We have a tokenizer which tokenizes a text file .The logic followed is quite weird but necessary in our context.
An email such as
xyz.zyx#gmail.com
will result in the following tokens :
xyz
.
zyx
#
gmail
I would like to know how can we recognize the field as email if we are allowed to use only these tokens. No regex is allowed. We are allowed only to use the tokens and their surrounding tokens to figure out if the field is an email field
ok.. try some (bad) logic like this...
int i=0,j=0;
if(str.contains(".") && str.contains("#"))
{
if((i=str.indexOf(".") < (j=str.indexOf("#"))
{
if(i!=0 && i+1!=j) //ignore Strings like .# , abc.#
return true;
}
}
return false
Logically split an e-mail address into 3 parts:
A user name (or resource name), for this explanation let's call it the user name
The # character.
A host name, consisting of any number of "word dot" sequences + a final top level domain string.
Do a walk like this:
while token can be part of a user name
fetch next token;
if there no more -> no e-mail;
check if the next token is #
if not -> no e-mail
while there are tokens
while token can be part of a host name subpart (the "word" above)
fetch next token;
if there are no more -> might be a valid e-mail address
check if the next token is a dot
if not -> might be a valid e-mail address
set a flag that you found at least one dot
check if the next token can be part of a host name subpart
if not -> no valid e-mail address (or maybe you ignore a trailing dot and take what was found so far)
Add further checks if there are more tokens where needed. You also may have to post process the found tokens to ensure a valid e-mail address and you may have to rewind your tokenizer (or cache the fetched tokens) in case you did not find a valid e-mail address and need to feed the same input to some other recognition process.
Check if a list of tokens is an email:
list contains exactly one token #
index of token # != 0
at least 3 tokens after #
at least 1 . token after #, but not immediately after
starts and ends with character tokens
Additional checks:
no two . subsequent tokens
no special characters
length of character tokens after # is at least 2
total length of all character tokens before # is at least 3
Related
I have the following regular expression that I'm using to remove the dev. part of my URL.
String domain = "dev.mydomain.com";
System.out.println(domain.replaceAll(".*\\.(?=.*\\.)", ""));
Outputs: mydomain.com but this is giving me issues when the domains are in the vein of dev.mydomain.com.pe or dev.mydomain.com.uk in those cases I am getting only the .com.pe and .com.uk parts.
Is there a modifier I can use on my regex to make sure it only takes what is before the first . (dot included)?
Desired output:
dev.mydomain.com -> mydomain.com
stage.mydomain.com.pe -> mydomain.com.pe
test.mydomain.com.uk -> mydomain.com.uk
You may use
^[^.]+\.(?=.*\.)
See the regex demo and the regex graph:
Details
^ - start of string
[^.]+ - 1 or more chars other than dots
\. - a dot
(?=.*\.) - followed with any 0 or more chars other than line break chars as many as possible and then a ..
Java usage example:
String result = domain.replaceFirst("^[^.]+\\.(?=.*\\.)", "");
Following regex will work for you. It will find first part (if exists), captures rest of the string as 2nd matching group and replaces the string with 2nd matching group. .*? is non-greedy search that will match until it sees first dot character.
(.*?\.)?(.*\..*)
Regex Demo
sample code:
String domain = "dev.mydomain.com";
System.out.println(domain.replaceAll("(.*?\\.)?(.*\\..*)", "$2"));
domain = "stage.mydomain.com.pe";
System.out.println(domain.replaceAll("(.*?\\.)?(.*\\..*)", "$2"));
domain = "test.mydomain.com.uk";
System.out.println(domain.replaceAll("(.*?\\.)?(.*\\..*)", "$2"));
domain = "mydomain.com";
System.out.println(domain.replaceAll("(.*?\\.)?(.*\\..*)", "$2"));
output:
mydomain.com
mydomain.com.pe
mydomain.com.uk
mydomain.com
For Simple Java Mail I'm trying to deal with a somewhat free-format of delimited email addresses. Note that I'm specifically not validating, just getting the addresses out of a list of addresses. For this use case the addresses can be assumed to be valid.
Here is an example of a valid input:
"name#domain.com,Sixpack, Joe 1 <name#domain.com>, Sixpack, Joe 2 <name#domain.com> ;Sixpack, Joe, 3<name#domain.com> , nameFoo#domain.com,nameBar#domain.com;nameBaz#domain.com;"
So there are two basic forms "name#domain.com" and "Joe Sixpack ", which can appear in a comma / semicolon delimited string, ignoring white space padding. The problem is that the names can contains delimiters as valid characters.
