substr at Amazon Athena - 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.

Related

Extract multiple substrings of numbers of a specific length from string in Google Sheets

I'd need to split or extract only numbers made of 8 digits from a string in Google Sheets.
I've tried with SPLIT or REGEXREPLACE but I can't find a way to get only the numbers of that length, I only get all the numbers in the string!
For example I'm using
=SPLIT(lower(N2),"qwertyuiopasdfghjklzxcvbnm`-=[]\;' ,./!:##$%^&*()")
but I get all the numbers while I only need 8 digits numbers.
This may be a test value:
00150412632BBHBBLD 12458 32354 1312548896 ACT inv 62345471
I only need to extract "62345471" and nothing else!
Could you please help me out?
Many thanks!
Please use the following formula for a single cell.
Drag it down for more cells.
=INDEX(TRANSPOSE(QUERY(TRANSPOSE(IF(LEN(SPLIT(REGEXREPLACE(A2&" ","\D+"," ")," "))=8,
SPLIT(REGEXREPLACE(A2&" ","\D+"," ")," "),"")),"where Col1 is not null ",0)))
Functions used:
QUERY
INDEX
TRANSPOSE
IF
LEN
SPLIT
REGEXREPLACE
If you only need to do this for one cell (or you have your heart set on dragging the formula down into individual cells), use the following formula:
=REGEXEXTRACT(" "&N2&" ","\s(\d{8})\s")
However, I suspect you want to process the eight-digit number out of all cells running N2:N. If that is the case, clear whatever will be your results column (including any headers) and place the following in the top cell of that otherwise cleared results column:
=ArrayFormula({"Your Header"; IF(N2:N="",,IFERROR(REGEXEXTRACT(" "&N2:N&" ","\s(\d{8})\s")))})
Replace the header text Your Header with whatever you want your actual header text to be. The formula will show that header text and will return all results for all rows where N2:N is not null. Where no eight-digit number is found, null will be returned.
By prepending and appending a space to the N2:N raw strings before processing, spaces before and after string components can be used to determine where only eight digits exist together (as opposed to eight digits within a longer string of digits).
The only assumption here is that there are, in fact, spaces between string components. I did not assume that the eight-digit number will always be in a certain position (e.g., first, last) within the string.
Try this, take a look at Example sheet
=FILTER(TRANSPOSE(SPLIT(B2," ")),LEN(TRANSPOSE(SPLIT(B2," ")))=8)
Or this to get them all.
=JOIN(" ,",FILTER(TRANSPOSE(SPLIT(B2," ")),LEN(TRANSPOSE(SPLIT(B2," ")))=8))
Explanation
SPLIT with the dilimiter set to " " space TRANSPOSE and FILTER TRANSPOSE(SPLIT(B2," ") with the condition1 set to LEN(TRANSPOSE(SPLIT(B2," "))) is = 8
JOIN the outputed column whith " ," to gat all occurrences of number with a length of 8
Note: to get the numbers with the length of N just replace 8 in the FILTER function with a cell refrence.
Using this on a cell worked just fine for me:
(cell_with_data)=REGEXEXTRACT(A1,"[0-9]{8}$")

Retrieving the 12th through 14th characters from a long strong using ONLY regex - Grafana variable

I have a small issue, I am trying to get specific characters from a long string using regex but I am having trouble.
Workflow
Prometheus --> Grafana --> Variable (using regex)
I can't use anything other than Regex expressions to achieve this result
I am currently using this expression to grab the long string from some json output:
.*channel_id="(.*?)".*
FROM THIS
{account_id="XXXXXXX-xxxx-xxxx-xxxx-xxxxxxxxxx",account_name="testalpha",channel_id="s0022110430col0901241usa",channel_abbr="s0022109430col}
This returns a string that's ALWAYS 24 characters long:
s0022110430col0901241usa
PROBLEM:
I need to grab the 3 letters 'col' and 'usa' as they are the two teams that are playing, ideally I would be able to pipe the results from the first regex to get these values (the position is key, since the first value will ALWAYS be the 12-14th characters and the second value is the last 3 characters) if I could output these values in uppercase with the string "vs" in between to create a string such as:
COL vs USA
or
ARG vs BRA
I am open to any and every suggestion anyone may have
Thank you!
PS - The uppercase thing is 'nice to have' BUT not needed
I'm still learning RegEx, so this is all I could come up with:
For the col (first team):
(?<=(channel_id=".{11}))\w{3}
For the usa (second team):
(?<=(channel_id=".{21}))\w{3}
Can you define the channel_id?
It begins with 's' and then there are many numbers. If they are always numbers, you can use this regex:
channel_id=".[0-9]+([a-z]+)[0-9]+([a-z]+)
You will get 2 groups, one with "col" and the other with "usa".
Edit:
Or if you just know, that you have always the same size, you can use something like:
channel_id=".{11}([a-z]+).{7}([a-z]+)

