I need the username to be two or more characters of a-z, 0-9, all downcase. This is the current regex I am using
USER_REGEX = /\A[a-z0-9][-a-z0-9]{1,19}\z/i
With this regex, users are able to use uppercase charters in their username. How do I modify the current regex to avoid that?
The regular expression to filter for two to twenty lower-case characters or digits is
/^[a-z0-9]{2,20}$/
which means:
^ at the front of input
a-z accept lower-case 'a' through 'z'
0-9 accept '0' through '9'
{2,20} accept 2 to 20 elements from preceding [] block
$ until the end of input
You can make a regular expression case-insensitive with trailing i, as in your example; that appears to be the root of problem. That said, I don't know Ruby's peculiarities with respect to regular expressions.
If you must keep the RegEx - remove the "i" from the end
USER_REGEX = /\A[a-z0-9][-a-z0-9]{1,19}\z/i
USER_REGEX = /\A[a-z0-9][-a-z0-9]{1,19}\z/
the "i" tells the RegEx to be a case-insensitive RegEx.
but you want it to be case-sensitive and only match on lowercase letters.
Related
I want to select all the commas in a string that do not have any white space around. Suppose I have this string:
"He,she, They"
I want to select only the comma between he and she. I tried this in rubular and came up with this regex:
(,[^(,\s)(\s,)])
This selects the comma that I want, but also selects an s which is a character after it.
In your regex (,[^(,\s)(\s,)]) you capture a comma followed by a negated character class that matches not any of the specified characters, which could also be written as (,[^)(,\s]) which will capture for example ,s in a group,
What you could do is use a positive lookahead and a positve lookbehind to check what is on the left and what is on the right is not a \S whitespace character:
(?<=\S),(?=\S)
Regex demo
In Ruby, you may use [[:space:]] to match any (Unicode) whitespace and [^[:space:]] to match any char other than whitespace. Using these character classes inside lookarounds solves the problem:
/(?<=[^[:space:]]),(?=[^[:space:]])/
See the Rubular demo
Here,
(?<=[^[:space:]]) - a positive lookbehind that matches a location that is immediately preceded with a non-whitespace char (if the string start position should also be matched, replace with (?<![[:space:]]))
, - a comma
(?=[^[:space:]]) - a positive lookahead that matches a location that is immediately followed with a non-whitespace char (if the string end position should also be matched, replace with (?![[:space:]])).
Check the regex below and use the code hope it will help you!
re = /[^\s](,)[^\s]/m
str = 'check ,my,domain, qwe,sd'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Check LIVE DEMO HERE
How do I express in regex the letter "s" whose next non-space character is not a "/"?
These should match: "s", "str"
These should not: "s/m", "s /n"
I tried this
"str" =~ /s[^[[:space:]]]^\// #=> nil
but it does not even match the simple use case.
It seems you need to match any s that is not followed with any 0+ whitespace chars and a / after them.
Use
/s(?![[:space:]]*\/)/
See the Rubular demo.
Details
s - the letter s
(?![[:space:]]*\/) - a negative lookahead that fails the match if, immediately to the right of the current location, there are
[[:space:]]* - 0+ whitespaces
\/ - a /.
If you merely want to know the number of 's' characters that are not followed by zero or more spaces and then a forward slash (as opposed to their indices in the string), you don't have to use a regular expression.
"sea shells /by the sea s/hore".delete(" ").gsub("s/", "").count("s")
#=> 3
If you only want to know if there is at least one such 's' you could replace count("s") with include?("s").
I'm not arguing that this is preferable to the use of a regular expression.
I am currently working on a ruby program to calculate terms. It works perfectly fine except for one thing: brackets. I need to filter the content or at least, to put the content into an array, but I have tried for an hour to come up with a solution. Here is my code:
splitted = term.split(/\(+|\)+/)
I need an array instead of the brackets, for example:
"1-(2+3)" #=>["1", "-", ["2", "+", "3"]]
I already tried this:
/(\((?<=.*)\))/
but it returned:
Invalid pattern in look-behind.
Can someone help me with this?
UPDATE
I forgot to mention, that my program will split the term, I only need the content of the brackets to be an array.
If you need to keep track of the hierarchy of parentheses with arrays, you won't manage it just with regular expressions. You'll need to parse the string word by word, and keep a stack of expressions.
Pseudocode:
Expressions = new stack
Add new array on stack
while word in string:
if word is "(": Add new array on stack
Else if word is ")": Remove the last array from the stack and add it to the (next) last array of the stack
Else: Add the word to the last array of the stack
When exiting the loop, there should be only one array in the stack (if not, you have inconsistent opening/closing parentheses).
