I am trying to check if a string contains at least one lowercase letter, uppercase letter, and a number, but not punctuation (including spaces).
For example
4aBc8Fk3 should match
4aBc 8.;3 should not match
I tried the following, but it matches spaces:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9]).{6,}[^[:punct:]]$
Any ideas how to not match strings containing punctuation including spaces?
The regular expression you have got there does the following for as far as I understand (I'm not familiar with the ruby variety, and still quite new to regex myself; this will give you an idea, but may not be 100% correct):
Go to the beginning of the string
Ensure the string matches any number of any characters followed by a lowercase letter, e.g. --a
Ensure the string matches any number of any characters followed by an uppercase letter, e.g.--aA
Ensure the string matches any number of any characters followed by a number, e.g. --aA0
If that is all true, make sure the beginning of the string is followed by at least 6 random characters, e.g.--aA0-
Ensure that is followed by a single non-punctuation character (although this is the part I'm not sure about, as I haven't used character classes before, and don't know if it's [^[:punct:]] or [^:punct:]), e.g. --aA0-c
Ensure that is followed directly by the end of the string
Now, the lookaheads would also allow a different order of occurrences, e.g. 0---Aa, as long as the string contains any characters followed by what they are looking for.
What you probably want is ^[a-zA-Z0-9]{6,}$, i.e. at least six characters, with the characters being letters and numbers (though that would also allow aaaaaa, for example).
Maybe try ^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])[a-zA-Z0-9]{6,}$ to make sure each group is present, and to get alpha-numerical characters (at least six of them) only.
I always use a tool such as http://www.regexpal.com/ to slowly build up my regex and to see where I go wrong, deconstructing a "bad" regex until I get to a "good" one, then slowly adding to it again.
Hope that helps. :)
P.S.: I'm still a bit unclear how many characters you want to match in total, i.e. if the string is fixed length or not...?
Related
I have a special use case where I want to discard all the contractions from the string and select only words followed by alphabets which do not contain any special character.
For eg:
string = "~ ASAP ASCII Achilles Ada Stackoverflow James I'd I'll I'm I've"
string.scan(/\b[A-z][a-z]+\b/)
#=> ["Achilles", "Ada", "Stackoverflow", "James", "ll", "ve"]
Note: It's not discarding the whole word I'll and I've
Can someone please help how to discard the whole word which contains contractions?
Try this Regex:
(?:(?<=\s)|(?<=^))[a-zA-Z]+(?=\s|$)
Explanation:
(?:(?<=\s)|(?<=^)) - finds the position immediately preceded by either start of the line or by a white-space
[a-zA-Z]+ - matches 1+ occurrences of a letter
(?=\s|$) - The substring matched above must be followed by either a whitespace or end of the line
Click for Demo
Update:
To make sure that not all the letters are in upper case, use the following regex:
(?:(?<=\s)|(?<=^))(?=\S*[a-z])[a-zA-Z]+(?=\s|$)
Click for Demo
The only thing added here is (?=\S*[a-z]) which means that there must be atleast one lowercase letter
I know that there's an accepted answer already, but I'd like to give my own shot:
(?<=\s|^)\w+[a-z]\w*
You can test it here. This regex is shorter and more efficient (157 steps against 315 from the accepted answer).
The explanation is rather simple:
(?<=\s|^)- This is a positive look behind. It means that we want strings preceded by a whitespace character or the start of the string.
\w+[a-z]\w* - This one means that we want strings composed by letters only (word characters) containing least one lowercase letter, thus discarding words which are whole uppercase. Along with the positive look behind, the whole regex ends up discarding words containing special characters.
NOTE: this regex won't take into account one-letter words. If you want to accomplish that, then you should use \w*[a-z]\w* instead, with a little efficiency cost.
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
However, it matches all headings which does not contain chars as Č, Š, Ž(slovenian characters).
So I'm guessing [A-Z] only matches ASCII characters? How could I get utf8?
You are right in that when you define the ASCII range A-Z, the match is made literally only for those characters. This is to do with the history of characters on computers, more and more characters have been added over time, and they are not always structured in an encoding in ways that are easy to use.
You could make a larger character class that matches the slovenian characters you need, by listing them.
But there is a shortcut. Someone else has already added necessary data to the Unicode data so that you can write shorter matches for "all uppercase characters": /[[:upper:]]/. See http://ruby-doc.org//core-2.1.4/Regexp.html for more.
Altering your regular expression with just this adjustment:
^[[:upper:]]{2,}\s?([[:upper:]]{2,}\s?)*$
You may need to adjust it further, for instance it would not match the heading "I AM A HEADING" due to the match insisting each word is at least two letters long.
Without seeing all your examples, I would probably simplify the group matching and just allow spaces anywhere:
^[[:upper:]\s]+$
You can use unicode upper case letter:
\p{Lu}
Your regex:
\b\p{Lu}{2,}(?:\s*\p{Lu}{2,})\b
RegEx Demo
The regex below:
EMAIL_REGEX = /\A[\w+\-.]+#[a-z\d\-.]+\.[a-z]+\z/i
is what I initially used to validate email format. After finding that the format "name#email...com" was passing my tests, I copy/pasted a different piece of regex that limits the amount of periods. This looks like:
EMAIL_REGEX = /\A[\w+\-.]+#[a-z\d\-]+(?:\.[a-z\d\-]+)*\.[a-z]+\z/i
The main difference is the piece of regex below:
(?:\.[a-z\d\-]+)
I can't quite figure out how this bit works. Can someone break it down for me?
