simple symbol regex solution - ruby

The problem I'm looking at says only inputs with '+' symbols covering any letters in the string is true so like "+d++" or "+d+==+a+" but not
"f++d+"
"3+a=+b+"
"++d+=c+"
I tried to solve this using regex since it's kind of a string pattern matching problem. /(+[a-z][^+])|([^+.][a-z]+)/ but this does not cover patterns where the letters are at the beginning or end of the string. I need help something more comprehensive.


You should try following
/^\+{0,2}[a-z0-9]+\+{0,2}(=*\+{0-2}[a-z0-9]+\+{0,2})*$/


You could use the below regex.
^(?:[^\w\n]*\+[a-z]+\+)+[^\w\n]*$
DEMO
If you want to match +f+g+ also, then put the following + inside a positive lookahead assertion.
^(?:[^\w\n]*\+[a-z]+(?=\+))+[^\w\n]*$
DEMO

Related

Writing RegEx to locate a certain word in a string

I have the following code
b.element(:id, "SearchFor").send_keys "843842983"
b.element(:xpath, "//input[#value='Search']").click
b.text_field(:id, "AmountPaid").set "343"
b.element(:id, "paytype0").click
b.element(:css, "label.radio-custom-label.ng-binding").click
b.element(:id, "xxx").send_keys "11"
I want to locate b.element if .send_keys has been followed
for an example,I want to locate b.element when the string is
b.element(:id, "xxx").send_keys "11"
not when
b.element(:id, "paytype0").click
I am able to locate the whole string b.element().send_keys by writing the following regular expression b.element\(.*\).send_keys but I could not locate b.element.
Can anyone help me how to do that?

You are looking for asserting rather than matching. Asserting doesn't consume characters but matching does. For assertions you have to use lookarounds here:
b\.element(?=\([^)]*\)\.send_keys)
This is nearly your own solution but put in a positive lookahead.
Live demo
P.S. escape dots when they literally mean a dot character

Avoid combination of hyphen and space using a regex

I'm currently writing a very specific regex for a firstname field, that has several requirements. One of them is that spaces are not allowed before or after hyphens. For this, I have used a negative lookahead:
(?!.*(\s\-))
as part of the regex:
^(?!ß)(?!.*(\s\-))(?!(.)\1{2})(?!.*\s{2})(?!.*\'{2})(?!.*\-{2})[a-zA-ZßöüäÜÖÄ\s\-\']{2,30}(?<![\s\-])$
It does return a mismatch for:
asdf -asdf
but not for:
asdf- asdf
The latter also need to return an error. What am I missing?

You have to assert the other combination of hyphens and whitespaces absent in your string also:
(?!.*(\s\-))(?!.*(\-\s))

You can rewrite your pattern in a more simple way that avoids many problems and makes your pattern more efficient, example:
^(?=.{2,30}$)(?!(.)\1{2})[a-zA-ZöüäÜÖÄ]+(?:[-'\s][a-zA-ZßöüäÜÖÄ]+)*$

Simplest is probably a negative lookahead right after the ^:
/^(?!.*(\s-|-\s))#{main_pattern}/

ruby regex match any character besides a specific one

I am looking for a way to match any character besides, for example, a "#."
It would look something like...
gsub(/^foo.*foo$/)
But I'd want it to match
"foofdfdfdfoo"
But not
"fooddgdgd#fdfoo"
Thanks.

^[^#]+$
http://rubular.com/r/glijo99dU9
gsub is for substitution. If you just want to match, the .match method

To expand on Explosion Pills answer, a caret (^) will negate the match in a regex. This means that it will not match if the characters following it are found in the expression. You can read more about it in the documentation.

Ruby Regular Expressions: Matching if substring doesn't exist

I'm having an issue trying to capture a group on a string:
"type=gist\nYou need to gist this though\nbecause its awesome\nright now\n</code></p>\n\n<script src=\"https://gist.github.com/3931634.js\"> </script>\n\n\n<p><code>Not code</code></p>\n"
My regex currently looks like this:
/<code>([\s\S]*)<\/code>/
My goal is to get everything in between the code brackets. Unfortunately, it's matching up to the 2nd closing code bracket Is there a way to match everything inside the code brackets up until the first occurrence of ending code bracket?

All repetition quantifiers in regular expressions are greedy by default (matching as many characters as possible). Make the * ungreedy, like this:
/<code>([\s\S]*?)<\/code>/
But please consider using a DOM parser instead. Regex is just not the right tool to parse HTML.

And I just learned that for going through multiple parts, the
String.scan( /<code>(.*?)<\/code>/ ){
puts $1
}
is a very nice way of going through all occurences of code - but yes, getting a proper parser is better...

How can I simplify this regular expression?

The format I'm trying to match is:
# (Apple push notification codes)
"11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7"
The simplest expression I can think of is: /((\w{8}\s){7}\w{8})/i
Can anyone think of a simpler one?
(I'm using Ruby regular expressions)
UPDATE - thanks to user1096188, I've removed \d - this is included in \w

You can detect a word boundary using \b, and use (?: to prevent capturing groups
/(?:\w{8}\b\s?){8}/

You could do this if the end of the match is the end of the whole string.
(\w{8}(:?\s|$)){7}

Taking #zapthedingbat's solution one stage further, it looks like the code only contains hexadecimal characters (0-9 and a-f) and spaces. So you could possibly sacrifice a little simplicity for accuracy.
I'm making an assumption, but I suspect letters g to z are invalid.
If the format is hexadecimal only (you should check Apple's documentation to be sure), a tighter match would be:
/(?:[0-9a-f]{8}\b\s?){8}/
EDIT
In fact, in Ruby, it looks like you should be able to do:
/(?:\h{8}\b\s?){8}/

> "11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7".match(/((\w{8}\s)+)/)
> $&
=> "11a735e9 9f696c2f 700b2700 728042c6 137eeb7a 8442c27d 40e59d9e 3c7e0de7"

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