Free marker template: how to assign values having double quotes - freemarker

Suppose I have a String literal of value :Hello "Vivek" and I want to assign this to a value in free marker template. My question is how to do it.
<#assign var=" hello "vivek" ">
This seems illogical. need some escaping technique in ftl..

As you can use both " and ' for quoting string literals, the simplest solution is:
<#assign var='hello "vivek"'>
However, if you have both those characters inside the string literal, then you have to escape with \:
<#assign var='hello "vivek"\'s dog'>
or
<#assign var="hello \"vivek\"'s dog">

To escape a double quote in freemarker use the backwards slash. See the documentation for more information.
So your example would be
<#assign var=" \"vivek\" ">

Related

String literal without need to escape backslash

In C#, I can write backslashes and other special characters without escaping by using # before a string, but I have to escape double-quotes.
C#
string foo = "The integer division operator in VisualBASIC is written \"a \\ b\"";
string bar = #"The integer division operator in VisualBASIC is written \"a \ b\""
In Ruby, I could use the single-quote string literal, but I'd like to use this in conjuction with string interpolation like "text #{value}". Is there an equivalent in Ruby to # in C#?
There is somewhat similar thing available in Ruby. E.g.
foo = %Q(The integer division operator in VisualBASIC is written "a \\ b" and #{'interpolation' + ' works'})
You can also interpolate strings in it. The only caveat is, you would still need to escape \ character.
HTH
You can use heredoc with single quotes.
foo = <<'_'
The integer division operator in VisualBASIC is written "a \ b";
_
If you want to get rid of the newline character at the end, then chomp it.
Note that this does not work with string interpolation. If you want to insert evaluated expressions within the string, you can use % operation after you create the string.
foo = <<'_'
text %{value}
_
foo.chomp % {value: "foo"}

Replace text in brackets gsub

I would like to replace text inside of brackets, and that has a colon and a u.
For example, Here is a link [u:person]! would become Here is a link Person! I am not very experienced with regex, and I am having problems with \1 and $1
Here is the regex that I am using now:
string.gsub(/\[(\w*).*?\]/, "<a href='/user/\1'>\1</a>")
Make the regex /\[\w*:(.*?)\]/ so that person can be captured instead of u. Then use a single quoted string so that \1 isn't interpreted as \x01.
str = "Here is a link [u:person]!"
puts str.gsub(/\[\w*:(.*?)\]/, '\1')
# => Here is a link person!
I changed your regular expression to this, so that person is captured:
/\[\w*:(.*?)\]/
And then replaced it with this String:
"#{$1.capitalize}"
You were close with $1, it just needs to be evaluated as Ruby (using String interpolation, inside a block):
string.gsub(/\[\w*:(.*?)\]/) { "#{$1.capitalize}" }
You could use a regex like this:
/\[u\:([\S]+)\]/
and replace it with:
<a href='/user/#{$1}'>#{$1}</a>
Here's a breakdown of what it the regex does:
First, we have \[, which is just the literal [ character
Next, we have u and \:, which are the literal u and : character respectively
Next, we have ([\S]). The parentheses make a capturing group, which is what #{$1} will be filled in with in the replace part of the regex. [\S]+ looks for all non-whitespace characters.
Lastly, we have \], which is just the literal ] character.
Your code should look something like this:
string.gsub('/\[u\:([\S]+)\]/', '<a href='/user/#{$1}'>#{$1}</a>')
Here is a live test of the regex: https://regex101.com/r/vK0iO2
\[[^\]]*u:([^\]]*)\]
Try this.Replace by <a href='/user/\1'>\1</a>.See demo.
https://regex101.com/r/gX5qF3/13

