I am trying to extract the second date displayed in this string, however my code keeps extracting just the first date in gsheet:
String: BOT +1 1/1 CUSTOM IWM 100 12 SEP 22/7 SEP 22 184/184 PUT/CALL #6.13
This is my code: =REGEXEXTRACT(A3,"(\d{1,2}\s+[A-Za-z]+\s\d{2,4})")
my result: 12 SEP 22
Desired result should be: 7 SEP 22
Appreciate the help, thanks in advance!
Considering you have already a working formula for detecting dates, you can try adding first outside of the parentheses the same structure. So it will look for the first date, then .+ will consider that there will be some characters in between, and then your working pattern between parenthesis. Then only that last part will be extracted:
=REGEXEXTRACT(A3,"\d{1,2}\s+[A-Za-z]+\s\d{2,4}.+(\d{1,2}\s+[A-Za-z]+\s\d{2,4})")
Here's one approach to dynamically extract N number of dates within your string OR extract the 2nd or 3rd date pattern as per the requirement.
=index(if(len(A:A),lambda(y,regexextract(y,lambda(z,regexreplace(y,"(?i)("&z&")","($1)"))("\d{1,2}\s"&JOIN("\s\d{2}|\d{1,2}\s",INDEX(TEXT(SEQUENCE(12,1,DATE(2022,1,1),31),"MMM")))&"\s\d{2}")))(regexreplace(A:A,"[\(\)/+]","")),))
if its to pick specific number pattern, wrap the formula within index + number as shown in the screenshot
=index(formula,,pattern number)
To extract just the second date, you can modify the code as follows:
=REGEXEXTRACT(A3,"\d{1,2}\s+[A-Za-z]+\s\d{2,4}.*(\d{1,2}\s+[A-Za-z]+\s\d{2,4})")
This regular expression \d{1,2}\s+[A-Za-z]+\s\d{2,4}.*(\d{1,2}\s+[A-Za-z]+\s\d{2,4}) will match the first date and the second date in the string, and then extract just the second date.
Related
In Google BigQuery, I have to list all tables that contain hours between 02 till 23 hour in their names.
Aka match any hour in this range [02->23], so I created the following regex:
([0-2][2-9])
But the problem is it will skip the hour 10 and 11, and REGEX_EXTRACT returns null value for those “unparsable?” tables.
I could try to match all hours first, then exclude hour 00 and 01. But I could’t find a way in regex to add them as exceptions..
([0-2][0-9])
What would you recommend/suggest in this case? Given that I can’t split this into 2 different regexes.
Thank you.
You could write the pattern as
\b(?:0[2-9]|1\d|2[0-3])\b
Regex demo
I am new to regex and having difficulty obtaining values that are caught in between spaces.
I am trying to get the values "field 1" "abc/def try" from the sameple data below just using regex
Currently im using (^.{18}\s+) to skip the first 18 characters, but am at at loss of how to do grab values with spaces between.
A1234567890 field 1 abc/def try
02021051812 12 test test 12 pass
3333G132021 no test test cancel
any help/pointers will be appreciated.
If this text has fixed-width columns, you can match and trim the column values knowing the amount of chars between start of string and the column text.
For example, this regex will work for the text you posted:
^(.*?)\s*(?<=.{19})(.*?)\s*(?<=^.{34})(.*?)\s*(?<=^.{46})
See the regex demo.
So, Column 2 starts at Position 19, Column 3 starts at Position 34 and Column 4 (end of string here) is at Position 46.
However, this regex is not that efficient, and it would be really great if the data format is fixed on the provider's side.
Given the not knowing if the data is always the same length I created the following, which will provide you with a group per column you might want to use:
^((\s{0,1}\S{1,})*)(\s{2,})((\s{0,1}\S{1,})*)(\s{2,})((\s{0,1}\S{1,})*)
Regex demo
I need to validate with regex a date in format yyyy-mm-dd (2019-12-31) that should be within the range 2019-12-20 - 2020-01-10.
