Here's what I've tried so far. Is there something wrong with my Regex?
^(?:|0|[1-9]\\d*)(?:\\.\\d*)?.{10}$
What i did was values must be numbers only and with a maximum number of 10. I have no idea what my code doesn't catch when I input more than 10 numbers.
You can use following REGEX
/^(\+\d{1,3}[- ]?)?\d{10}$/
For more information use following links:
http://www.regular-expressions.info/numericranges.html
http://www.regexr.com/
You want only numeric values from 1 to maximum 10 so this is the REGEX for that:
[0-9]{1,10}
If this is not what you need then let us know a little bit more.
You can use https://www.regex101.com to test your REGEX
Use this regex, it also supports country code and spacing: /^(\+\d{1,3}[- ]?)?\d{10}$/
Do you want to have a maximum of 10 numbers only?
The below regex works for 8-10 digits in the number
if ($phone =~ /^[0-9]{8,10}$/) {
print "this is a valid phone number\n";
} else {
print "not a valid phone number \n";
}
Related
I've been looking for a good way to see if a string of items are all numbers, and thought there might be a way of specifying a range from 0 to 9 and seeing if they're included in the string, but all that I've looked up online has really confused me.
def validate_pin(pin)
(pin.length == 4 || pin.length == 6) && pin.count("0-9") == pin.length
end
The code above is someone else's work and I've been trying to identify how it works. It's a pin checker - takes in a set of characters and ensures the string is either 4 or 6 digits and all numbers - but how does the range work?
When I did this problem I tried to use to_a? Integer and a bunch of other things including ranges such as (0..9) and ("0..9) and ("0".."9") to validate a character is an integer. When I saw ("0-9) it confused the heck out of me, and half an hour of googling and youtube has only left me with regex tutorials (which I'm interested in, but currently just trying to get the basics down)
So to sum this up, my goal is to understand a more semantic/concise way to identify if a character is an integer. Whatever is the simplest way. All and any feedback is welcome. I am a new rubyist and trying to get down my fundamentals. Thank You.
Regex really is the right way to do this. It's specifically for testing patterns in strings. This is how you'd test "do all characters in this string fall in the range of characters 0-9?":
pin.match(/\A[0-9]+\z/)
This regex says "Does this string start and end with at least one of the characters 0-9, with nothing else in between?" - the \A and \z are start-of-string and end-of-string matchers, and the [0-9]+ matches any one or more of any character in that range.
You could even do your entire check in one line of regex:
pin.match(/\A([0-9]{4}|[0-9]{6})\z/)
Which says "Does this string consist of the characters 0-9 repeated exactly 4 times, or the characters 0-9, repeated exactly 6 times?"
Ruby's String#count method does something similar to this, though it just counts the number of occurrences of the characters passed, and it uses something similar to regex ranges to allow you to specify character ranges.
The sequence c1-c2 means all characters between c1 and c2.
Thus, it expands the parameter "0-9" into the list of characters "0123456789", and then it tests how many of the characters in the string match that list of characters.
This will work to verify that a certain number of numbers exist in the string, and the length checks let you implicitly test that no other characters exist in the string. However, regexes let you assert that directly, by ensuring that the whole string matches a given pattern, including length constraints.
Count everything non-digit in pin and check if this count is zero:
pin.count("^0-9").zero?
Since you seem to be looking for answers outside regex and since Chris already spelled out how the count method was being implemented in the example above, I'll try to add one more idea for testing whether a string is an Integer or not:
pin.to_i.to_s == pin
What we're doing is converting the string to an integer, converting that result back to a string, and then testing to see if anything changed during the process. If the result is =>true, then you know nothing changed during the conversion to an integer and therefore the string is only an Integer.
EDIT:
The example above only works if the entire string is an Integer and won’t properly deal with leading zeros. If you want to check to make sure each and every character is an Integer then do something like this instead:
pin.prepend(“1”).to_i.to_s(1..-1) == pin
Part of the question seems to be exactly HOW the following portion of code is doing its job:
pin.count("0-9")
This piece of the code is simply returning a count of how many instances of the numbers 0 through 9 exist in the string. That's only one piece of the relevant section of code though. You need to look at the rest of the line to make sense of it:
pin.count("0-9") == pin.length
The first part counts how many instances then the second part compares that to the length of the string. If they are equal (==) then that means every character in the string is an Integer.
Sometimes negation can be used to advantage:
!pin.match?(/\D/) && [4,6].include?(pin.length)
pin.match?(/\D/) returns true if the string contains a character other than a digit (matching /\D/), in which case it it would be negated to false.
One advantage of using negation here is that if the string contains a character other than a digit pin.match?(/\D/) would return true as soon as a non-digit is found, as opposed to methods that examine all the characters in the string.
If i have a string like "123123123" - Here 123 is repeated 3 times.
1. So how can i get only "123" in ruby?
2. So if the string is "12312312" - Here 123 is repeated 2 times and then just 12, so here still i need to get "123".
3. Even if string is 99123123123, still i need to get 123.
Is this possible in Ruby Regex?
EDIT: I want this to solve Project Euler Problem 26 . So here 123 can be anything. All i want is to extract 1 number of at-least 2 repeated numbers.
This regex will detect all repeating groups.