PostgreSQL - finding string using regular expression

What I am looking to do is to, within Postgres, search a column for a string (an account number). I have a log table, which has a parameters column that takes in parameters from the application. It is a paragraph of text and one of the parameters stored in the column is the account number.
The position of the account number is not consistent in the text and some rows in this table have nothing in the column (since no parameters are passed on certain screens). The account number has the following format: L1234567899. So for the account number, the first character is a letter and then it is followed by ten digits.
I am looking for a way to extract the account number alone from this column so I can use it in a view for a report.
So far what I have tried is getting it into an array, but since the position changes, I cannot count on it being in the same place.
select foo from regexp_split_to_array(
(select param from log_table where id = 9088), E'\\s+') as foo
You can use regexp_match() to achieve that result.
(regexp_match(foo,'[A-Z][0-9]{10}'))[1]
DBFiddle
Use substring to pull out the match group.
select substring ('column text' from '[A-Z]\d{10}')
Reference: PostgreSQL regular expression capture group in select

Pattern Matching in MS Access: Is there an "or" operator?

I've tried searching online for the answer to this, but my Google-fu has failed me.
I have an Access database containing records represented by a string. The first 3 characters of that string are a 3-digit representation of the 366-day calendar date on which the record was created (000-366...yes, leap days count).
I'm having trouble coming up with the correct pattern match to include in a query that matches a 3-digit substring that can be between 000 and 366, where you don't lose the significant figures.
I know the query would be something like:
SELECT * FROM myTable WHERE Field1 LIKE "^[0-2]## or 3[0-5]# or 36[0-6]*";
...but I can't find any resource that says, in MS Access, what the "or" operator is. I tried "||" (double pipe) and "|" (single pipe), neither of which worked.
Is there an "or" operator that can be used with a MS Access pattern match?
The LIKE operator in Access is pretty limited, and doesn't support most of the features more 'fully-fledged' regular expression engines provide.
Instead, use multiple conditions in your WHERE clause like this:
SELECT *
FROM myTable
WHERE Field1 LIKE "[0-2]##*" OR
Field1 LIKE "3[0-5]#*" OR
Field1 LIKE "36[0-6]*"
Another alternative is to simply extract the first 3 characters to a string, convert them to an integer and test to see if their value is within the acceptable range.
Why not just pull the first three characters?
SELECT * FROM myTable WHERE CInt(Left(Field1,3)) <= 366
http://www.techonthenet.com/access/functions/datatype/cint.php

extract number from string in Oracle

I am trying to extract a specific text from an Outlook subject line. This is required to calculate turn around time for each order entered in SAP. I have a subject line as below
SO# 3032641559 FW: Attached new PO 4500958640- 13563 TYCO LJ
My final output should be like this: 3032641559
I have been able to do this in MS excel with the formulas like this
=IFERROR(INT(MID([#[Normalized_Subject]],SEARCH(30,[#[Normalized_Subject]]),10)),"Not Found")
in the above formula [#[Normalized_Subject]] is the name of column in which the SO number exists. I have asked to do this in oracle but I am very new to this. Your help on this would be greatly appreciated.
Note: in the above subject line the number 30 is common in every subject line.
The last parameter of REGEXP_SUBSTR() indicates the sub-expression you want to pick. In this case you can't just match 30 then some more numbers as the second set of digits might have a 30. So, it's safer to match the following, where x are more digits.
SO# 30xxxxxx
As a regular expression this becomes:
SO#\s30\d+
where \s indicates a space \d indicates a numeric character and the + that you want to match as many as there are. But, we can use the sub-expression substringing available; in order to do that you need to have sub-expressions; i.e. create groups where you want to split the string:
(SO#\s)(30\d+)
Put this in the function call and you have it:
regexp_substr(str, '(SO#\s)(30\d+)', 1, 1, 'i', 2)
SQL Fiddle