Regexp: wildcard character with exceptions - ruby-on-rails

I am not very familiar with regex. I am trying to match routes in ruby. I have a situation where I have some /route/:id that can take the 'id' parameter. I would like to match the route to any string with parameters as long as there is no forward slash. So anything like /route/123 should match but anything like /route/subroute/123 should not since there is a forward slash after 'subroute'. This is my current regex pattern that matches the '/routes/' portion and allows any string to take place of the 'id' parameter: \A\/routes\/\z*. This works, but if a forward slash is present in the 'id' portion of the route, the match still succeeds. What can I do to allow any string as the 'id' as long as a forward slash is not present?

This ended up being the pattern that worked for my case:
^\/route(\/[A-Za-z0-9\-\_]*)?$
Since it is a route, I found it better to allow only valid url characters and I use the parentheses (...)? for cases of routes that do not take parameters at all so '/route', '/route/', and '/route/abc' will all work, but '/route/abc/' will not.

In Ruby, ^ marks the start of any line, not string, so instead of ^, you need \A. Same with $, to match a string end position, you need to use \z.
Also, to match a single path subpart (the string between two slashes here) you can use a negated character class [^\/]+ / [^\/]*. If you plan to restrict the chars in the subpart to alphanumeric, hyphen and underscore, you can replace [^\/] with [\w-].
So, you can also use
/\A\/route(?:\/[\w-]*)?\z/
Details:
\A - start of string
\/route - a literal /route string
(?:\/[\w-]*)? - an optional (due to the last ?) non-capturing group that matches an optional sequence of / and then zero or more alphanumeric, underscore (\w) or hyphen chars
\z - end of string.
See the Rubular demo (here, ^ and $ are used for demo purposes only since the input is a single multiline text).

Related

Ruby Convert string into undescore, avoid the "/" in the resulting string

I have a name spaced class..
"CommonCar::RedTrunk"
I need to convert it to an underscored string "common_car_red_trunk", but when I use
"CommonCar::RedTrunk".underscore, I get "common_car/red_trunk" instead.
Is there another method to accomplish what I need?
Solutions:
"CommonCar::RedTrunk".gsub(':', '').underscore
or:
"CommonCar::RedTrunk".sub('::', '').underscore
or:
"CommonCar::RedTrunk".tr(':', '').underscore
Alternate:
Or turn any of these around and do the underscore() first, followed by whatever method you want to use to replace "/" with "_".
Explanation:
While all of these methods look basically the same, there are subtle differences that can be very impactful.
In short:
gsub() – uses a regex to do pattern matching, therefore, it's finding any occurrence of ":" and replacing it with "".
sub() – uses a regex to do pattern matching, similarly to gsub(), with the exception that it's only finding the first occurrence (the "g" in gsub() meaning "global"). This is why when using that method, it was necessary to use "::", otherwise a single ":" would have been left. Keep in mind with this method, it will only work with a single-nested namespace. Meaning "CommonCar::RedTrunk::BigWheels" would have been transformed to "CommonCarRedTrunk::BigWheels".
tr() – uses the string parameters as arrays of single character replacments. In this case, because we're only replacing a single character, it'll work identically to gsub(). However, if you wanted to replace "on" with "EX", for example, gsub("on", "EX") would produce "CommEXCar::RedTrunk" while tr("on", "EX") would produce "CEmmEXCar::RedTruXk".
Docs:
https://apidock.com/ruby/String/gsub
https://apidock.com/ruby/String/sub
https://apidock.com/ruby/String/tr
This is a pure-Ruby solution.
r = /(?<=[a-z])(?=[A-Z])|::/
"CommonCar::RedTrunk".gsub(r, '_').downcase
#=> "common_car_red_trunk"
See (the first form of) String#gsub and String#downcase.
The regular expression can be made self-documenting by writing it in free-spacing mode:
r = /
(?<=[a-z]) # assert that the previous character is lower-case
(?=[A-Z]) # assert that the following character is upper-case
| # or
:: # match '::'
/x # free-spacing regex definition mode
(?<=[a-z]) is a positive lookbehind; (?=[A-Z]) is a positive lookahead.
Note that /(?<=[a-z])(?=[A-Z])/ matches an empty ("zero-width") string. r matches, for example, the empty string between 'Common' and 'Car', because it is preceeded by a lower-case letter and followed by an upper-case letter.
I don't know Rails but I'm guessing you could write
"CommonCar::RedTrunk".delete(':').underscore

Getting nil while matching the correct regex for the URL with a backlash and parameters Rails 6