What would be the regex for this?
Thanks
Regex only deal with characters. so we have to work out at each position in the date what are the valid characters.
The first part is easy. The first two characters have to be 20
Now it gets complicated the next character can be a 1 or a 2 but what follows depends on the value of that character so we split the rest of the regex into two sections the first if the third character matches 1 and the second if it matches 2
We know that if the third character is a 1 then what must follow is the characters 9-12- as the range starts at 2019-12-20 now for the day part. The 9th character is the tens for the day this can only be 2 or 3 as we are already in the last month and the minimum date is 20. The last character can be any digit 0-9. This gives us a day match of [23][0-9]. Putting this together we now have a pattern for years starting 2019 as 19-12-[23][0-9]
It the third character is a 2 then we can match up to the day part of the date a gain as the range ends in January. This gives us a partial match of 20-01- leaving us to work on the day part. Hear we know that the first character of the day can either be a 1 or 0 however if it's a 1 then the last character must be a 0 and if it's a 0 then the last character can only be in the range 1 to 9. This give us another alteration (?:0[1-9]|10) Putting the second part together we get 20-01-(?:0[1-9]|10).
Combining these together gives the final regex 20(?:19-12-[23][0-9]|20-01-(?:0[1-9]|10))
Note that I'm assuming that the date you are testing against is a validly formatted date.
Try this:
(2019|2020)\-(12|01)\-([0-3][0-9]|[0-9])
But be aware that this will allow number up to where the first digit is between zero and three and the second digit between zero and nine for the dd value. You could specify all numbers you want to allow (from 20 to 10) like this (20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10).
(2019|2020)\-(12|01)\-(20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10)
But honestly... Regular-Expressions are not the right tool for this. RegExp gives a mask to something, not a logical context. Use regex to extract the data/value from a string and validate those values using another language.
The above 2nd Regex will, f.e. match your dates, but also values outside of this range since there is no context between 2019|2020 and the second group 12|01 so they match values like 2019-12-11 but also 2020-12-11.
To only match the values you want this will be a really large regex like this (inner brackets only if you need them) ((2019)-(12)-(20)|(2019)-(12)-(21)|(2019)-(12)-(22)|...) and continue with all possible dates - and ask yourself: what would you do if you find such a regex in a project you have to work with ;)
Better solution (quick and dirty, there might be better solutions):
(?<yyyy>20[0-9]{2})\-(?<mm>[01][0-9]|[0-9])\-(?<dd>[0-3][0-9]|[0-9])
This way you have three named groups (yyyy, mm, dd) you can access and validate the matched values... The regex is smaller, you have a better association between code and regex and both are easier to maintain.
I am trying to use Regex to parse a series of strings to extract one or more text dates that may be in multiple formats. The strings will look something like the following:
24 Aug 2016: nno-emvirt010a/b; 16 Aug 2016 nnt-emvirt010a/b nnd-emvirt010a/b COSI-1.6.5
24.16 nno-emvirt010a/b nnt-emvirt010a/b nnd-emvirt010a/b EI.01.02.03\
9/23/16: COSI-1.6.5 Logs updated at /vobs/COTS/1.6.5/files/Status_2016-07-27.log, Status_2016-07-28.log, Status_2016-08-05.log, Status_2016-08-08.log
I am not concerned about validating the individual date fields; just extracting the date string. The part I am unable to figure out is how to not match on number sequences that match the pattern but aren’t dates (‘1.6.5’ in ex. (1) and 01.02.03 in ex. (2)) and dates that are part of a file name (2016-07-27 in ex. (3)). In each of these exception cases in my input data, the initial numbers are preceded by either a period(.), underscore (_) or dash (-), but I cannot determine how to use this to edit the pattern syntax to not match these strings.
The pattern I have that partially works is below. It will only ignore the non date matches if it starts with 1 digit as in example 1.
