This question already has answers here:
When or Why to use a "SET DEFINE OFF" in Oracle Database
(2 answers)
How to avoid variable substitution in Oracle SQL Developer
(5 answers)
Closed 1 year ago.
This is for Oracle, not MySql in case there might be a slight difference.
I am trying to get paths of this url
select regexp_substr(
'https://domain/l202108091338.csv.zip?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAQ6R%2F20210820%2Fca-central-1%2Fs3%2Faws4_request&X-Amz-Date=20210820T223109Z&X-Amz-Expires=300&X-Amz-Signature=bd3d40250afdc751c67c58302a2ca152c7&X-Amz-SignedHeaders=host',
'[^=]+',
1,
2
) from dual;
the above command outputs
AWS4-HMAC-SHA256-Amz-Credential
but the actually value is
AWS4-HMAC-SHA256&X-Amz-Credential
How can I make the output includes &X too?
Thanks in advance for any suggestions and help.
Related
This question already has answers here:
Perform calculation only if both cells are not blank with arrayformula? [duplicate]
(3 answers)
ArrayFormula and "AND" Formula in Google Sheets
(4 answers)
Closed 4 months ago.
So I am trying to get this Arrayformula to work so I can plot this formula instead of calculating every Y myself.
=arrayformula(Sum(IF(Z3:Z294>Y$1;IFS(AB3:AB294>0;0;Z3:Z294>Y$1;Y3:Y294-(F3:F294*Y$1));0)-(Sum(IF(and(J3:J294>0;Z3:Z294>Y$1);F3:F294;0)))))
It gives me the (correct) return of this part:
Sum(IF(Z3:Z294>Y$1;IFS(AB3:AB294>0;0;Z3:Z294>Y$1;Y3:Y294-(F3:F294*Y$1));0)
But it doesn't subtract the second part:
-(Sum(IF(and(J3:J294>0;Z3:Z294>Y$1);F3:F294;0)))))
I am quite new to extensive Excel/Sheets formulas so I have no idea how to get this to work, It is also quite weird that the second part doesn't add up even seperate from the first part. So this also doesn't work:
=arrayformula(Sum(IF(and(J3:J294>0;Z3:Z294>Y$1);F3:F294;0)))
I hope it makes some sense without any context, thanks in advance!
Have a great rest of your day,
P.S. Please ignore simple mistakes, I don't code that often in Sheets ;)
AND is not supported. instead of
and(J3:J294>0;Z3:Z294>Y$1)
do this:
(J3:J294>0)*(Z3:Z294>Y$1)
This question already has an answer here:
I get Formula parse error with all formulas
(1 answer)
Closed 4 months ago.
I am trying to use IF statement in a Google Sheet, but whatever I am typing in the IF I am always getting a formula parse error. Do I have to activate something, or am I missing something ?
A1 & B1 have the same type (integer)
I have no idea why.
Thanks in advance
instead of , use ; - this should solve your issue asap
This question already has answers here:
Regex to match Date
(7 answers)
Closed 2 years ago.
I made a google forms which i asked a date of birth like dd/mm/yyyy.
I'm looking for a RegEx that allow every date from 01/01/1900 to 31/12/2015 but refuse every date who contains this 5 years 2016, 2017, 2018, 2019, 2020.
Does someone have an idea ?
Thanks for help.
If you really only want to check dates for years 1900-2015, it suffices to code
\b(\d{1,2}/\d{1,2}/(19\d{2}|200\d|201[0-5]))\b
The \b...\b bound is less restrictive than ^...$
Because the previous answer did not specify any year bounds, they need to be added, for example, 1900…2099 (excluding 2016-2020)
\b(?!2016|2017|2018|2019|2020)(\d{1,2}/\d{1,2}/(19|20)\d{2})\b
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 4 years ago.
I have a series of dates that I am trying to separate into years, months and days. These dates are in the yyyy-mm-dd format. I'm not very familiar with RegEx, but I have tried (\dddd)\-(\dd)\-(\dd).
Any help is appreciated.
Accepted answer would also match invalid dates like 1234-56-78, or if mm & dd were in the wrong positions.
([1-2][0-9]{3})\-([0-1][0-9])\-([0-3][0-9])
Does a little more validation.
This question already has answers here:
How to query MongoDB with "like"
(45 answers)
Closed 6 years ago.
how to check the second character match in mongoDB. in sql we are using like '_a%'. how to use in mongoDB??
you should use syntax like below:-
db.users.find({"name": /.a./})
or like >> db.users.find({"name": /a/})
if you want to search for character "a"