When using FreeMarker, I want to replace some words in the template, but replace function does not handle word boundary, so my output is messed. Is it possible handle word boundary in FreeMarker? Thanks!
Edit:
The word boundary question is solved, but I have another question about backreference.
I just found I should use the third optional flag 'r' to tell FreeMarker that I am using regular expression. For my purpose, I use something like this:
block?replace("\\b${arg}\\b", "__${arg}", "r")
Note we must use \\b for the word boundary matching.
Related
I have an XSL template that takes 2 parameters (text and separator) and calls tokenize($text, $separator).
The problem is, the separator is supposed to be a regex. I have no idea what I get passed as separator string.
In Java I would call Pattern.quote(separator) to get a pattern that matches the exact string, no matter what weird characters it might contain, but I could not find anything like that in XPath.
I could iterate through the string and escape any character that I recognize as a special character with regard to regex.
I could build an iteration with substring-before and do the tokenize that way.
I was wondering if there is an easier, more straightforward way?
You could escape your separator tokens using the XPath replace function to find any character that requires escaping, and precede each occurrence with a \. Then you could pass such an escaped token to the XPath tokenize function.
Alternatively you could just implement your own tokenize stylesheet function, and use the substring-before and substring-after functions to extract substrings, and recursion to process the full string.
I'm completely new to Ruby so I was just wondering if someone could help me out.
I have the following String:
"<planKey><key>OR-J8U</key></planKey>"
What is the regex I have to write to get the center part OR-J8U?
Use the following:
str = "<planKey><key>OR-J8U</key></planKey>"
str[/(?<=\<key\>).*(?=\<\/key\>)/]
#=> "OR-J8U"
This captures anything in between opening and closing 'key' tags using lookahead and lookbehinds
If you want to get the string OR-J8U then you could simply use that string in the regular expression; the - character has to be escaped:
/OR\-J8U/
Though, I believe you want any string that is enclosed within <planKey><key> and </key></planKey>. In that case ice's answer is useful if you allow for an empty string:
/(?<=\<key\>).*(?=\<\/key\>)/
If you don't allow for an empty string, replace the * with +:
/(?<=\<key\>).*(?=\<\/key\>)/
If you prefer a more general approach (any string enclosed within any tags), then I believe the common opinion is not to use a regular expression. Instead consider using an HTML parser. On SO you can find some questions and answers in that regard.
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
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...
I want to transform the following text
This is a , here is another 
into
This is a , here is another 
In other words I want to find all the image paths that are enclosed between brackets (the text is in Markdown syntax) and replace them with other paths. The string containing the new path is returned by a separate real_path function.
I would like to do this using String#gsub in its block version. Currently my code looks like this:
re = /!\[.*?\]\((.*?)\)/
rel_content = content.gsub(re) do |path|
real_path(path)
end
The problem with this regex is that it will match  instead of just foto.jpeg. I also tried other regexen like (?>\!\[.*?\]\()(.*?)(?>\)) but to no avail.
My current workaround is to split the path and reassemble it later.
Is there a Ruby regex that matches only the path inside the brackets and not all the contextual required characters?
Post-answers update: The main problem here is that Ruby's regexen have no way to specify zero-width lookbehinds. The most generic solution is to group what the part of regexp before and the one after the real matching part, i.e. /(pre)(matching-part)(post)/, and reconstruct the full string afterwards.
In this case the solution would be
re = /(!\[.*?\]\()(.*?)(\))/
rel_content = content.gsub(re) do
$1 + real_path($2) + $3
end
A quick solution (adjust as necessary):
s = 'This is a '
s.sub!(/!(\[.*?\])\((.*?)\)/, '\1(/folder1/\2)' )
p s # This is a [foto](/folder1/foto.jpeg)
You can always do it in two steps - first extract the whole image expression out and then second replace the link:
str = "This is a , here is another "
str.gsub(/\!\[[^\]]*\]\(([^)]*)\)/) do |image|
image.gsub(/(?<=\()(.*)(?=\))/) do |link|
"/a/new/path/" + link
end
end
#=> "This is a , here is another "
I changed the first regex a bit, but you can use the same one you had before in its place. image is the image expression like , and link is just the path like foto.jpeg.
[EDIT] Clarification: Ruby does have lookbehinds (and they are used in my answer):
You can create lookbehinds with (?<=regex) for positive and (?<!regex) for negative, where regex is an arbitrary regex expression subject to the following condition. Regexp expressions in lookbehinds they have to be fixed width due to limitations on the regex implementation, which means that they can't include expressions with an unknown number of repetitions or alternations with different-width choices. If you try to do that, you'll get an error. (The restriction doesn't apply to lookaheads though).
In your case, the [foto] part has a variable width (foto can be any string) so it can't go into a lookbehind due to the above. However, lookbehind is exactly what we need since it's a zero-width match, and we take advantage of that in the second regex which only needs to worry about (fixed-length) compulsory open parentheses.
Obviously you can put real_path in from here, but I just wanted a test-able example.
I think that this approach is more flexible and more readable than reconstructing the string through the match group variables
In your block, use $1 to access the first capture group ($2 for the second and so on).
From the documentation:
In the block form, the current match string is passed in as a parameter, and variables such as $1, $2, $`, $&, and $' will be set appropriately. The value returned by the block will be substituted for the match on each call.
As a side note, some people think '\1' inappropriate for situations where an unconfirmed number of characters are matched. For example, if you want to match and modify the middle content, how can you protect the characters on both sides?
It's easy. Put a bracket around something else.
For example, I hope replace a-ruby-porgramming-book-531070.png to a-ruby-porgramming-book.png. Remove context between last "-" and last ".".
I can use /.*(-.*?)\./ match -531070. Now how should I replace it? Notice
everything else does not have a definite format.
The answer is to put brackets around something else, then protect them:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1.')
# => "a-ruby-porgramming-book.png"
If you want add something before matched content, you can use:
"a-ruby-porgramming-book-531070.png".sub(/(.*)(-.*?)\./, '\1-2019\2.')
# => "a-ruby-porgramming-book-2019-531070.png"