Ruby string sub without regex back references

I'm trying to do a simple string sub in Ruby.
The second argument to sub() is a long piece of minified JavaScript which has regular expressions contained in it. Back references in the regex in this string seem to be effecting the result of sub, because the replaced string (i.e., the first argument) is appearing in the output string.
Example:
input = "string <!--tooreplace--> is here"
output = input.sub("<!--tooreplace-->", "\&")
I want the output to be:
"string \& is here"
Not:
"string & is here"
or if escaping the regex
"string <!--tooreplace--> is here"
Basically, I want some way of doing a string sub that has no regex consequences at all - just a simple string replace.
To avoid having to figure out how to escape the replacement string, use Regex.escape. It's handy when replacements are complicated, or dealing with it is an unnecessary pain. A little helper on String is nice too.
input.sub("<!--toreplace-->", Regexp.escape('\&'))
You can also use block notation to make it simpler (as opposed to Regexp.escape):
=> puts input.sub("<!--tooreplace-->") {'\&'}
string \& is here
Use single quotes and escape the backslash:
output = input.sub("<!--tooreplace-->", '\\\&') #=> "string \\& is here"
Well, since '\\&' (that is, \ followed by &) is being interpreted as a special regex statement, it stands to reason that you need to escape the backslash. In fact, this works:
>> puts 'abc'.sub 'b', '\\\\&'
a\&c
Note that \\\\& represents the literal string \\&.

Escaping single and double quotes in a string in ruby?

How can I escape single and double quotes in a string?
I want to escape single and double quotes together. I know how to pass them separately but don't know how to pass both of them.
e.g: str = "ruby 'on rails" " = ruby 'on rails"
My preferred way is to not worry about escaping and instead use %q, which behaves like a single-quote string (no interpolation or character escaping), or %Q for double quoted string behavior:
str = %q[ruby 'on rails" ] # like single-quoting
str2 = %Q[quoting with #{str}] # like double-quoting: will insert variable
See https://docs.ruby-lang.org/en/trunk/syntax/literals_rdoc.html#label-Strings and search for % strings.
Use backslash to escape characters
str = "ruby \'on rails\" "
Here is a complete list:
From http://learnrubythehardway.org/book/ex10.html
You can use Q strings which allow you to use any delimiter you like:
str = %Q|ruby 'on rails" " = ruby 'on rails|
>> str = "ruby 'on rails\" \" = ruby 'on rails"
=> "ruby 'on rails" " = ruby 'on rails"
I would go with a heredoc if I'm starting to have to worry about escaping. It will take care of it for you:
string = <<MARKER
I don't have to "worry" about escaping!!'"!!
MARKER
MARKER delineates the start/end of the string. start string on the next line after opening the heredoc, then end the string by using the delineator again on it's own line.
This does all the escaping needed and converts to a double quoted string:
string
=> "I don't have to \"worry\" about escaping!!'\"!!\n"
I would use just:
str = %(ruby 'on rails ")
Because just % stands for double quotes(or %Q) and allows interpolation of variables on the string.
Here is an example of how to use %Q[] in a more complex scenario:
%Q[
<meta property="og:title" content="#{#title}" />
<meta property="og:description" content="#{#fullname}'s profile. #{#fullname}'s location, ranking, outcomes, and more." />
].html_safe
One caveat:
Using %Q[] and %q[] for string comparisons is not intuitively safe.
For example, if you load something meant to signify something empty, like "" or '', you need to use the actual escape sequences. For example, let's say qvar equals "" instead of any empty string.
This will evaluate to false
if qvar == "%Q[]"
As will this,
if qvar == %Q[]
While this will evaluate to true
if qvar == "\"\""
I ran into this issue when sending command-line vars from a different stack to my ruby script. Only Gabriel Augusto's answer worked for me.

using regular expressions in ruby to find a string in quotations

I am trying to construct a regex to find a string in ruby
str = "foo"
I want to be able to stop trying to find the string after it finds the closing quotation mark. I also want to keep the quotation marks so I can output the string I found as:
puts "the string is:" + str
=> the string is: "foo"
I am pretty new to using regular expressions.
Here is a start:
/".*?"/
Explanation:
" Match a literal double quote.
.*? Match any characters, as few as possible (non-greedy)
" Match a second literal double quote.
Rubular
Note that this won't work if the string contains escaped quotes or is quoted with single quotes.

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