Ruby Regular Expressions: Matching if substring doesn't exist - ruby

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...

Related

simple symbol regex solution

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

Content Inside Parenthesis Regular Expression Ruby

I'm trying to take out the the content inside the parenthesis. For example, if the string is "(blah blah) This is stack(over)flow", I want to just take out "(blah blah)" but leave "(over)" alone. I'm trying
/\A\(.*\)/
but returns "(blah blah) This is stack(over)", and I'm sure why it's returning that.
Easiest fix:
/\A\(.*?\)/
Normally, * will try to match as much as it possibly can, so it'll match all the way to the last ) in the line. This is called "greedy" matching. Putting ? after +/*/? makes them non-greedy, and they'll match the shortest possible string.
But note that this won't work for nested parentheses. That's rather more complicated. Given your example, I assume this is for a pretty simple ad-hoc format where nesting isn't a concern.

Regex Tag-Within-Tag

I have a fairly simple regex problem for a little personal experiment that I haven't quite figured out.
In a string, I might have several <tag>[some characters here] that I need to match. The obvious way to do it would be with a /<tag>\[.*?\]/ regex, to match any characters after the <tag>[ and before the ].
I'd like to be able to have <tag>s within <tag>s, however. This causes a problem. If I had the following:
<tag>[some characters <tag>[in here] to match]
the regex would stop matching as soon as it reached the first closing-bracket, and completely fail to match the last part of the statement. I've tried to solve the problem by telling the regex to ignore any internal <tag>s, so I can do a match on the stripped contents later. I haven't quite gotten it working. The closest I've come is:
/<tag>\[(.*?(?:<tag>\[.*?\])*?.*?)\]/
which doesn't quite work. I would hope that it would match any number of characters, and any inner tags if they exist. It still has trouble with that first closing bracket, however.
Maybe somebody who's better at regular expressions knows a good solution to this.
Though you should probably drop regex and do this manually if the mini-language becomes more complex, you can use recursive regex.
Your regex would look something like this:
/(?<reg>(\w+\[([^\]\[]|\g<reg>)*\]))/
You can see it in action here: http://rubular.com/r/9F7isgZpj9
Here is the regex broken down to its parts:
(?<reg>( # start a regex named "reg"
\w+ # the tag name
\[ # open bracket
( # which can contain
[^\]\[] # non-bracket characters
| # or
\g<reg> # sub-tags (this is where the magic happens)
)* # zero or more times
\] # close the tag
)
)

how to use regex negation string

can any body tell me how to use regex for negation of string?
I wanna find all line that start with public class and then any thing except first,second and finally any thing else.
for example in the result i expect to see public class base but not public class myfirst:base
can any body help me please??
Use a negative lookahead:
public\s+class\s+(?!first|second).+
If Peter is correct and you're using Visual Studio's Find feature, this should work:
^:b*public:b+class:b+~(first|second):i.*$
:b matches a space or tab
~(...) is how VS does a negative lookahead
:i matches a C/C++ identifier
The rest is standard regex syntax:
^ for beginning of line
$ for end of line
. for any character
* for zero or more
+ for one or more
| for alternation
Both the other two answers come close, but probably fail for different reasons.
public\s+class\s+(?:(?!first|second).)+
Note how there is a (non-capturing) group around the negative lookahead, to ensure it applies to more than just the first position.
And that group is less restrictive - since . excludes newline, it's using that instead of \S, and the $ is not necessary - this will exclude the specified words and match others.
No slashes wrapping the expression since those aren't required in everything and may confuse people that have only encountered string-based regex use.
If this still fails, post the exact content that is wrongly matched or missed, and what language/ide you are using.
Update:
Turns out you're using Visual Studio, which has it's own special regex implementation, for some unfathomable reason. So, you'll be wanting to try this instead:
public:b+class:b+~(first|second)+$
I have no way of testing that - if it doesn't work, try dropping the $, but otherwise you'll have to find a VS user. Or better still, the VS engineer(s) responsible for this stupid non-standard regex.
Here is something that should work for you
/public\sclass\s(?:[^fs\s]+|(?!first|second)\S)+(?=\s|$)/
The second look a head could be changed to a $(end of line) or another anchor that works for your particular use case, like maybe a '{'
Edit: Try changing the last part to:
(?=\s|$)

Problem With Regular Expression to Remove HTML Tags

In my Ruby app, I've used the following method and regular expression to remove all HTML tags from a string:
str.gsub(/<\/?[^>]*>/,"")
This regular expression did just about all I was expecting it to, except it caused all quotation marks to be transformed into “
and all single quotes to be changed to ”
.
What's the obvious thing I'm missing to convert the messy codes back into their proper characters?
Edit: The problem occurs with or without the Regular Expression, so it's clear my problem has nothing to do with it. My question now is how to deal with this formatting error and correct it. Thanks!
Use CGI::unescapeHTML after you perform your regular expression substitution:
CGI::unescapeHTML(str.gsub(/<\/?[^>]*>/,""))
See http://www.ruby-doc.org/core/classes/CGI.html#M000547
In the above code snippet, gsub removes all HTML tags. Then, unescapeHTML() reverts all HTML entities (such as <, &#8220) to their actual characters (<, quotes, etc.)
With respect to another post on this page, note that you will never ever be passed HTML such as
<tag attribute="<value>">2 + 3 < 6</tag>
(which is invalid HTML); what you may receive is, instead:
<tag attribute="<value>">2 + 3 < 6</tag>
The call to gsub will transform the above to:
2 + 3 < 6
And unescapeHTML will finish the job:
2 + 3 < 6
You're going to run into more trouble when you see something like:
<doohickey name="<foobar>">
You'll want to apply something like:
gsub(/<[^<>]*>/, "")
...for as long as the pattern matches.
This regular expression did just about
all I was expecting it to, except it
caused all quotation marks to be
transformed into “ and all
single quotes to be changed to ”
.
This doesn't sound as if the RegExp would be doing this. Are you sure it's different before?
See this question here for information about the problem, it has got an excellent answer:
Get non UTF-8 form fields as UTF-8 in php.
I've run into a similar problem with character changes, this happened when my code ran through another module that enforced UTF-8 encoding and then when it came back, I had a different file (slurped array of lines) on my hands.
You could use a multi-pass system to get the results you are looking for.
After running your regular expression, run an expression to convert &8220; to quotes and another to convert &8221; to single quotes.

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