Notice that in this subexpression:
(?:\.[a-z\d\-]+)
The character class [a-z\d-] does not contain a period. The expression requires there to be at least one (+) of those characters after the period (\.) in order to match. Therefore, a series of periods with no letters or digits or hyphens between them won't match the repetition of the subexpression.
The problem with your regular expression here is that you're allowing for multiple dots:
/[a-z\.]+\.[a-z]+\z/
To fix this you need to make your repeating pattern more specific in terms of structure:
/(?:[a-z]+\.)+[a-z]+\z/
That means you can have one or more repeating groups of letters plus dot. That will exclude multiple dots in a row.
Do keep in mind that email addresses are getting increasingly insane with the introduction of new GTLDs that are often used without any sort of prefix. That is, example#google may be a valid address in the future. You can't expect there to be a dot in the domain.
You have [a-z\d\-]+(?:\.[a-z\d\-]+)*. The [a-z\d\-]+ part ensures that this part of the string starts with a sequence of at least one non-period character. A period is only allowed one per (?:\.[a-z\d\-]+) structure. In each (?:\.[a-z\d\-]+), the period \. is necessarily followed by [a-z\d\-]+, which includes at least one non-period character. This ensures that whenever a period appears, it has at least one non-period character on the left and on the right. In other words, consecutive periods are not allowed.
I have the following
address.gsub(/^\d*/, "").gsub(/\d*-?\d*$/, "").gsub(/\# ?\d*/,"")
Can this be done in one gsub? I would like to pass a list of patterns rather then just one pattern - they are all being replaced by the same thing.
You could combine them with an alternation operator (|):
address = '6 66-666 #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/, "")
# " 66-666 "
address = 'pancakes 6 66-666 # pancakes #99 11-23'
address.gsub(/^\d*|\d*-?\d*$|\# ?\d*/,"")
# "pancakes 6 66-666 pancakes "
You might want to add little more whitespace cleanup. And you might want to switch to one of:
/\A\d*|\d*-?\d*\z|\# ?\d*/
/\A\d*|\d*-?\d*\Z|\# ?\d*/
depending on what your data really looks like and how you need to handle newlines.
Combining the regexes is a good idea--and relatively simple--but I'd like to recommend some additional changes. To wit:
address.gsub(/^\d+|\d+(?:-\d+)?$|\# *\d+/, "")
Of your original regexes, ^\d* and \d*-?\d*$ will always match, because they don't have to consume any characters. So you're guaranteed to perform two replacements on every line, even if that's just replacing empty strings with empty strings. Of my regexes, ^\d+ doesn't bother to match unless there's at least one digit at the beginning of the line, and \d+(?:-\d+)?$ matches what looks like an integer-or-range expression at the end of the line.
Your third regex, \# ?\d*, will match any # character, and if the # is followed by a space and some digits, it'll take those as well. Judging by your other regexes and my experience with other questions, I suspect you meant to match a # only if it's followed by one or more digits, with optional spaces intervening. That's what my third regex does.
If any of my guesses are wrong, please describe what you were trying to do, and I'll do my best to come up with the right regex. But I really don't think those first two regexes, at least, are what you want.
EDIT (in answer to the comment): When working with regexes, you should always be aware of the distinction between a regex the matches nothing and a regex that doesn't match. You say you're applying the regexes to street addresses. If an address doesn't happen to start with a house number, ^\d* will match nothing--that is, it will report a successful match, said match consisting of the empty string preceding the first character in the address.
That doesn't matter to you, you're just replacing it with another empty string anyway. But why bother doing the replacement at all? If you change the regex to ^\d+, it will report a failed match and no replacement will be performed. The result is the same either way, but the "matches noting" scenario (^\d*) results in a lot of extra work that the "doesn't match" scenario avoids. In a high-throughput situation, that could be a life-saver.
The other two regexes bring additional complications: \d*-?\d*$ could match a hyphen at the end of the string (e.g. "123-", or even "-"); and \# ?\d* could match a hash symbol anywhere in string, not just as part of an apartment/office number. You know your data, so you probably know neither of those problems will ever arise; I'm just making sure you're aware of them. My regex \d+(?:-\d+)?$ deals with the trailing-hyphen issue, and \# *\d+ at least makes sure there are digits after the hash symbol.
I think that if you combine them together in a single gsub() regex, as an alternation,
it changes the context of the starting search position.
Example, each of these lines start at the beginning of the result of the previous
regex substitution.
s/^\d*//g
s/\d*-?\d*$//g
s/\# ?\d*//g
and this
s/^\d*|\d*-?\d*$|\# ?\d*//g
resumes search/replace where the last match left off and could potentially produce a different overall output, especially since a lot of the subexpressions search for similar
if not the same characters, distinguished only by line anchors.
I think your regex's are unique enough in this case, and of course changing the order
changes the result.
I need a regex for a password which meets following constraints in my rails project:
have a minimum of 8 and a maximum of 16 characters
be alphanumeric only
contain at least one letter and one number.
My current regex is:
/^(?=.*\d)(?=.*([a-z]|[A-Z])).{8,16}$/
This allows me all the restrictions but the special characters part is not working. What is it that I am doing wrong. Can someone please correct this regex?
Thanks in advance.
/^(?=.*\d)(?=.*[a-zA-Z])[0-9a-zA-Z]{8,16}$/
The last part of your regex, .{8,16}, allows any character with a dot.
The lookahead only makes sure that there's at least one digit and one letter - it doesn't say anything about other characters. Also, note that I've updated your letter matching part - you don't need two character classes.
Disallowing special characters in a password is totally counter intuitive. Why are you doing that?