I am wondering if someone can help me get this formula right in google spreadsheets.
After a 2 week event I do get a spreadsheet with more that 2000 rows of comments which include phone numbers here and there. I am trying to extract the phone numbers from those strings.
example string: call at 228-219-4241 after
formula: =IFERROR(REGEXEXTRACT(V133,"^(?(?:\d{3}))?[-.]?(?:\d{3})[-.]?(?:\d{4})$"),"NOT FOUND!!!")
and I do get "NOT FOUND!!!!
image from gsheet... NOT FOUND!!!
But it works only in this case..
just the number
Cheers.
Your regex is too complicated and your restricting it to a rule that says the number is the first thing in the string, change to this:
=iferror(regexextract(A1,"\d{3}\-\d{3}\-\d{4}"))
In your example the '^' sign means beginning of the line and '$' means the end so your saying the first thing in your string will always be 3 numbers and the last will always be 4
Related
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}$")
I have a directory with hundreds of phone numbers with the form "xx.xx.xx.xx.xx".
where "x" are numbers from 0 to 9. Groups of 2 numbers are seaparated by a dot.
I'm looking for a regex that would integrate the last 9 numbers in a syntax like this
xx.xx.xx.xx.xx
The +33 is the French International extension that replaces the first number that always starts by 0.
I would like to keep the original format between the tags if possible.
Try this regex, if you want to check official france-number:
^(?:(?:\+|00)33|0)\.[1-9](?:[\.]\d{2}){4}$
So you can enter a number with +33.6.11.22.33.44, or with 0033.6.11 22.33.44 and both will matches.
I hope this is what you are searching
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 a noob at regex and i've been trying to select 6 numbers from within a file and then replace those 6 numbers with the same numbers plus , new line (making a CSV obviously).
Anyway sample data is simply nonsense like this:
fafksadjlkgtjafglkj210000adsfaklgjadklgjag3600001skfjaklaj093i393593390000002sadfljafkjgakjgasafksadjlkgtjafglkj£94.00 489438adsfaklgjadklgjag7700001skfjaklaj093i393593390000002ssafksa djlkgtjafglkj000000adsfaklgjadklgjag0000001skfj aklaj093i393593£39.00900002ssafksadjlk gtjafglkj000000adsfaklgjadklgjag0000001skfjaklaj093i3935£933.90000002s
Note some of the numbers are attached to currency values as well (and some are next to it but contain a space before hand) but the end will always be 6 numbers (consider them to be random as I can't see a pattern).
So I basically need to select strings matching numerics that are six digits long or longer, if longer then it just uses the last 6 digits.
Then I will replace it with itself and a comma and new line.
I hope that makes sense, i've tried a few things without success..
Thanks, edit the closest I have is:
(\d)\d{6}(?!\d)
In the Find what: text field, type in (\d{6})(\D). In the Replace with: text field, type in $1\r\n$2. Make sure that the regular expression radio button is selected. For your input, that should yield this:
fafksadjlkgtjafglkj210000
adsfaklgjadklgjag3600001
skfjaklaj093i393593390000002
sadfljafkjgakjgasafksadjlkgtjafglkj£94.00 489438
adsfaklgjadklgjag7700001
skfjaklaj093i393593390000002
ssafksa djlkgtjafglkj000000
adsfaklgjadklgjag0000001
skfj aklaj093i393593
£39.00900002
ssafksadjlk gtjafglkj000000
adsfaklgjadklgjag0000001
skfjaklaj093i3935£933.90000002
s
You want
\d{6}(?=\D*$)
Read more about anchors here.
i've been trying to select 6 numbers from within a file and then replace those 6 numbers with the same numbers plus , new line
So you're basically trying to do this, right?:
Find:
(\d{6})(\D)
Replace:
\1\n\2
[Online example]
How about:
Find what: (\d{6,})(?:\D*)$
Replace with: $1,\n
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