First off, I'm sorry if this has already been asked somewhere else, it's just I could not find it. If it has been, I apologize deeply.
I am terrible at Regular Expressions and generally avoid them but I know the problem I have can be simply solved using them so I have come here for help.
I have a text field containing some information about a company (name, address, identifier, etc), but not all information always appears on the field and the order the information appears in is not set.
What I need is the company identifier which is a 14-digit number which can or cannot be formatted as such: XX.XXX.XXX/XXXX-XX
What expression could I use that would identify if there are either 14 digits in a row or the number formatted in the manner described above?
/[0-9]{2}[.]{1}[0-9]{3}[.]{1}[0-9]{3}[\/]{1}[0-9]{4}[-]{1}[0-9]{2}/ for XX.XXX.XXX/XXXX-XX
/[0-9]{14}/ for 14 digits
There's probably some edge cases in here somewhere.
There's also probably a way to do both of these in one, but I don't have the patience nor the time to try and figure it out.
Try Regex: \b(?:\d{14}|\d{2}\.\d{3}\.\d{3}\/\d{4}-\d{2})\b
Demo
Related
I'm trying to validate the format of a street address in Google Forms using regex. I won't be able to confirm it's a real address, but I would like to at least validate that the string is:
[numbers(max 6 digits)] [word(minimum one to max 8 words with
spaces in between and numbers and # allowed)], [words(minimum one to max four words, only letters)], [2
capital letters] [5 digit number]
I want the spaces and commas I left in between the brackets to be required, exactly where I put them in the above example. This would validate
123 test st, test city, TT 12345
That's obviously not a real address, but at least it requires the entry of the correct format. The data is coming from people answering a question on a form, so it will always be just an address, no names. Plus they're all address is one area South Florida, where pretty much all addresses will match this format. The problem I'm having is people not entering a city, or commas, so I want to give them an error if they don't. So far, I've found this
^([0-9a-zA-Z]+)(,\s*[0-9a-zA-Z]+)*$
But that doesn't allow for multiple words between the commas, or the capital letters and numbers for zip. Any help would save me a lot of headaches, and I would greatly appreciate it.
There really is a lot to consider when dealing with a street address--more than you can meaningfully deal with using a regular expression. Besides, if a human being is at a keyboard, there's always a high likelihood of typing mistakes, and there just isn't a regex that can account for all possible human errors.
Also, depending on what you intend to do with the address once you receive it, there's all sorts of helpful information you might need that you wouldn't get just from splitting the rough address components with a regex.
As a software developer at SmartyStreets (disclosure), I've learned that regular expressions really are the wrong tool for this job because addresses aren't as 'regular' (standardized) as you might think. There are more rigorous validation tools available, even plugins you can install on your web form to validate the address as it is typed, and which return a wealth of of useful metadata and information.
Try Regex:
\d{1,6}\s(?:[A-Za-z0-9#]+\s){0,7}(?:[A-Za-z0-9#]+,)\s*(?:[A-Za-z]+\s){0,3}(?:[A-Za-z]+,)\s*[A-Z]{2}\s*\d{5}
See Demo
Accepts Apt# also:
(^[0-9]{1,5}\s)([A-Za-z]{1,}(\#\s|\s\#|\s\#\s|\s)){1,5}([A-Za-z]{1,}\,|[0-9]{1,}\,)(\s[a-zA-Z]{1,}\,|[a-zA-Z]{1,}\,)(\s[a-zA-Z]{2}\s|[a-zA-Z]{2}\s)([0-9]{5})
I'm attempting to block a long string of unnecessary text that's on every page of a document.
Ex: "36075 This is another page and this is the date March 4 2013"
I know this must be very simple, but I'm hoping there is a way to block text verbatim. Is the only way to block this text by using a lot of /d/s/w+/+ etc or is there is a way to say, "match 36075 This is another page and this is the date March 4 2013".
This would be SO HELPFUL to know. Thank you for helping!
From what you wrote I assume you need to get leading numbers from string, to do it you just need to use this pattern: ^\d+ which from this input:
36075 This is another page and this is the date March 4 2013
will return this:
36075
For future, in case of such questions please provide example string and expected output. As well as what you have tried.
I realized the issue I was having. I didn't need to use RegEx. The program I was using has the functionality to match specific words or groups of words and pronounce them differently. What I discovered is that it will not match the words unless the word groups are input exactly the way the program typically reads them.