I'm testing a regex that allows me to validate whether or not the current string matches the url https:// or http:// http://it_tv.sharepoint.com. Its working to the extent that it does check if the url matches when I do https://it_tv.sharepoint.com. What I would like to happen, is the ability to include http://it_tv.sharepoint.com/resources..../22 in that sense. I need it to be considered valid when the url has a resource or extension attached to it. http://it_tv.sharepoint.com/resources should match true as well. Is there a way to add a wild card for the / portion of the regex expression?
url_validator.rb
if "https://it_tv.sharepoint.com/".match(%r{^(http[s]?://|)(www.)?it_tv.sharepoint.com($|/$)})
# => returns true
if "https://it_tv.sharepoint.com/resource/1005".match(%r{^(http[s]?://|)(www.)?it_tv.sharepoint.com($|/$)})
# => returns false/nil
You may use
%r{\A(?:https?://)?(?:www\.)?it_tv\.sharepoint\.com(?:/.*)?\z}
See the regex demo
Details
\A - start of string
(?:https?://)? - an optional substring, https:// or http://
(?:www\.)? - an optional www. substring
it_tv\.sharepoint\.com - an it_tv.sharepoint.com substring
(?:/.*)? - an optional / followed with any 0+ chars other than line break chars, as many as possible, char sequence
\z - end of string.
If you want to require at least one non-/ char between /s, you may use
%r{\A(?:https?://)?(?:www\.)?it_tv\.sharepoint\.com(?:/[^/]+)*/?\z}
where (?:/[^/]+)*/? matches 0 or more occurrences of / and then 1+ chars other than /, and then an optional / (that is at the end of the string).

Validate name to have no tabs or backslashes - Rails [duplicate]

I need a regular expression able to match everything but a string starting with a specific pattern (specifically index.php and what follows, like index.php?id=2342343).
Regex: match everything but:
a string starting with a specific pattern (e.g. any - empty, too - string not starting with foo):
Lookahead-based solution for NFAs:
^(?!foo).*$
^(?!foo)
Negated character class based solution for regex engines not supporting lookarounds:
^(([^f].{2}|.[^o].|.{2}[^o]).*|.{0,2})$
^([^f].{2}|.[^o].|.{2}[^o])|^.{0,2}$
a string ending with a specific pattern (say, no world. at the end):
Lookbehind-based solution:
(?<!world\.)$
^.*(?<!world\.)$
Lookahead solution:
^(?!.*world\.$).*
^(?!.*world\.$)
POSIX workaround:
^(.*([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.])|.{0,5})$
([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.]$|^.{0,5})$
a string containing specific text (say, not match a string having foo):
Lookaround-based solution:
^(?!.*foo)
^(?!.*foo).*$
POSIX workaround:
Use the online regex generator at www.formauri.es/personal/pgimeno/misc/non-match-regex
a string containing specific character (say, avoid matching a string having a | symbol):
^[^|]*$
a string equal to some string (say, not equal to foo):
Lookaround-based:
^(?!foo$)
^(?!foo$).*$
POSIX:
^(.{0,2}|.{4,}|[^f]..|.[^o].|..[^o])$
a sequence of characters:
PCRE (match any text but cat): /cat(*SKIP)(*FAIL)|[^c]*(?:c(?!at)[^c]*)*/i or /cat(*SKIP)(*FAIL)|(?:(?!cat).)+/is
Other engines allowing lookarounds: (cat)|[^c]*(?:c(?!at)[^c]*)* (or (?s)(cat)|(?:(?!cat).)*, or (cat)|[^c]+(?:c(?!at)[^c]*)*|(?:c(?!at)[^c]*)+[^c]*) and then check with language means: if Group 1 matched, it is not what we need, else, grab the match value if not empty
a certain single character or a set of characters:
Use a negated character class: [^a-z]+ (any char other than a lowercase ASCII letter)
Matching any char(s) but |: [^|]+
Demo note: the newline \n is used inside negated character classes in demos to avoid match overflow to the neighboring line(s). They are not necessary when testing individual strings.
Anchor note: In many languages, use \A to define the unambiguous start of string, and \z (in Python, it is \Z, in JavaScript, $ is OK) to define the very end of the string.
Dot note: In many flavors (but not POSIX, TRE, TCL), . matches any char but a newline char. Make sure you use a corresponding DOTALL modifier (/s in PCRE/Boost/.NET/Python/Java and /m in Ruby) for the . to match any char including a newline.
Backslash note: In languages where you have to declare patterns with C strings allowing escape sequences (like \n for a newline), you need to double the backslashes escaping special characters so that the engine could treat them as literal characters (e.g. in Java, world\. will be declared as "world\\.", or use a character class: "world[.]"). Use raw string literals (Python r'\bworld\b'), C# verbatim string literals #"world\.", or slashy strings/regex literal notations like /world\./.
You could use a negative lookahead from the start, e.g., ^(?!foo).*$ shouldn't match anything starting with foo.
You can put a ^ in the beginning of a character set to match anything but those characters.
[^=]*
will match everything but =
Just match /^index\.php/, and then reject whatever matches it.
In Python:
>>> import re
>>> p='^(?!index\.php\?[0-9]+).*$'
>>> s1='index.php?12345'
>>> re.match(p,s1)
>>> s2='index.html?12345'
>>> re.match(p,s2)
<_sre.SRE_Match object at 0xb7d65fa8>
Came across this thread after a long search. I had this problem for multiple searches and replace of some occurrences. But the pattern I used was matching till the end. Example below
import re
text = "start![image]xxx(xx.png) yyy xx![image]xxx(xxx.png) end"
replaced_text = re.sub(r'!\[image\](.*)\(.*\.png\)', '*', text)
print(replaced_text)
gave
start* end
Basically, the regex was matching from the first ![image] to the last .png, swallowing the middle yyy
Used the method posted above https://stackoverflow.com/a/17761124/429476 by Firish to break the match between the occurrence. Here the space is not matched; as the words are separated by space.
replaced_text = re.sub(r'!\[image\]([^ ]*)\([^ ]*\.png\)', '*', text)
and got what I wanted
start* yyy xx* end