The following array shows the data needed (trailing spaces or delimiters would not be a big problem):
["name#domain.com",
"Sixpack, Joe 1 <name#domain.com>",
"Sixpack, Joe 2 <name#domain.com>",
"Sixpack, Joe, 3<name#domain.com>",
"nameFoo#domain.com",
"nameBar#domain.com",
"nameBaz#domain.com"]
I can't think of a clean way to deal with this. Any suggestion how I can reliably recognize whether a comma is part of a name or is a delimiter?
Final solution (variation on the accepted answer):
var string = "name#domain.com,Sixpack, Joe 1 <name#domain.com>, Sixpack, Joe 2 <name#domain.com> ;Sixpack, Joe, 3<name#domain.com> , nameFoo#domain.com,nameBar#domain.com;nameBaz#domain.com;"
// recognize value tails and replace the delimiters there, disambiguating delimiters
const result = string
.replace(/(#.*?>?)\s*[,;]/g, "$1<|>")
.replace(/<\|>$/,"") // remove trailing delimiter
.split(/\s*<\|>\s*/) // split on delimiter including surround space
console.log(result)
Or in Java:
public static String[] extractEmailAddresses(String emailAddressList) {
return emailAddressList
.replaceAll("(#.*?>?)\\s*[,;]", "$1<|>")
.replaceAll("<\\|>$", "")
.split("\\s*<\\|>\\s*");
}
since you are not validating, i assume that the email addresses are valid.
Based on this assumption, i will look up an email address followed by ; or , this way i know its valid.
var string = "name#domain.com,Sixpack, Joe 1 <name#domain.com>, Sixpack, Joe 2 <name#domain.com> ;Sixpack, Joe, 3<name#domain.com> , nameFoo#domain.com,nameBar#domain.com;nameBaz#domain.com;"
const result = string.match(/(.*?#.*?\..*?)[,;]/g)
console.log(result)
This pattern works for your provided examples:
([^#,;\s]+#[^#,;\s]+)|(?:$|\s*[,;])(?:\s*)(.*?)<([^#,;\s]+#[^#,;\s]+)>
([^#,;\s]+#[^#,;\s]+) # email defined by an # with connected chars except ',' ';' and white-space
| # OR
(?:$|\s*[,;])(?:\s*) # start of line OR 0 or more spaces followed by a separator, then 0 or more white-space chars
(.*?) # name
<([^#,;\s]+#[^#,;\s]+)> # email enclosed by lt-gt
PCRE Demo
Using Java's replaceAll and split functions (mimicked in javascript below), I would say lock onto what you know ends an item (the ".com"), replace separator characters with a unique temp (a uuid or something like <|>), and then split using your refactored delimiter.
Here is a javascript example, but Java's repalceAll and split can do the same job.
var string = "name#domain.com,Joe Sixpack <name#domain.com>, Sixpack, Joe <name#domain.com> ;Sixpack, Joe<name#domain.com> , name#domain.com,name#domain.com;name#domain.com;"
const result = string.replace(/(\.com>?)[\s,;]+/g, "$1<|>").replace(/<\|>$/,"").split("<|>")
console.log(result)
Basically i am trying to restrict a user not to input characters that are not allowed in username box. i found that this could be implement by String.matches()
String UcharSet = "[a-zA-Z0-9-~!##().]+";
boolean UMORN = "Username.is#example.com".matches(UcharSet);
if(UMORN != true)
UNotAllowedCharEC = "0x00000030";
as you can see i have string of characters to be allowed in my username box but somehow when i input # it return false although i have it in my allowed string list.
and do tell should i add any other characters to be allowed for my username box.
I just tested this and '#' results in true. You problem probably lies elsewhere.
I'm working on GUI validation...
Please see the problem below...
How to validate an email with a specific format? at least one digit before the # and one digit after and at least two letters after the dot.
String EmailFormat = "m#m.co";
Pattern patternEmail = Pattern.compile("\\d{1,}#\\d{1,}.\\d{2,}");
Matcher matcherName = patternEmail.matcher(StudentEmail);
Don't write your own validator. Email has been around for decades and there are many standard libraries which work, address parts of the standard you may not know about, and are well tested by many other developers.
Apache Commons Email Validator is a good example. Even if you use a standard validator you need to be aware of the limitations or gotchas in validating an email address. Here are the javadocs for Commons EmailValidator which state, "This implementation is not guaranteed to catch all possible errors in an email address. For example, an address like nobody#noplace.somedog will pass validator, even though there is no TLD "somedog"" . So you can use a good email validator to determine if an address is valid, but you will have to do extra work to guarantee that the domain exists, accepts email, and accepts email fro that address.
If you require good addresses you will need a secondary mechanism. A confirmation email is a good mechanism. You send a link to the given address and the user must visit that link to verify that email can be sent to that address.
This the regex pattern for emails
String pt = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*#[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
You can try it like this
List email = Arrays.asList("xyzl#gmail.com", "#", "sxd");
Predicate<String> validMail = (n) -> n.matches(pt);
email.stream().filter(validMail).forEach((n) -> System.out.println(n));
This is the description you can change it according to your need.