/[^_\.\(\/]\d{1,4}[/\-\.\s*]([1-9]|0[1-9]|[12][0-9]|3[01]|[a-z]{3})[/\-\.\s*]\d{1,4}/ig`
I am not sure about vba check if this works . seems they have given so much options : https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9781449327453/ch04s04.html
^(?:(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])|↵
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9]))/(?:[0-9]{2})?[0-9]{2}$
^(?:
# m/d or mm/dd
(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])
|
# d/m or dd/mm
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9])
)
# /yy or /yyyy
/(?:[0-9]{2})?[0-9]{2}$
According to the test strings you've presented, you can use the following regex
See this regex in use here
(?<=[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
This regex ensures that specific date formats are met and are preceded by nothing (beginning of the string) or by a non-word character (specifically a-z, A-Z, 0-9) or dot .. The date formats that will be matched are:
24 Aug 2016
24.16
9/23/16
The regex could be further manipulated to ensure numbers are in the proper range according to days/month, etc., however, I don't feel that is really necessary.
Edits
Edit 1
Since VBA doesn't support lookbehinds, you can use the following. The date is in capture group 1.
(?:[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
Edit 2
As per bulbus's comment below
(?:[^\w.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d{2,4})|(?:(?:\d{1,2}\/){2}\d{2,4})|(?:\d{2,4}(?:-\d{2}){2})|\d{2}\.\d{2})
Took liberty to edit that a bit.
replaced [^a-zA-Z\d.] with [^\w.], comes with added advantage of excluding dates with _2016-07-28.log
Due to 1 removed trailing condition (?=[^a-zA-Z\d.]).
Forced year digits from \d+ to \d{2,4}
Edit 3
Due to added conditions of the regex, I've made the following edits (to improve upon both previous edits). As per the OP:
The edited pattern above works in all but 2 cases:
it does not find dates with the year first (ex. 2016/07/11)
if the date is contained within parenthesis in the string, it returns the left parenthesis as part of the date (ex. match = (8/20/2016)
Can you provide the edit to fix these?
In the below regexes, I've changed years to \d+ in order for it to work on any year greater than or equal to 0.
See the code in use here
(?:[^\w.]|^)((?:\d{1,2}\s+[A-Z][a-z]{2}\s+\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:\/\d{1,2}){2})|(?:\d+(?:-\d{2}){2})|\d{2}\.\d+)
This regex adds the possibility of dates in the XXXX/XX/XX format where the date may appear first.
The reason you are getting ( as a match before the regex is the nature of the Full Match. You need to, instead, grab the value of the first capture group and not the whole regex result. See this answer on how to grab submatches from a regex pattern in VBA.
Also, note that any additional date formats you need to catch need to be explicitly set in the regex. Currently, the regex supports the following date formats:
\d{1,2}\s+[A-Z][a-z]{2}\s+\d+
12 Apr 17
12 Apr 2017
(?:\d{1,2}\/){2}\d+
1/4/17
01/04/17
1/4/2017
01/04/2017
\d+(?:\/\d{1,2}){2}
17/04/01
2017/4/1
2017/04/01
17/4/1
\d+(?:-\d{2}){2}
17-04-01
2017-04-01
\d{2}\.\d+ - Although I'm not sure what this date format is even used for and how it could be considered efficient if it's missing month
24.16
I'd like to be able to match this entire line (to highlight this sort of thing in vim): Fri Mar 18 14:10:23 ICT 2011. I'm trying to do it by finding a line that contains ICT 20 (first two digits of the year of the year), like this: syntax match myDate /^*ICT 20*$/, but I can't get it working. I'm very new to regex. Basically what I want to say: find a line that contains "ICT 20" and can have anything on either side of it, and match that whole line. Is there an easy way to do this?
.*ITC 20.*
should do the trick. . is a wildcard that matches any character, and * means you can have 0 or more of the pattern it follows. (i.e. ba(na)* will match ba, banana, bananananana and so on)