I am trying to make a regular expression, that allow to create string with the small and big letters + numbers - a-zA-z0-9 and also with the chars: .-_
How do I make such a regex?
The following regex should be what you are looking for (explanation below):
\A[-\w.]*\z
The following character class should match only the characters that you want to allow:
[-a-zA-z0-9_.]
You could shorten this to the following since \w is equivalent to [a-zA-z0-9_]:
[-\w.]
Note that to include a literal - in your character class, it needs to be first character because otherwise it will be interpreted as a range (for example [a-d] is equivalent to [abcd]). The other option is to escape it with a backslash.
Normally . means any character except newlines, and you would need to escape it to match a literal period, but this isn't necessary inside of character classes.
The \A and \z are anchors to the beginning and end of the string, otherwise you would match strings that contain any of the allowed characters, instead of strings that contain only the allowed characters.
The * means zero or more characters, if you want it to require one or more characters change the * to a +.
/\A[\w\-\.]+\z/
\w means alphanumeric (case-insensitive) and "_"
\- means dash
\. means period
\A means beginning (even "stronger" than ^)
\z means end (even "stronger" than $)
for example:
>> 'a-zA-z0-9._' =~ /\A[\w\-\.]+\z/
=> 0 # this means a match
UPDATED thanks phrogz for improvement
Related
I want to use this regex to match any block comment (c-style) in a string.
But why the below does not?
rblockcmt = Regexp.new "/\\*[.\s]*?\\*/" # match block comment
p rblockcmt=~"/* 22/Nov - add fee update */"
==> nil
And in addition to what Sir Swoveland posted, a . matches any character except a newline:
The following metacharacters also behave like character classes:
/./ - Any character except a newline.
https://ruby-doc.org/core-2.3.0/Regexp.html
If you need . to match a newline, you can specify the m flag, e.g. /.*?/m
Options
The end delimiter for a regexp can be followed by one or more
single-letter options which control how the pattern can match.
/pat/i - Ignore case
/pat/m - Treat a newline as a character matched by .
...
https://ruby-doc.org/core-2.3.0/Regexp.html
Because having exceptions/quirks like newline not matching a . can be painful, some people specify the m option for every regex they write.
It appears that you intend [.\s]*? to match any character or a whitespace, zero or more times, lazily. Firstly, whitespaces are characters, so you don't need \s. That simplifies your expression to [.]*?. Secondly, if your intent is to match any character there is no need for a character class, just write .. Thirdly, and most importantly, a period within a character class is simply the character ".".
You want .*? (or [^*]*).
I have a string that looks something like '\\u001E\\u001Csome_text\u001F' - where the first two characters are escaped with two backslashes and the last one only has one backslash.
I want to convert that string so that all the unicode literals have two backslashes in them, so the output would look like '\\u001E\\u001Csome_text\\u001F'. What ways could I go about doing this?
If the goal is to find instances of '\u' not preceded by a '\' character, then a negative lookbehind should suit the need to match:
(?<!\\)\\u
If you want to match the whole character, then you can also use a positive lookahead to verify the following characters:
(?<!\\)\\u(?=[0-9A-Z]{4})
Both of these will return the instances of '\u', which you can replace with '\\u'
I have a string with newline characters that I want to gsub out for white space.
"hello I\r\nam a test\r\n\r\nstring".gsub(/[\\r\\n]/, ' ')
something like this ^ only my regex seems to be replacing the 'r' and 'n' letters as well. the other constraint is sometimes the pattern repeats itself twice and thus would be replaced with two whitespaces in a row, although this is not preferable it is better than all the text being cut apart.
If there is a way to only select the new line characters. Or even better if there a more rubiestic way of approaching this outside of going to regex?
If you have mixed consecutive line breaks that you want to replace with a single space, you may use the following regex solution:
s.gsub(/\R+/, ' ')
See the Ruby demo.
The \R matches any type of line break and + matches one or more occurrences of the quantified subpattern.