username regex in rails

I am trying to find a regex to limit what a person can use for a username on my site. I don't need to have it check to see how many characters there are in it, as another validation does this. Basically all I need to make it do is make sure that it allows: letters (capital and lowercase) numbers, dashes and underscores.
I came across this: /^[-a-z]+$/i
But it doesn't seem to allow numbers.
What am I missing?
The regex you're looking for is
/\A[a-z0-9\-_]+\z/i
Meaning one or more characters of range a-z, range 0-9, - (needs to be escaped with a backslash) and _, case insensitive (the i qualifier)
Use
/\A[\w-]+\z$/
\w is shorthand for letters, digits and underscore.
\A matches at the start of the string, \z matches at the end of the string. These tokens are called anchors, and Ruby is a bit special with regard to them: Most regex engines use ^ and $ as start/end-of-string anchors by default, whereas in Ruby they can also match at the start/end of lines (which matters if you're working with multiline strings). Therefore, it's safer (as #JustMichael pointed out) to use \A and \z because there is no such ambiguity.
Your regular expression contains a character class [-a-z] that allows the characters - (dash) and a through z. In order to expand the range of characters allowed by this character class, you will need to add more characters within the [].
Please see Character Classes or Character Sets for further information and examples.

Regex to check consecutive occurrence of period symbol in username

I have to validate username in my app so that it cannot contain two consecutive period symbols. I tried the following.
username.match(/(..)/)
but found out that this matches "a." and "a..". I expected to get nil as the output of the match operation for input "a.". Is my approach right ?
You need to put a \ in front of the periods, because period is a reserved character for "any character except newline".
So try:
username.match(/(\.\.)/)
Short answer
You can use something like this (see on rubular.com):
username.match(/\.{2}/)
The . is escaped by preceding with a backslash, {2} is exact repetition specifier, and the brackets are removed since capturing is not required in this case.
On metacharacters and escaping
The dot, as a pattern metacharacter, matches (almost) any character. To match a literal period, you have at least two options:
Escape the dot as \.
Match it as a character class singleton [.]
Other metacharacters that may need escaping are | (alternation), +/*/?/{/} (repetition), [/] (character class), ^/$ (anchors), (/) (grouping), and of course \ itself.
References
regular-expressions.info/Literals and metacharacters, Dot: ., Character Class: […], Anchors: ^$, Repetition: *+?{…}, Alternation: |, Optional: ?, Grouping: (…)
On finite repetition
To match two literal periods, you can use e.g. \.\. or [.][.], i.e. a simple concatenation. You can also use the repetition construct, e.g. \.{2} or [.]{2}.
The finite repetition specifier also allows you write something like x{3,5} to match at least 3 but at most 5 x.
Note that repetition has a higher precedence that concatenation, so ha{3} doesn't match "hahaha"; it matches "haaa" instead. You can use grouping like (ha){3} to match "hahaha".
On grouping
Grouping (…) captures the string it matches, which can be useful when you want to capture a match made by a subpattern, or if you want to use it as a backreference in the other parts of the pattern.
If you don't need this functionality, then a non-capturing option is (?:…). Thus something like (?:ha){3} still matches "hahaha" like before, but without creating a capturing group.
If you don't actually need the grouping aspect, then you can just leave out the brackets altogether.

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