(\d+)(?=.*\1)
Demo
Works great with ruby too.
result = '9912341234123'.scan(/(\d+)(?=.*\1)/)
#gets group with largest length
longestRepeatingGroup = result.max_by{|arr| arr[0].length}
puts longestRepeatingGroup
puts longestRepeatingGroup[0].length
Try this
99123123123.scan(/123/).count
12312312.scan(/123/).count
Trying to work out how to parse out phone numbers that are left in a string.
e.g.
"Hi Han, this is Chewie, Could you give me a call on 02031234567"
"Hi Han, this is Chewie, Could you give me a call on +442031234567"
"Hi Han, this is Chewie, Could you give me a call on +44 (0) 203 123 4567"
"Hi Han, this is Chewie, Could you give me a call on 0207-123-4567"
"Hi Han, this is Chewie, Could you give me a call on 02031234567 OR +44207-1234567"
And be able to consistently replace any one of them with some other item (e.g. some text, or a link).
Am assuming it's a regex type approach (I'm already doing something similar with email which works well).
I've got to
text.scan(/([^A-Z|^"]{6,})/i)
Which leaves me a leading space I can't work out how to drop (would appreciate the help there).
Is there a standard way of doing this that people use?
It also drops things into arrays, which isn't particularly helpful
i.e. if there were multiple numbers.
[["02031234567"]["+44207-1234567"]]
as opposed to
["02031234567","+44207-1234567"]
Adding in the third use-case with spaces is difficult. I think the only way to successfully meet that acceptance criteria would be to chain a #gsub call on to your #scan.
Thus:
text.gsub(/\s+/, "").scan(/([^A-Z|^"|^\s]{6,})/i)
The following code will extract all the numbers for you:
text.scan(/(?<=[ ])[\d \-+()]+$|(?<=[ ])[\d \-+()]+(?=[ ]\w)/)
For the examples you supplied this results in:
["02031234567"]
["+442031234567"]
["+44 (0) 203 123 4567"]
["0207-123-4567"]
["02031234567", "+44207-1234567"]
To understand this regex, what we are matching is:
[\d \-+()]+ which is a sequence of one or more digits, spaces, minus, plus, opening or closing brackets (in any order - NB regex is greedy by default, so it will match as many of these characters next to each other as possible)
that must be preceded by a space (?<=[ ]) - NB the space in the positive look-behind is not captured, and therefore this makes sure that there are no leading spaces in the results
and is either at the end of the string $, or | is followed by a space then a word character (?=[ ]\w) (NB this lookahead is not captured)
This pattern will get rid of the space but not match your third case with spaces:
/([^A-Z|^"|^\s]{6,})/i
This is what I came to in the end in case it helps somebody
numbers = text.scan(/([^A-Z|^"]{6,})/i).collect{|x| x[0].strip }
That gives me an array of
["+442031234567", "02031234567"]
I'm sure there is a more elegant way of doing this and possibly you'd want to check the numbers for likelihood of being phonelike - e.g. using the brilliant Phony gem.
numbers = text.scan(/([^A-Z|^"]{6,})/i).collect{|x| x[0].strip }
real_numbers = numbers.keep_if{|n| Phony.plausible? PhonyRails.normalize_number(n, default_country_code: "GB")}
Which should help exclude serial numbers or the like from being identified as numbers. You'll obviously want to change the country code to something relevant for you.
I have a Cisco ASA 8.4 VPN Concentrator. I am trying to use Lua to extract digits from a certificate string coming in and use them in a LDAP lookup with AD for authorization. I found a string that works...sometimes.
The string comes in with the format:
LAST_NAME.FIRST_NAME.MIDDLE_NAME.1234567890
My LDAP only wants to see the digits and #domainname. The script I am currently us is: return string.gsub(cert.subject.cn, "^(%w+)%.(%w+)%.(%w+)%.(%w+)$", "%4#domain")
This script works fine in most cases (80-90% of the time). When it doesn't work is when people have no middle name, 4 names instead of 3, etc.
My question is how can I get it to output only the 10 digits, regardless of what comes before it. Seems too easy with a return string.match, but so far I can't get it to work. Any ideas?
You can use the pattern .*(%d%d%d%d%d%d%d%d%d%d)$:
local str = 'LAST_NAME.FIRST_NAME.MIDDLE_NAME.1234567890'
print(str:match('.*(' .. ('%d'):rep(10) .. ')$'))
or .*(%d+)$ if the number of digits is always 10.
If the 10 digits is always the last 10 characters, this works:
print(str:sub(-10, -1))
In my application I need to get the SIM card number. I am getting that with SIMCardInfo.getIMSI(), but in some other format.
My SIM number is 89919400002018929130, but I am getting: 404940.20.189291.3
I used the following code:
try
{
currentSimNo = GPRSInfo.imeiToString(SIMCardInfo.getIMSI());
}
catch (Exception e)
{
}
The imeiToString takes a IMEI not a IMSI
sampath says:
My SIM number is 89919400002018929130, but I am getting: 404940.20.189291.3
If I line up your two numbers and remove separators, it does look like they are roughly the same:
89919400002018929130
404940201892913
Where did you get the first one from?
In the Wikipedia article on IMSI, you can see that the first three digits are the MCC, then the second three are the MNC. So it looks like your SIM is from India, though I don't see a MNC that matches 940.
These values are frequently encoded as binary-coded decimal, which is why the imeiToString method needs to be used.
EDIT
The IMSI wikipedia article also mentions:
An IMSI is usually presented as a 15 digit long number, but can be shorter.
I think your code is returning the correct value as 404940201892913. Whatever number you are trying to compare it to (899194...) must be encoded differently.