 or 31/02) you'll still have the problem of accepting 29 02 on non leap years and about a single separator date format.
The problem is not easy if you want to really validate a date, check this forum thread.
For an example or a better way, in C#, check this link
If you are using another platform/language, let us know
Perl 6 version
rx{
^
$<month> = (\d ** 1..2)
{ $<month> <= 12 or fail }
'/'
$<day> = (\d ** 1..2)
{
given( +$<month> ){
when 1|3|5|7|8|10|12 {
$<day> <= 31 or fail
}
when 4|6|9|11 {
$<day> <= 30 or fail
}
when 2 {
$<day> <= 29 or fail
}
default { fail }
}
}
'/'
$<year> = (\d ** 4)
$
}
After you use this to check the input the values are available in $/ or individually as $<month>, $<day>, $<year>. ( those are just syntax for accessing values in $/ )
No attempt has been made to check the year, or that it doesn't match the 29th of Feburary on non leap years.
If you're going to insist on doing this with a regular expression, I'd recommend something like:
( (0?1|0?3| <...> |10|11|12) / (0?1| <...> |30|31) |
0?2 / (0?1| <...> |28|29) )
/ (19|20)[0-9]{2}
This might make it possible to read and understand.
/(([1-9]{1}|0[1-9]|1[0-2])\/(0[1-9]|[1-9]{1}|[12]\d|3[01])\/[12]\d{3})/
This would validate for following -
Single and 2 digit day with range from 1 to 31. Eg, 1, 01, 11, 31.
Single and 2 digit month with range from 1 to 12. Eg. 1, 01, 12.
4 digit year. Eg. 2021, 1980.
A slightly different approach that may or may not be useful for you.
I'm in php.
The project this relates to will never have a date prior to the 1st of January 2008. So, I take the 'date' inputed and use strtotime(). If the answer is >= 1199167200 then I have a date that is useful to me. If something that doesn't look like a date is entered -1 is returned. If null is entered it does return today's date number so you do need a check for a non-null entry first.
Works for my situation, perhaps yours too?