Note that in case you have to deal with an older version of Ruby, you will need to use the negated character class [\r\n] that matches either \r or \n:
.gsub(/[\r\n]+/, ' ')
or - add all possible linebreaks:
/gsub(/(?:\u000D\u000A|[\u000A\u000B\u000C\u000D\u0085\u2028\u2029])+/, ' ')
This should work for your test case:
"hello I\r\nam a test\r\n\r\nstring".gsub(/[\r\n]/, ' ')
If you don't want successive \r\n characters to result in duplicate spaces you can use this instead:
"hello I\r\nam a test\r\n\r\nstring".gsub(/[\r\n]+/, ' ')
(Note the addition of the + after the character class.)
As Wiktor mentioned, you're using \\ in your regex, which inside the regex literal /.../ actually escapes a backslash, meaning you're matching a literal backslash \, r, or n as part of your expression. Escaping characters works differently in regex literals, since \ is used so much, it makes no sense to have a special escape for it (as opposed to regular strings, which is a whole different animal).
A string must begin with 3 or 4 letters (not numbers), and a ":" symbol should follow these letters, and after the colon there should be three more characters, like AAA. For example, AAAA:AAA or AAA:AAA.
I`m starting to build this, but regex is so much pain for me, can anyone help me with this?
Here is what I have now:
^[a-zA-Z]{3,4}(:)$
Your regex is almost there: you need to add [a-zA-Z]{3}.
I prefer the [[:alpha:]] POSIX class in Ruby to match letters though.
/[[:alpha:]]/ - Alphabetic character
POSIX bracket expressions are also similar to character classes. They provide a portable alternative to the above, with the added benefit that they encompass non-ASCII characters.
So, here is a possible regex:
\A[[:alpha:]]{3,4}:[[:alpha:]]{3}\z
See demo
The regex matches:
\A - start of string (in RoR, you have to use \A instead of ^, or you will get errors)
[[:alpha:]]{3,4} - 3 or 4 letters
: - literal :
[[:alpha:]]{3} - 3 letters
\z - end of string (in RoR, you have to use \z instead of $, or you will get errors)
To allow just AAA or AAAA, you need to introduce an optional (? quantifier) non-capturing group ((?:...) construction):
\A[[:alpha:]]{3,4}(?::[[:alpha:]]{3})?\z
^^^ ^^
See another demo
Try using this (quotes if regex in your dialect must be passed as a string)
"^[a-zA-Z]{3,4}:[a-zA-Z]{3}$"
In the documentation I read:
Use \A and \z to match the start and end of the string, ^ and $ match the start/end of a line.
I am going to apply a regular expression to check username (or e-mail is the same) submitted by user. Which expression should I use with validates_format_of in model? I can't understand the difference: I've always used ^ and $ ...
If you're depending on the regular expression for validation, you always want to use \A and \z. ^ and $ will only match up until a newline character, which means they could use an email like me#example.com\n<script>dangerous_stuff();</script> and still have it validate, since the regex only sees everything before the \n.
My recommendation would just be completely stripping new lines from a username or email beforehand, since there's pretty much no legitimate reason for one. Then you can safely use EITHER \A \z or ^ $.
According to Pickaxe:
^
Matches the beginning of a line.
$
Matches the end of a line.
\A
Matches the beginning of the string.
\z
Matches the end of the string.
\Z
Matches the end of the string unless the string ends with a "\n", in which case it matches just before the "\n".
So, use \A and lowercase \z. If you use \Z someone could sneak in a newline character. This is not dangerous I think, but might screw up algorithms that assume that there's no whitespace in the string. Depending on your regex and string-length constraints someone could use an invisible name with just a newline character.
JavaScript's implementation of Regex treats \A as a literal 'A' (ref). So watch yourself out there and test.
Difference By Example
/^foo$/ matches any of the following, /\Afoo\z/ does not:
whatever1
foo
whatever2
foo
whatever2
whatever1
foo
/^foo$/ and /\Afoo\z/ all match the following:
foo
The start and end of a string may not necessarily be the same thing as the start and end of a line. Imagine if you used the following as your test string:
my
name
is
Andrew
Notice that the string has many lines in it - the ^ and $ characters allow you to match the beginning and end of those lines (basically treating the \n character as a delimeter) while \A and \Z allow you to match the beginning and end of the entire string.