^ #start of the line
[_A-Za-z0-9-\\+]+ # must start with string in the bracket [ ], must contains one or more (+)
( # start of group #1
\\.[_A-Za-z0-9-]+ # follow by a dot "." and string in the bracket [ ], must contains one or more (+)
)* # end of group #1, this group is optional (*)
# # must contains a "#" symbol
[A-Za-z0-9-]+ # follow by string in the bracket [ ], must contains one or more (+)
( # start of group #2 - first level TLD checking
\\.[A-Za-z0-9]+ # follow by a dot "." and string in the bracket [ ], must contains one or more (+)
)* # end of group #2, this group is optional (*)
( # start of group #3 - second level TLD checking
\\.[A-Za-z]{2,} # follow by a dot "." and string in the bracket [ ], with minimum length of 2
) # end of group #3
$ #end of the line
Split email into two parts using # as delimiter:
String email = "some#email.com";
String[] parts = email.split("#"); // parts = [ "some", "email.com" ]
Validate each part separately, using multiple checks if necessary:
// validate username
String username = parts[0];
if (username.matches("\\d")) {
// ok
}
// validate domain
String domain = parts[1];
if (domain.matches("\\d") && domain.matches("\\.[a-z]{2,4}$")) {
// ok
}
Note that this is a very poor email validator and it shouldn't be used standalone.
I have tried to use the following kind of regex
([_a-z0-9-]+(\.[_a-z0-9-]+)*#[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4}))|(FakeEmail:)|(Email:)|(\1\2)|(\1\3)
(pretend the \1 is the email regex group, and \2 is FakeEmail: and \3 is Email: because I didnt count the parens to figure out the real grouping)
What I am trying to do is say "Find the word email: and if you find it, pick up any email address following the word."
That email regex I got off some other question on stack overflow.
my test string could be something like
"This guy is spamming me from
FakeEmail: fakeemailAdress#someplace.com
but here is is real info:
Email: testemail#someplace.com"
Any tips? Thanks
I'm either quite confused as to what you're trying to do, or your Regex is just very wrong. In particular:
Why do you have Email: at the end, instead of the beginning - to match your example?
Why do you have both your Email: and your \1\2 separated by pipe characters, almost as if they're in fields? This is compiling the pattern as ORs. (Find the email pattern, OR the word "Email:", OR whatever \1\2 will end up meaning as it is out of context here.)
If all you're trying to do is match something like Email: testemail#someplace.com, you don't need any backtracking.
Something like this is probably all you need:
Email:\s+([_a-z0-9-]+(\.[_a-z0-9-]+)*#[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4}))
Also, I'd strongly advise against trying to validate an email address so strictly. You may want to read http://haacked.com/archive/2007/08/21/i-knew-how-to-validate-an-email-address-until-i.aspx . I'd simplify the pattern to something more along the lines of:
Email:\s+(\S+)*#(\S+\.\S+)
Try:
(Fake)?Email: *([_a-z0-9-]+(\.[_a-z0-9-]+)*#[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4}))
And captured group \1 will be empty if it's a real email and contain "Fake" if it's a fake email, while \2 will be the email itself.
Do you actually want to capture it if it's FakeEmail though? If you want to capture all Email but ignore all FakeEmail then do:
\bEmail: *([_a-z0-9-]+(\.[_a-z0-9-]+)*#[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4}))
The word boundary prevents the Email bit from matching "FakeEmail".
UPDATE: note your regex only matches lowercase since it's got a-z in the [] everywhere but not [A-Z]. Make sure you feed your regex into the java match function with the ignore case switch. i.e.:
Pattern.compile("(Fake)?Email: .....", Pattern.CASE_INSENSITIVE)
You can use following code to match all type of email address:
String text = "This guy is spamming me from\n" +
"FakeEmail: fakeemail+Adress#someplace.com\n" +
"fakeEmail: \n" +
"fakeemail#someplace.com" +
"but here is is real info:\n" +
"Email: test.email+info#someplace.com\n";
Matcher m = Pattern.compile("(?i)(?s)Email:\\s*([_a-z\\d\\+-]+(\\.[_a-z\\d\\+-]+)*#[a-z\\d-]+(\\.[a-z\\d-]+)*(\\.[a-z]{2,4}))").matcher(text);
while(m.find())
System.out.printf("Email is [%s]%n", m.group(1));
This will match email text:
appearing on different lines by using (?s)
ignoring case comparison by using (?i)
Email address with a period . in it
Email address with a plus sign + in it
OUTPUT: From above code is
Email is [fakeemail+Adress#someplace.com]
Email is [fakeemail#someplace.comb]
Email is [test.email+info#someplace.com]