Note: If your ultimate goal is to evaluate the expression, you could save time and parse the string in Postfix aka Reverse-Polish Notation.
Also consider using off-the-shelf libraries.
A solution depends on the pattern you expect between the parentheses, which you have not specified. (For example, for "(st12uv)" you might want ["st", "12", "uv"], ["st12", "uv"], ["st1", "2uv"] and so on). If, as in your example, it is a natural number followed by a +, followed by another natural number, you could do this:
str = "1-( 2+ 3)"
r = /
\(\s* # match a left parenthesis followed by >= 0 whitespace chars
(\d+) # match one or more digits in a capture group
\s* # match >= 0 whitespace chars
(\+) # match a plus sign in a capture group
\s* # match >= 0 whitespace chars
(\d+) # match one or more digits in a capture group
\s* # match >= 0 whitespace chars
\) # match a right parenthesis
/x
str.scan(r0).first
=> ["2", "+", "3"]
Suppose instead + could be +, -, * or /. Then you could change:
(\+)
to:
([-+*\/])
Note that, in a character class, + needn't be escaped and - needn't be escaped if it is the first or last character of the class (as in those cases it would not signify a range).
Incidentally, you received the error message, "Invalid pattern in look-behind" because Ruby's lookarounds cannot contain variable-length matches (i.e., .*). With positive lookbehinds you can get around that by using \K instead. For example,
r = /
\d+ # match one or more digits
\K # forget everything previously matched
[a-z]+ # match one or more lowercase letters
/x
"123abc"[r] #=> "abc"
I have a string:
app_copy--28.ipa
The result I want is:
app_copy
The number after -- could be of variable length, so I want to match everything including and after --.
I've tried a few patterns, but none are matching for some reason:
gsub("--\*", "")
gsub("--*", "")
gsub("--*.ipa", "")
gsub("--\[0-9].ipa", "")
What am I missing?
Let's take a look at your test patterns:
"--\*" is actually equivalent to "--*" (since the \* is an escape sequence).
"--*" will match a single - character, followed by zero or more - characters.
"--*.ipa" will match a single - character, followed by zero or more - characters, followed by any single character, followed by a literal ipa.
"--\[0-9].ipa" is actually equivalent to "--[0-9].ipa" (since the \[ is an escape sequence), which will match a literal --, followed by a single decimal digit, followed by any single character, followed by a literal ipa.
However, none of these patterns would work as you used them because gsub will not treat it as a regular expression:
The pattern is typically a Regexp; if given as a String, any regular expression metacharacters it contains will be interpreted literally…
You'd need to wrap type convert your pattern to a Regexp (using Regexp.new), or use a regular expression literal.
Try this pattern
--.*
This pattern will find any literal --, followed by zero or more of any character.
For example:
"app_copy--28.ipa".gsub(/--.*/, "") # app_copy
Don't use gsub to try to change the string, simply use a pattern to match the part you want:
"app_copy--28.ipa"[/^(.+?)--/, 1] # => "app_copy"
String's [] takes a lot of different types of parameters. You can pass in a pattern, and the index of the capture that you want, to extract just that part. From the documentation:
str[regexp, capture] → new_str or nil
If a Regexp is supplied, the matching portion of the string is returned. If a capture follows the regular expression, which may be a capture group index or name, follows the regular expression that component of the MatchData is returned instead.
How is this ?
str = "app_copy--28.ipa"
str[0..str.index("-")-1]
# => "app_copy"
str = "app_copy--28.ipa"
str.split("--").first
# => "app_copy"
I want to remove any leading and trailing non-alphabetic character in my string.
for eg. ":----- pt-br:-" , i want "pt-br"
Thanks
result = subject.gsub(/\A[\d_\W]+|[\d_\W]+\Z/, '')
will remove non-letters from the start and end of the string.
\A and \Z anchor the regex at the start/end of the string (^/$ would also match after/before a newline which is probably not what you want - but that might not matter in this case);
[\d_\W]+ matches one or more digits, the underscore or anything else that is not an alphanumeric character, leaving only letters.
| is the alternation operator.
In ruby 1.9.1 :
":----- pt-br:-".partition( /[a-zA-Z](...)[a-zA-Z]/ )[1]
partition searches the pattern in the string and returns the part before it, the match, and the part after it.
result = subject.gsub(/^[^a-zA-Z]+/, '').gsub(/[^a-zA-Z]+$/, '')