I’m printing a parameter returned from a query that’s a string of letters and underscores.
The label prints just the letters without the underscores, and I’m not sure how to fix it.
^FD<String>^FS
^FH^FD<String>^FS
Thank you very much.
(Removing the FH Only reads to the first underscore.
The ^FH command without parameter defaults to underscore as the hexidecimal escape character. Either remove the ^FH or specify a different escape character like backslash using ^FH\^FD<String>^FS.
Related
I have the following line in a plugin to display page views on my Jekyll site:
html = pv.to_s.reverse.gsub(/...(?=.)/,'\& ').reverse
It adds space between thousands, for example 23 678.
How can I add hair space instead of regular space in this string?
In HTML is a so-called decimal numeric character reference:
The ampersand must be followed by a "#" (U+0023) character, followed by one or more ASCII digits, representing a base-ten integer that corresponds to a Unicode code point that is allowed according to the definition below. The digits must then be followed by a ";" (U+003B) character.
Ruby has the \u escape sequence. However it expects the following characters to represent a hexadecimal (base-sixteen) integer. That's 200A. You also have to use a double-quoted string literal which means now the \ character needs to be escaped with another one:
"\\&\u200A"
Alternatively just use it directly:
'\& '
So I want to split a string in java on any non-alphanumeric characters.
Currently I have been doing it like this
words= Str.split("\\W+");
However I want to keep apostrophes("'") in there. Is there any regular expression to preserve apostrophes but kick the rest of the junk? Thanks.
words = Str.split("[^\\w']+");
Just add it to the character class. \W is equivalent to [^\w], which you can then add ' to.
Do note, however, that \w also actually includes underscores. If you want to split on underscores as well, you should be using [^a-zA-Z0-9'] instead.
For basic English characters, use
words = Str.split("[^a-zA-Z0-9']+");
If you want to include English words with special characters (such as fiancé) or for languages that use non-English characters, go with
words = Str.split("[^\\p{L}0-9']+");
I have an html file that I need to replace some characters with html entities. Right now I'm trying to replace — with — but when I use the Replace All button, the result is that all of those instances of — are replaced with —mdash;
I thought maybe escaping the "&" will work, so I changed the Replace with value to \— but that just results in \—mdash;
The strange thing is that if I go to each, one by one, i.e., click Next, then click Replace, and so on, then it replaces it correctly.
Is this a bug in MacVim? Or am I missing something?
Enter into command line:
:%s/—/\—/g
Also it's possible to get character code. Place your cursor on the character and press ga. Use decimal, hex or octal code into replacement string:
\%d match specified decimal character
\%x match specified hex character
\%o match specified octal character
\%u match specified multibyte character
\%U match specified large multibyte character
:%s/\%d8212/\$mdash;/g
I am using Ruby on Rails 3.0.9 and I would like to validate a string that can contain only characters (case insensitive characters), blank spaces and numbers.
More:
special characters are not allowed (eg: !"£$%&/()=?^) except - and _;
accented characters are allowed (eg: à, è, é, ò, ...);
The regex that I know from this question is ^[a-zA-Z\d\s]*$ but this do not validate special characters and accented characters.
So, how I should improve the regex?
I wrote the ^(?:[^\W_]|\s)*$ answer in the question you referred to (which actually would have been different if I'd known you wanted to allow _ and -). Not being a Ruby guy myself, I didn't realize that Ruby defaults to not using Unicode for regex matching.
Sorry for my lack of Ruby experience. What you want to do is use the u flag. That switches to Unicode (UTF-8), so accented characters are caught. Here's the pattern you want:
^[\w\s-]*$
And here it is in action at Rubular. This should do the trick, I think.
The u flag works on my original answer as well, though that one isn't meant to allow _ or - characters.
Something like ^[\w\s\-]*$ should validate characters, blank spaces, minus, and underscore.
Validation string only for not allowed characters. In this case |,<,>," and &.
^[^|<>\"&]*$
I am using nokogiri to screen scrape some HTML. In some occurrences, I am getting some weird characters back, I have tracked down the ASCII code for these characters with the following code:
#parser.leads[0].phone_numbers[0].each_byte do |c|
puts "char=#{c}"
end
The characters in question have an ASCII code of 194 and 160.
I want to somehow strip these characters out while parsing.
I have tried the following code but it does not work.
#parser.leads[0].phone_numbers[0].gsub(/160.chr/,'').gsub(/194.chr/,'')
Can anyone tell me how to achieve this?
I found this question while trying to strip out invisible characters when "trimming" a string.
s.strip did not work for me and I found that the invisible character had the ord number 194
None of the methods above worked for me but then I found "Convert non-breaking spaces to spaces in Ruby " question which says:
Use /\u00a0/ to match non-breaking spaces: s.gsub(/\u00a0/, ' ') converts all non-breaking spaces to regular spaces
Use /[[:space:]]/ to match all whitespace, including Unicode whitespace like non-breaking spaces. This is unlike /\s/, which matches only ASCII whitespace.
So glad I found that! Now I'm using:
s.gsub(/[[:space:]]/,'')
This doesn't answer the question of how to gsub specific character codes, but if you're just trying to remove whitespace it seems to work pretty well.
Your problem is that you want to do a method call but instead you're creating a Regexp. You're searching and replacing strings consisting of the string "160" followed by any character and then the string "chr", and then doing the same except with "160" replaced with "194".
Instead, do gsub(160.chr, '').
Update (2018): This code does not work in current Ruby versions. Please refer to other answers.
You can also try
s.gsub(/\xA0|\xC2/, '')
or
s.delete 160.chr+194.chr
First thought would be should you be using gsub! instead of gsub
gsub returns a string and gsub! performs the substitution in place
I was getting "invalid multibyte escape" error while trying the above solution, but for a different situation. Google was return \xA0 when the number is greater than 999 and I wanted to remove it. So what I did was use return_value.gsub(/[\xA0]/n,"") instead and it worked perfectly fine for me.