I am working on Google spreadsheet. I am trying to make a validation rule on the text length.
For example the text length must be equal to 12 characters which includes number, on dash and one capital character.
Example:
123-56C89112
I tried to search in google groups but no result! can you help in this please
You can use Data > Validation > Custom Formula is
=REGEXMATCH(B3, "^(?=.{12}$)(?=(^[^A-Z]*[A-Z]{1}[^A-Z]*$))(?=(^[^-]*[-]{1}[^-]*$)).*")
I'm not an expert on Regex so this may be a bit over the top, but it appears to work:
^(?=.{12}$)(?=(^[^A-Z]*[A-Z]{1}[^A-Z]*$))(?=(^[^-]*[-]{1}[^-]*$)).*
Debuggex Demo
This allows any string with exactly 12 characters, 1 upper case letter and 1 dash.
EDIT: The (Lookahead) Regex doesn't seem to work in Spreadsheets (see comment)
EDIT2: Using a custom function like the following seems to work, but only if you use warnings (instead of rejections) for some reason (bug?)
function test(myString) {
return myString.match("^(?=.{12}$)(?=(^[^A-Z]*[A-Z]{1}[^A-Z]*$))(?=(^[^-]*[-]{1}[^-]*$)).*")!=null;
}
Related
emails
vera#mail.com
estebangarrido#mail.c
hurtado#mail com
jmariano2mail.com
How can I pass a fuction which correct all domains to #mail.com. I know I have to use =RIGHT(,9) but when you reach the last error it does not apply
Try below formula-
=ArrayFormula(IF(A2:A="",,QUERY(SPLIT(SUBSTITUTE(SUBSTITUTE(A2:A,"mail","|"),"#",""),"|"),"select Col1",0)&"#mail.com"))
This should also work.
=INDEX(IF(LEN(A2:A),QUERY(SPLIT(SUBSTITUTE(SUBSTITUTE(A2:A,"mail","|"),"#",""),"|"),"select Col1")&"#mail.com",""))
Answer
The following formula should produce the results you desire. It assumes that the data you provide is in cells A2:A5 of your spreadsheet. If this is not the case, adjust the A2:A5 portion of the formula appropriately.
=ARRAYFORMULA(REGEXREPLACE(A2:A5,"[#|2].*","#mail.com"))
Explanation
This formula uses REGEXREPLACE to get rid of all rogue characters and replace them with #mail.com. The first argument of REGEXREPLACE is the string to be evaluated. In this case, that is the range from A2 through A5. The second argument is which characters to look for. In this case that is all characters (done using .*) that follow either an at-sign or a numeral two (done using [#|2]). The third argument is which new string to replace the found characters with. In this case that is #mail.com, the correct domain without typos.
The REGEXREPLACE is wrapped in =ARRAYFORMULA because normally REGEXREPLACE can only be used with a single cell rather than a range of cells.
Please note that this solution relies on the assumption you stated that "Everything before # or 2 is correct."
I am using the following formula to extract the substring venue01 from column C, the problem is that when value string in column C is shorter it only extracts the value 1 I need it to extract anything straight after the - (dash) no matter the length of the value text in column c
={"VenueID";ARRAYFORMULA(IF(ISBLANK(A2:A),"",RIGHT(C2:C,SEARCH("-",C2:C)-21)))}
There is a much simpler solution using regular expressions.
=REGEXEXTRACT(A1,".*-(.*)")
In case you are no familiar with Regular Expressions what this means is, get me every string of characters ((.*)) after a dash (-).
Example
Reference
REGEXTRACT
Test regular expressions
Cheat sheet for regular expressions
To answer bomberjackets question in the comment of Raserhin:
To select the part of the string before the "-"
=REGEXEXTRACT(A1,"(.*)-.*")
EXAMPLE
example of code
Adding to your original formula. I think if you'd use RIGHT and inside it reverse the order of the string with ARRAY then that may work.
=Right(A1,FIND("-",JOIN("",ARRAYFORMULA(MID(A1,LEN(A1)-ROW(INDIRECT("1:"&LEN(A1)))+1,1))))-1)
It takes string from the right side up to X number of characters.
Number of character is fetched from reversing the text, then finding
the dash "-".
It adds one more +1 of the text as it will take out so it accounts
for the dash itself, if no +1 is added, it will show the dash on
the extracted string.
The REGEX on the other answer works great too, however, you can control a number of character to over or under trim. E.g. if there is a space after the dash and you would like to always account for one more char.
I have a sheet containing my weekly schedule. Only school cells have a room number in it, so how do I format the cells to color only the ones that contain a number.
note: Actually, the room number is a number in range(A:E) followed by a three digits number in range(000:499). Ex.:(A433, B166, D254)
I tried: Text contains"(A:F)(000:444)" but it didn't work.
EDIT:
For some reason, "=REGEXMATCH(B2, "[A-F][0-9]{3}")" worked. Could anyone tell me why? I tried replacing B2 by B1, but then it didn't work. Does it have anything to do with the fact that B1 is a weekday, and so does not contain REGEXP(B1,"[A-F][0-9]{3}) returned false.
What seemed more logical to me was "=REGEXMATCH(B2:F22, "[A-F][0-9]{3}")" To apply this function in range B2 to F22. What am I missing here?
In order to match patterns, you'll need to use regular expressions. Since the standard Conditional Formatting options don't include regular expressions, you'll need to choose "Custom formula is" and then use REGEXMATCH, which returns a Boolean value.
