I am using Yahoo Pipes to rewrite a URL. Everything is fine but the last 6 characters of the URL need to be removed from all the links.
Ex.
http://www.mysite.com/blahblah/34567
needs to be rewritten to
http://www.mysite.com/blahblah
The number at the end is always changing, so I am hoping to just chop the last 6 characters off of each url.
You'll want the Sub string module. Presuming it functions in a similar way to other languages, you'll want an undefined From value, and a Length value of -6, which should cut 6 characters off the end of the URL.
Seems that a regex in (.*)(.....) and takinq only the first match should do the trick
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 am facing an issue in creating a ODATA URL.
For the Following URL,
https://xxxxxxx.xxx.xxxxxxxx.com/sap/c4c/odata/v1/c4codataapi/CorporateAccountHasContactPersonCollection?$filter=AccountID eq '1000024'
Result :
- <m:properties>
<d:ObjectID>00163E10AD0B1ED686EF458B4E8C51D5</d:ObjectID>
<d:ParentObjectID>00163E10AD201EE5A4F0B592DE751AE8</d:ParentObjectID
<d:AccountID>1000024</d:AccountID>
<d:ContactID>1002636</d:ContactID>
<d:FunctionCode>Z021</d:FunctionCode>
<d:Mobile>+33 123456789</d:Mobile>
<d:Phone>+33 987654321</d:Phone>
</m:properties>
Same for result when i change filter to FunctionCode
https://xxxxxxx.xxx.xxxxxxxx.com/sap/c4c/odata/v1/c4codataapi/CorporateAccountHasContactPersonCollection?$filter=FunctionCode eq 'Z021'
But When I search with filter Phone
https://xxxxxxx.xxx.xxxxxxxx.com/sap/c4c/odata/v1/c4codataapi/CorporateAccountHasContactPersonCollection?$filter=Phone eq '+33 123456789'
URL doesnt work at all. Is it becuase of special character "+" in Phone Number
I tried with $filter=endswith(Phone, '123456789'), It worked fine. But this is not solution i am looking for.
Can any one suggest another ways?
Thank you
Regards
Prat
The reason is that these special characters has different meaning when used in URLs. The JavaScript “encodeUri” or “encodeUriComponent” does not solve this problem.
Here is the list of the special characters that needs to be replaced when used in the OData queries:
https://msdn.microsoft.com/en-us/library/aa226544(SQL.80).aspx
The special character + Indicates a space (and spaces cannot be used in a URL) so you need to replace the character's with its hexadecimal value, in this case %2B.
Your corrected filter should then be $filter=Phone eq '%2B33 123456789'.
Find a good article here.
I'm attempting to construct a regex string in Swift 4 that gets characters at the start of a line where some are known and others aren't.
Let's say I've got a text file with line breaks for each word that reads as follows:
pucker
tuckered
duckerdinger
sucker punch
I'd like to get every word that contains "cker" in it that's 1 to 8 characters long.
I'm attempting to use this statement ^..cker..{1,8} as my RegEx string. All I'm getting is a partial match in Patterns (a Mac App), but Regex101.com's saying no match, and most importantly, Xcode says I'm using an invalid regex. I've also tried ^(..cker..) and a bazillion other variations.
What am I screwing up and how do I fix it? What I'm trying to do seems like it would be super simple, but I've wasted more time than I care to admit fiddling with it.
Update:
This has been the best I've been able to get so far...
"\\b..cker..", but I'm only able to get words that are exactly 8 characters long. I'd like to capture words that contain "cker" that are the 3rd, 4th, 5th, and 6th letters while capturing words up to 8 characters long.
Try this regex:
\b(?=.*cker)[a-zA-Z]{1,8}\b
Click for Demo
Explanation:
\b - matches a word boundary
(?=.*cker) - Positive Lookahead to make sure our string should contain the character sequence cker
[a-zA-Z]{1,8} - Matches 1 to 8 occurrences of a letter
\b - matches a word boundary
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/
text.scan(/\"[\d\w\s\+\-\*\/]*\"/)
I'm simply looking to find any thing within quotations that can contain letters, numbers, spaces, plus, minus, star, or forward slash. Everything works great in console. Each of the following works in a browser:
"abc"
"123"
"x-1" or "x - 1"
"x/1" or "x / 1"
But the plus sign and star fail in a browser (despite working fine in console with the same regex). Does anyone have any ideas?
Edit #1: I'm performing a quick gsub to add some formatting to the results of the scan. If the quotations have a plus or star in them, they don't even get picked up by the scan. The same code and text pasted in console works just fine.
Edit #2: I figured out a better way to frame this question without extraneous details and got the answer. "Why can't I perform a gsub on each of the results from a scan if the result contains regex special characters?"
Turned out that this problem was related to regexp string insertion (/#{whatever}/) not escaping special characters - manually escaping clears it up (/#{Regexp.escape(whatever)}/). See this question for a full example/explanation.
I don't know what do you mean "work in browser" but I'm making an assumption that you're trying to parse an URL. In URL the + & * signs can be converted to %2B & %2A respectively.
Try this regexp:
/"[(\d\w\s\+\-\*\/|%2B|%2A)]+"/
...or decode URL before parsing.