i have a string that contains an international phone number e.g. +44 9383 33333 but bizzarely when I attempt to regex match (note the regex is correctly 'escaped') the regex match fails
e.g.
"001144 9383 33333".match(/(001144|004|0|\\+44)/) # works
"+44 9383 33333".match(/(001144|004|0|\\+44)/) # DOES NOT WORK
i've tried escaping the input string e.g. +, \+ etc. etc but to no avail.
i must be doing something really stupid here!?
You've got a double backslash, which is telling the regex parser to look for a literal backslash. Since it's also followed by a + the regex parser is then looking for 1 or more backslashes. Try just \+ (so the whole thing should be: /(001144|004|0|\+44)/
Related
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'
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 $, or you will get errors)
To allow just AAA or AAAA, you need to introduce an optional (? quantifier) non-capturing group ((?:...) construction):
\A[[:alpha:]]{3,4}(?::[[:alpha:]]{3})?\z
^^^ ^^
See another demo
Try using this (quotes if regex in your dialect must be passed as a string)
"^[a-zA-Z]{3,4}:[a-zA-Z]{3}$"
I often see the gsub function being called with the pattern parameter enclosed in forward slashes. For example:
>> phrase = "*** and *** ran to the ###."
>> phrase.gsub(/\*\*\*/, "WOOF")
=> "WOOF and WOOF ran to the ###."
I thought maybe it had something to do with escaping asterisks, but using single quotes and double quotes works just as well:
>> phrase = "*** and *** ran to the ###."
>> phrase.gsub('***', "WOOF")
=> "WOOF and WOOF ran to the ###."
>> phrase.gsub("***", "WOOF")
=> "WOOF and WOOF ran to the ###."
Is it just convention to use forward slash? What am I missing?
Use forward slashes if you need to use regular expressions.
If you use a string argument with gsub, it will just do a plain character match.
In your example, you need backslashes to escape the asterisks when using a regular expression, because asterisks have a special meaning in regex (optionally match something any number of times). They are not necessary when using a string, because they are just matched exactly.
In your example, you probably don't need to use a regular expression, since it is a simple pattern. However, if you wanted to match *** only when it was at the beginning of a string (e.g. the first bunch in your example), then you would want to use a regex, for example:
phrase.gsub(/^\*{3}/, "WOOF")
For more information on regular expressions, see: http://www.regular-expressions.info/.
For more information on using regular expressions in Ruby, see: http://ruby-doc.org/core-2.2.0/Regexp.html.
To play with regular expressions as they work in Ruby, try: http://rubular.com/.
You are missing reading the documentation:
The pattern is typically a Regexp; if given as a String, any regular expression metacharacters it contains will be interpreted literally, e.g. '\d' will match a backlash followed by ‘d’, instead of a digit.
http://ruby-doc.org/core-2.1.4/String.html#method-i-gsub
In other words, you can give a string or a regular expression. Regular expressions can be delimited several ways:
Regexps are created using the /.../ and %r{...} literals, and by the Regexp::new constructor.
http://ruby-doc.org/core-2.2.2/Regexp.html
The benefit of %r and of the alternate %r delimiters is you can usually find a delimiter that doesn't collide with characters in the pattern, which would force escaping them, as in your example.
* has to be escaped because it has special meaning in a regex, but in a string it does not.
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
I am trying to write a regex in Ruby to test a string such as:
"GET \"anything/here.txt\""
the point is, everything can be in the outer double quote, but all double quotes in the outer double quotes must be escaped by back slash(otherwise it doesnt match). So for example
"GET "anything/here.txt""
this will not be a proper line.
I tried many ways to write the regex but doest work. can anyone help me with this? thank you
You can use positive lookbehind:
/\A"((?<=\\)"|[^"])*"\z/
This does exactly what you asked for: "if a double quote appears inside the outer double quotes without a backslash prefixed, it doesn't match."
Some comments:
\A,\z: These match only at the beginning and end of the string. So the pattern has to match against the whole string, not a part of it.
(?<=): This is the syntax for positive lookbehind; it asserts that a pattern must match directly before the current position. So (?<=\\)" matches "a double quote which is preceded by a backslash".
[^"]: This matches "any character which is not a backslash".
One point about this regex, is that it will match an inner double quote which is preceded by two backslashes. If that is a problem, post a comment and I'll fix it.
If your version of Ruby doesn't have lookbehind, you could do something like:
/\A"(\\.|[^"\\])*"\z/
Note that unlike the first regexp, this one does not count a double backslash as escaping a quote (rather, the first backslash escapes the second one), so "\\"" will not match.
This works:
/"(?<method>[A-Z]*)\s*\\\"(?<file>[^\\"]*)\\""/
See it on Rubular.
Edit:
"(?<method>[A-Z]*)\s(?<content>(\\\"|[a-z\/\.]*)*)"
See it here.
Edit 2: without (? ...) sequence (for Ruby 1.8.6):
"([A-Z]*)\s((\\\"|[a-z\/\.]*)*)"
Rubular here.
Tested this on Rubular successfully:
\"GET \\\".*\\\"\"
Breakdown:
\" - Escape the " for the regex string, meaning the literal character "
GET - Assuming you just want GET than this is explicit
\\" - Escape \ and " to get the literal string \"
.* - 0 or more of any character other than \n
\\"\" - Escapes for the literal \""
I'm not sure a regex is really your best tool here, but if you insist on using one, I recommend thinking of the string as a sequence of tokens: a quote, then a series of things that are either \\, \" or anything that isn't a quote, then a closing quote at the end. So this:
^"(\\\\|\\"|[^"])*"$