I have a - in my regular expression. When I run it, I get warning: character class has '-' without escape: /[^a-zA-Z0-9-_\\.]/
How can I silence this warning?
At this place [^a-zA-Z0-9-_\\.] there is a possibility of an error. Usage of '-' either means the new range(with mistyping) or just a symbol. Either move single characters in front of ranges or prefix with '\'
verbosity = $VERBOSE
$VERBOSE = nil
/[^a-zA-Z0-9-_\\.]/ #=> /[^a-zA-Z0-9-_\\.]/
$VERBOSE = verbosity
Add a backslash (\) to escape the lonely dash. It appears the offending dash is immediately after [^a-zA-Z0-9. Add \- instead of just the dash character -. The full regex including the minor change is
/[^a-zA-Z0-9\-_\\.]/
This will remove the warning. Escaping it with the backslash clearly instructs the regex engine that you mean to search for that character and not that you forgot to complete a range like [a-z]
Related
When I use
``# ``
in my Sphinx documentation I get the following warning:
WARNING: Inline literal start-string without end-string.
Trying
:samp:`# `
leads to
WARNING: Inline interpreted text or phrase reference start-string without end-string.
The problem seems to be the trailing whitespace however I couldn't figure out a way of getting around this problem. Escaping the whitespace with a backslash (\) doesn't help either (for the first example the warning persists and for the second example the whitespace is omitted in the generated docs).
This answer doesn't work because the inline code section interprets the |space| as a literal string.
Experienced with Sphinx 1.6.2.
A workaround is to use a no-break space character (U+00A0) instead of a regular space (U+0020) for the trailing whitespace.
There are several ways to insert a literal no-break space character. See https://en.wikipedia.org/wiki/Non-breaking_space#Keyboard_entry_methods.
Use a "literal" role__ with an escaped space after the intended trailing space::
:literal:`# \ `
__https://docutils.sourceforge.io/docs/ref/rst/roles.html#literal
I want to use this regex to match any block comment (c-style) in a string.
But why the below does not?
rblockcmt = Regexp.new "/\\*[.\s]*?\\*/" # match block comment
p rblockcmt=~"/* 22/Nov - add fee update */"
==> nil
And in addition to what Sir Swoveland posted, a . matches any character except a newline:
The following metacharacters also behave like character classes:
/./ - Any character except a newline.
https://ruby-doc.org/core-2.3.0/Regexp.html
If you need . to match a newline, you can specify the m flag, e.g. /.*?/m
Options
The end delimiter for a regexp can be followed by one or more
single-letter options which control how the pattern can match.
/pat/i - Ignore case
/pat/m - Treat a newline as a character matched by .
...
https://ruby-doc.org/core-2.3.0/Regexp.html
Because having exceptions/quirks like newline not matching a . can be painful, some people specify the m option for every regex they write.
It appears that you intend [.\s]*? to match any character or a whitespace, zero or more times, lazily. Firstly, whitespaces are characters, so you don't need \s. That simplifies your expression to [.]*?. Secondly, if your intent is to match any character there is no need for a character class, just write .. Thirdly, and most importantly, a period within a character class is simply the character ".".
You want .*? (or [^*]*).
I want to create a regular expression such as:
/\\\s*\\\s*$/
I am trying it in this way:
Regexp.new('\\\s*\\\s*$') # => /\\s*\\s*$/
What am I doing wrong?
Well (\\) matches a single backslash. Backslash serves as an escape character for Regexp.
rgx = Regexp.new('\\\\\\s*\\\\\\s*$')
A more verbose way of doing this would be the following as #Cary Swoveland stated.
rgx = Regexp.new('\\{3}s*\\{3}s*$')
Using literal notation avoids some confusion. This compiles to what you said you want:
/\\\s*\\\s*$/
Though, to be clear, this still matches a single backslash, optional whitespace a single backslash and more optional whitespace. Backslashes are escaped when you inspect a regexp.
I have deprecated "eregi" problem.
if (eregi("data/cheditor4[^<>]*\.(gif|jp[e]?g|png|bmp)", $edit_img, $tmp))
So I changed into this,
if (preg_match("/data/cheditor4[^<>]*\.(gif|jp[e]?g|png|bmp)/i", $edit_img,$tmp))
But I got new warmingmessage,
Warning: preg_match() [function.preg-match]: Unknown modifier 'c'
Please let me know what is wrong.
Thanks in advance.
You have a '/' inside your regex ('data/cheditor') but you are also using '/' as the regex delimiter ('/myregex/flags'): you can either escape the internal '/', or use a different regex delimiter.
E.g. first option:
preg_match('/data\/cheditor4[^<>]*\.(gif|jp[e]?g|png|bmp)/i',...
or (second option, I chose '#' as the delimiter):
preg_match('#data/cheditor4[^<>]*\.(gif|jp[e]?g|png|bmp)#i',...
Also note I changed the " around your regex to ' because otherwise you need to double the backslashes within double-quotes in PHP.
I am trying to write a regex in Ruby to test a string such as:
"GET \"anything/here.txt\""
the point is, everything can be in the outer double quote, but all double quotes in the outer double quotes must be escaped by back slash(otherwise it doesnt match). So for example
"GET "anything/here.txt""
this will not be a proper line.
I tried many ways to write the regex but doest work. can anyone help me with this? thank you
You can use positive lookbehind:
/\A"((?<=\\)"|[^"])*"\z/
This does exactly what you asked for: "if a double quote appears inside the outer double quotes without a backslash prefixed, it doesn't match."
Some comments:
\A,\z: These match only at the beginning and end of the string. So the pattern has to match against the whole string, not a part of it.
(?<=): This is the syntax for positive lookbehind; it asserts that a pattern must match directly before the current position. So (?<=\\)" matches "a double quote which is preceded by a backslash".
[^"]: This matches "any character which is not a backslash".
One point about this regex, is that it will match an inner double quote which is preceded by two backslashes. If that is a problem, post a comment and I'll fix it.
If your version of Ruby doesn't have lookbehind, you could do something like:
/\A"(\\.|[^"\\])*"\z/
Note that unlike the first regexp, this one does not count a double backslash as escaping a quote (rather, the first backslash escapes the second one), so "\\"" will not match.
This works:
/"(?<method>[A-Z]*)\s*\\\"(?<file>[^\\"]*)\\""/
See it on Rubular.
Edit:
"(?<method>[A-Z]*)\s(?<content>(\\\"|[a-z\/\.]*)*)"
See it here.
Edit 2: without (? ...) sequence (for Ruby 1.8.6):
"([A-Z]*)\s((\\\"|[a-z\/\.]*)*)"
Rubular here.
Tested this on Rubular successfully:
\"GET \\\".*\\\"\"
Breakdown:
\" - Escape the " for the regex string, meaning the literal character "
GET - Assuming you just want GET than this is explicit
\\" - Escape \ and " to get the literal string \"
.* - 0 or more of any character other than \n
\\"\" - Escapes for the literal \""
I'm not sure a regex is really your best tool here, but if you insist on using one, I recommend thinking of the string as a sequence of tokens: a quote, then a series of things that are either \\, \" or anything that isn't a quote, then a closing quote at the end. So this:
^"(\\\\|\\"|[^"])*"$