I'm trying to basically trying to remove the . in the the string .extension
So I'm using
'.extension'[1..10] #gives extension
Of course this will work for 10 characters, how would I make it work for any length. I dont know what to search for because I dont know what this array style [x..y] is called.
You can use negative indexes. This will offset them from the end. -1 is the last element, -2 is the second last and so on.
'.extension'[1..-1] # => extension
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.
I'm using Rails and Nokogiri and I'm trying to parse some website.
This is where I'm stuck:
doc.css('#example > li:nth-child(1)').each do |node|
money = node.xpath('//*ul/li/div/span').text
end
It returns something like:
$100,000£230,000$40,000$9,000€600$800,000
I want to split those items, save them to the database and finally hand them to the view.
So, in the view, I want it to appear like:
(1)$100,000
(2)£230,000
(3)$40,000
(4)$9,000
(5)€600
(6)$800,000
I tried to split those items by this code below.
money = node.xpath('//*ul/li/div/span').text.split(/[$€£]/)
but the result looks like this:
["", "100,000", "230,000", "40,000", "9,000", "600", "800,000"]
And I don't know which item is in Dollar, Euro, or Pond.
Is there any good way to solve this problem?
you're almost there,
just use the positive lookahead :)
irb(main):005:0> "$100,000£230,000$40,000$9,000€600$800,000".split(/(?=[$£€])/)
=> ["$100,000", "£230,000", "$40,000", "$9,000", "€600", "$800,000"]
It needs a regular expression. This works:
"$100,000£230,000$40,000$9,000$600$800,000".scan(/([^\d][0-9,]+)/)
=> [["$100,000"],
["£230,000"],
["$40,000"],
["$9,000"],
["$600"],
["$800,000"]]
The regex contains these parts:
[^\d]: A character class matching a single non-digit. This will match the currency symbol.
`[0-9,]+': Another character class, this time repeating (the '+'). It matches the numeric part (0-9) plus the thousand's separator.
Hi I've been struggling with this for the last hour and am no closer. How exactly do I strip everything except numbers, commas and decimal points from a rails string? The closest I have so far is:-
rate = rate.gsub!(/[^0-9]/i, '')
This strips everything but the numbers. When I try add commas to the expression, everything is getting stripped. I got the aboves from somewhere else and as far as I can gather:
^ = not
Everything to the left of the comma gets replaced by what's in the '' on the right
No idea what the /i does
I'm very new to gsub. Does anyone know of a good tutorial on building expressions?
Thanks
Try:
rate = rate.gsub(/[^0-9,\.]/, '')
Basically, you know the ^ means not when inside the character class brackets [] which you are using, and then you can just add the comma to the list. The decimal needs to be escaped with a backslash because in regular expressions they are a special character that means "match anything".
Also, be aware of whether you are using gsub or gsub!
gsub! has the bang, so it edits the instance of the string you're passing in, rather than returning another one.
So if using gsub! it would be:
rate.gsub!(/[^0-9,\.]/, '')
And rate would be altered.
If you do not want to alter the original variable, then you can use the version without the bang (and assign it to a different var):
cleaned_rate = rate.gsub!(/[^0-9,\.]/, '')
I'd just google for tutorials. I haven't used one. Regexes are a LOT of time and trial and error (and table-flipping).
This is a cool tool to use with a mini cheat-sheet on it for ruby that allows you to quickly edit and test your expression:
http://rubular.com/
You can just add the comma and period in the square-bracketed expression:
rate.gsub(/[^0-9,.]/, '')
You don't need the i for case-insensitivity for numbers and symbols.
There's lots of info on regular expressions, regex, etc. Maybe search for those instead of gsub.
You can use this:
rate = rate.gsub!(/[^0-9\.\,]/g,'')
Also check this out to learn more about regular expressions:
http://www.regexr.com/
I am using ruby on rails
I have
article.id = 509969989168Q000475601
I would like the output to be
article.id = 68Q000475601
basically want to get rid of all before it gets to 68Q
the numbers in front of the 68Q can be various length
is there a way to remove up to "68Q"
it will always be 68Q and Q is always the only Letter
is there a way to say remove all characters from 2 digits before "Q"
I'd use:
article.id[/68Q.*/]
Which will return everything from 68Q to the end of the string.
article.id.match(/68Q.+\z/)[0]
You can do this easily with the split method:
'68Q' + article.id.split('68Q')[1]
This splits the string into an array based on the delimiter you give it, then takes the second element of that array. For what it's worth though, #theTinMan's solution is far more elegant.
How do I check a value starting with a decimal
is_a_number(value) .... works for 12, 12.0, 12.2, 0.23 but not .23
Basically I'm doing a validation in a form, and I want to allow values starting with . i.e .23
but obviously pop up a flag (false) when its not a number
".23" isn't really a number, in my book.
If you want to treat it like one, check if the first character is a decimal point, if it is, prepend a "0" and try again.
Actually, you could probably prepend a zero regardless. It shouldn't affect the value of any "legitimate" number. (EDIT: As long as you can explicitly specify base 10 when actually converting to a number)
Read your input into a string and dynamically add the zero if needed. For example:
if (inputvar[0] == '.')
inputvar = "0#{inputvar}"
end
The resulting value can be converted into a number by .to_i, .to_f, etc.