If you really want to look for the specific room number format you mentioned, then you would use the formula:
=REGEXMATCH(A1, "[A-E][0-9]{3}")
But if you just want to look for any numbers, you can use
=REGEXMATCH(A1, "[0-9]+")
In both cases, the text you're checking is in cell A1
You might try Conditional Formatting with a custom formula rule of the type:
=if(isnumber(A1),1,regexmatch(A1,"\d"))
The above was an attempt to respond to:
Google spreadsheets conditional formatting if text contains numbers
A more particular fit for the stated room number style would be:
=REGEXMATCH(A1,"[A-F]\d\d\d")
where the first character is any of the first six letters of the alphabet, if capitalised, followed by three instances of any number.
I did some searching and in openoffice and excel it looks like you can simply add an * at the beginning or end of a character to delete everything before and after it, but in Google spreadsheet this isn't working. Does it support this feature? So if I have:
keyword USD 0078945jg .12 N N 5748 8
And I want to remove USD and everything after it what do I use? I have tried:
USD* and (USD*) with regular expressions checked
But it doesn't work. Any ideas?
The * quantifier just needs to be applied to a dot (.) which will match any character.
To clarify: the * wildcard used in certain spreadsheet functions (eg COUNTIF) has a different usage to the * quantifier used in regular expressions.
In addition to options that would be available in Excel (LEFT + FIND) pointed out by pnuts, you can use a variety of regex tools available in Google Sheets for text searching / manipulation
For example, RegexReplace:
=REGEXREPLACE(A1,"(.*)USD.*","$1")
(.*) <- capture group () with zero or more * of any character .
USD.* <- exact match on USD followed by zero or more * of any character .
$1 <- replace with match in first capture group
Please try:
and also have a look at.
For spaces within keyword I suggest a helper column with a formula such as:
=left(A1,find("USD",A1)-1)
copied down to suit. The formula could be converted to values and the raw data (assumed to be in ColumnA) then deleted, if desired.
To add to the answers here, you can get into trouble when there are special characters in the text (I have been struggling with this for years).
You can put a frontslash \ in front of special characters such as ?, + or . to escape them. But I still got stuck when there were further special characters in the text. I finally figured it out after reading find and replace in google sheets with regex.
Example: I want to remove the number, period and space from the beginning of a question like this: 1. What is your name?
Go to Edit → Find and replace
In the Find field, enter the following: .+\. (note: this includes a space at the end).
Note: In the Find and replace dialogue box, be sure to check "Search using regular expressions" and "match case". Leave the Replace field blank.
The result will be this text only: What is your name?
I have a column in open office like this:
abc-23
abc-32
abc-1
Now, I need to get only the sum of the numbers 23, 32 and 1 using a formula and regular expressions in calc.
How do I do that?
I tried
=SUMIF(F7:F16,"([:digit:].)$")
But somehow this does not work.
Starting with LibreOffice 6.4, you can use the newly added REGEX function to generically extract all numbers from a cell / text using a regular expression:
=REGEX(A1;"[^[:digit:]]";"";"g")
Replace A1 with the cell-reference you want to extract numbers from.
Explanation of REGEX function arguments:
Arguments are separated by a semicolon ;
A1: Value to extract numbers from. Can be a cell-reference (like A1) or a quoted text value (like "123abc"). The following regular expression will be applied to this cell / text.
"[^[:digit:]]": Match every character which is not a decimal digit. See also list of regular expressions in LibreOffice
The outer square brackets [] encapsulate the list of characters to search for
^ adds a NOT, meaning that every character not included in the search list is matched
[:digit:] represents any decimal digit
"": replace matching characters (every non-digit) with nothing = remove them
"g": replace all matches (don't stop after the first non-digit character)
Unfortunately Libre-Office only supports regex in find/replace and in search.
If this is a once-only deal, I would copy column A to column to B, then use [data] [text to columns] in B and use the - as a separator, leaving you with all the text in column B and the numbers in column C.
Alternatively, you could use =Right(A1,find("-",A1,1)+1) in column B, then sum Column C.
I think that this is not exactly what do you want, but maybe it can help you or others.
It is all about substring (in Calc called [MID][1] function):
First: Choose your cell (for example with "abc-23" content).
Secondly: Enter the start length ("british" --> start length 4 = tish).
After that: To print all remaining text, you can use the [LEN][2] function (known as length) with your cell ("abc-23") in parameter.
Code now looks like this:
D15="abc-23"
=MID(D15; 5; LEN(D15))
And the output is: 23
When you edit numbers (in this example 23), no problem. However, if you change anything before (text "abc-"), the algorithm collapses because the start length is defined to "5".
Paste the string in a cell, open search and replace dialog (ctrl + f) extended search option mark regular expression search for ([\s,0-9])([^0-9\s])+ and replace it with $1
adjust regex to your needs
I didn't figure out how to do this in OpenOffice/LibreOffice directly. After frustrations in searching online and trying various formulas, I realised my sheet was a simple CSV format, so I opened it up in vim and used vim's built-in sed-like feature to find/replace the text in vim command mode:
:%s/abc-//g
This only worked for me because there were no other columns with this matching text. If there are other columns with the same text, then the solution would be a bit more complex.
If your sheet is not a CSV, you could copy the column out to a text file and use vim to find/replace, and then paste the data back into the spreadsheet. For me, this was a lot less frustrating than trying to figure this out in LibreOffice...
I won't bother with a solution without knowing if there really is interest, but, you could write a macro to do this. Extract all the numbers and then implement the sum by checking for contained numbers in the text.