Ergo --> The channel saw
the end of the British hold over
Would have to be listed as one group for, "The channel saw" and a second group for "the end of the British hold over"
In addition, there were some numbers --> 11960_30_o_ho_
and if the program naturally read 119 and then 60_3 and then _o_ho_ then three strings would need to be input for each section.
A few frustrating hours later, problem solved :) Thank you for your assistance.
I am attempting to write a MVC model validation that verifies that there is 10 or more words in a string. The string is being populated correctly, so I did not include the HTML. I have done a fair bit of research, and it seems that something along the lines of what I have tries should work, but, for whatever reason, mine always seem to fail. Any ideas as to what I am doing wrong here?
(using System.ComponentModel.DataAnnotations, in a mvc 4 vb.net environment)
Have tried ([\w]+){10,}, ((\\S+)\s?){10,}, [\b]{20,}, [\w+\w?]{10,}, (\b(\w+?)\b){10,}, ([\w]+?\s){10}, ([\w]+?\s){9}[\w], ([\S]+\s){9}[\S], ([a-zA-Z0-9,.'":;$-]+\s+){10,} and several more varaiations on the same basic idea.
<Required(ErrorMessage:="The Description of Operations field is required"), RegularExpression("([\w]+){20,}", ErrorMessage:="ERROZ")>
Public Property DescOfOperations As String = String.Empty
Correct Solution was ([\S]+\s+){9}[\S\s]+
EDIT Moved accepted version to the top, removing unused versions. Unless I am wrong and the whole sequence needs to match, then something like (also accounting for double spaces):
([\S]+\s+){9}[\S\s]+
Or:
([\w]+?\s+){9}[\w]+
Give this a try:
([a-zA-Z0-9,.'":;$-]+\s){10,}
I have a string in date format 06/09/2011 03:00 PM. I want to remove all of the forward slashes, and if the first digit of the month (06) is a zero, remove it as well as the first digit of the day (09) remove it as well. Any body who can help me out?
thanks!
The usual way to do this is by taking an available date parser where you hand in the input format and output it to a different output format.
Patterns differ, Implementations etc differ also. It is not convenient neither practicable to do date parsing via regex.
Something like that
0([1-9]+)/0([1-9]+)/([0-9]+)
Of course, it will only work in valid dates; it does not parse the date or anything.
BTW: I find better (more readable, detects errors in a more meaningful manner) fyr's answer. This is just to show that it can be done with regex, if fyr's solution is not available in your platform.
This might be a hard one (if not impossible), but can anyone think of a regular expression that will find a person's name, in say, a resume? I know this won't be 100% accurate, but I can't come up with something.
Let's assume the name only shows up once in the document.
No, you can't use regular expressions for this. The only chance you have is if the document is always in the same format and you can find the name based on the context surrounding it. But this probably isn't the case for you.
If you are asking your applicants to submit their résumé online you could provide a separate field for them to enter their name and any other information you need instead of trying to automatically parse résumés.
Forget it - seriously.
Or expect to get a lot of applications from a Mr C Vitae
In my experience, having written something very similar (but a very long time ago), about 95% of resumes have the person's name as the very first line. You could probably have a pretty loose regex checking for alpha, hyphens, periods, and assume that's the name.
Obviously there's no way to do this 100% accurately, as you said, but this would be close.
Unless you wanted to build an expression that contained every possible name, or-ed together, the expression you are referring to is not "Regular," with a capital R. A good guess might be to go looking for the largest-font words in the document. If they follow a pattern that looks like firstname-lastname, name-initial-name, etc., you could call it a good guess...
That's a really hairy problem to tackle. The regex has to match two words that could be someone's name. The problem with that is that some people, of Hispanic origin, for example, might have a name that's more than 2 words. Also, how would you define two words to match for a name? Would you use a database of common first and last name fields? That might work unless someone has an uncommon name.
I'm reminded of a story of a COBOL teacher in college told me about an individual of Asian origin who's name would break every rule the programmers defined for a bank's internal system. His first name was "O." just the letter O.
The only remotely dependable way to nail down the regex would be if you had something to set off your search with; maybe if a line of text in the resume began with "Name: " then you'd know where to start looking.
tl;dr: People's names and individual resumes are too heavily varied for a regular expression to pick apart.
You could do something like Amazon does for book overviews: SIPs. This would require some after-the-fact double checking by humans but you might find the person's name(s) in there.