I have a string like
rule = ".radio-inline,\n.checkbox-inline {\n display: inline-block;\n}"
The string contains special characters like \n. How can I return the part of the string before {}?
For the above string it should return ".radio-inline,\n.checkbox-inline "
I tried to write it using
Regexp.escape(rule).match(/(.*){/)[1]
The problem is that it returns:
"\\.radio\\-inline,\\n\\.checkbox\\-inline\\ \\"
This should do it:
rule = ".radio-inline,\n.checkbox-inline {\n display: inline-block;\n}"
rule[/.*(?={)/m]
#=> ".radio-inline,\n.checkbox-inline "
.* matches zero or more characters from the beginning of the line. Being greedy by default, it matches as many characters as it can.
(?={), a positive lookahead, requires the character { to immediately follow the matched string, but it is not part of the match.
m stipulates that the match is to be made over multiple lines.
This uses the form of the method String#[] that takes a regex as an argument.
This regex gets you the part befor {}:
[^"{}]+(?={.*})
Try this demo
However it doesn't select you the " at the and and beggining, I hope this will work for you
Related
Beginner here, obviously. I need to add a sum and a string together and from the product, I have to replace every 4th character with underscore, the end product should look something like this: 160_bws_np8_1a
I think .gsub is the way, but I can find a way to format the first part in .gsub where I have to specify every 4th character.
total = (1..num).sum
final_output = "#{total.to_s}" + "06bwsmnp851a"
return final_output.gsub(//, "_")
This would work:
s = '12345678901234'
s.gsub(/(...)./, '\1_')
#=> "123_567_901_34"
The regex matches 3 characters (...) that are captured (parentheses) followed by another character (.). Each match is replaced by the first capture (\1) and a literal underscore (_).
s = "12345678901234"
Here are two ways to do that. Both return
"123_567_901_34"
Match every four-character substring and replace the match with the first three characters of the match followed by an underscore
s.gsub(/.{4}/) { |s| s[0,3] << '_' }
Chain the enumerator s.gsub(/./) to Enumerator#with_index and replace every fourth character with an underscore
s.gsub(/./).with_index { |c,i| i%4 == 3 ? '_' : c }
See the form of String#gsub that takes a single argument and no block.
We have a string which contains address in it like below:
"first-name, last-name, email, address\n Ashok, G, \"Hyderabad\nTelangana\n India\"\n John, M, \"Mayur Vihar\nNew Delhi\n110096, India\"\n"
and the requirement is to replace all the newline characters ("\n") characters with "" from the address string only (inside \" \")
The Expected output should be like:
"first-name, last-name, email, address\n Ashok, G, \"Hyderabad Telangana India\"\n John, M, \"Mayur Vihar, New Delhi 110096, India\"\n "
\\n(?=(?:(?!\\").)*\\"(?:(?:(?!\\").)*\\"(?:(?!\\").)*\\")*(?:(?!\\").)*$)
Try this.Replace by empty string.See demo.
https://www.regex101.com/r/rG7gX4/7
I suggest you do it as follows:
str.gsub(/(?<=\").*?(?=\")/) { |s| s.gsub(/\n/,' ') }
#=> "first-name, last-name, email, address\n Ashok, G, \"heyderabad |
Telangana India\" ABCD, L, \"Guntur AP 500505, India\"\n"
This matches each string bracketed by \", which in turn is passed to the block for removal of all \n's. (?<=\") is a positive lookbehind; (?=\") is a postive lookahead. ? is needed to make .* non-greedy, so the match stops before the first matching postive lookahead.
This doesn't give quite the spacing contained in your desired output. That spacing seems somewhat inconsistent, however. For example, where did the single space at the end of the string come from? You said you wanted to replace \n between pairs of \", but you didn't say what you want to replace it with. (I assumed one space.) If you want different spacing, you could adjust the regex used by gsub inside the block. For example, you might have /\s*\n\s*/.
Can someone tell me, what's wrong in this code:
if ((!preg_match("[a-zA-Z0-9 \.\s]", $username)) || (!preg_match("[a-zA-Z0-9 \.\s]", $password)));
exit("result_message=Error: invalid characters");
}
??
Several things are wrong. I assume that the code you are looking for is:
if (preg_match('~[^a-z0-9\h.]~i', $username) || preg_match('~[^a-z0-9\h.]~i', $password))
exit('result_message=Error: invalid characters');
What is wrong in your code?
the pattern [a-zA-Z0-9 \.\s] is false for multiple reasons:
a regex pattern in PHP must by enclosed by delimiters, the most used is /, but as you can see, I have choosen ~. Example: /[a-zA-Z \.\s]/
the character class is strange because it contains a space and the character class \s that contains the space too. IMO, to check a username or a password, you only need the space and why not the tab, but not the carriage return or the line feed character! You can remove \s and let the space, or you can use the \h character class that matches all horizontal white spaces. /[a-zA-Z\h\.]/ (if you don't want to allow tabs, replace the \h by a space)
the dot has no special meaning inside a character class and doesn't need to be escaped: /[a-zA-Z\h.]/
you are trying to verify a whole string, but your pattern matches a single character! In other words, the pattern checks only if the string contains at least an alnum, a space or a dot. If you want to check all the string you must use a quantifier + and anchors for the start ^ and the end $ of the string. Example ∕^[a-zA-Z0-9\h.]+$/
in fine, you can shorten the character class by using the case-insensitive modifier i: /^[a-z0-9\h.]+$/i
But there is a faster way, instead of negate with ! your preg_match assertion and test if all characters are in the character range you want, you can only test if there is one character you don't want in the string. To do this you only need to negate the character class by inserting a ^ at the first place:
preg_match('/[^a-z0-9\h.]/i', ...
(Note that the ^ has a different meaning inside and outside a character class. If ^ isn't at the begining of a character class, it is a simple literal character.)
Apparently I still don't understand exactly how it works ...
Here is my problem: I'm trying to match numbers in strings such as:
910 -6.258000 6.290
That string should gives me an array like this:
[910, -6.2580000, 6.290]
while the string
blabla9999 some more text 1.1
should not be matched.
The regex I'm trying to use is
/([-]?\d+[.]?\d+)/
but it doesn't do exactly that. Could someone help me ?
It would be great if the answer could clarify the use of the parenthesis in the matching.
Here's a pattern that works:
/^[^\d]+?\d+[^\d]+?\d+[\.]?\d+$/
Note that [^\d]+ means at least one non digit character.
On second thought, here's a more generic solution that doesn't need to deal with regular expressions:
str.gsub(/[^\d.-]+/, " ").split.collect{|d| d.to_f}
Example:
str = "blabla9999 some more text -1.1"
Parsed:
[9999.0, -1.1]
The parenthesis have different meanings.
[] defines a character class, that means one character is matched that is part of this class
() is defining a capturing group, the string that is matched by this part in brackets is put into a variable.
You did not define any anchors so your pattern will match your second string
blabla9999 some more text 1.1
^^^^ here ^^^ and here
Maybe this is more what you wanted
^(\s*-?\d+(?:\.\d+)?\s*)+$
See it here on Regexr
^ anchors the pattern to the start of the string and $ to the end.
it allows Whitespace \s before and after the number and an optional fraction part (?:\.\d+)? This kind of pattern will be matched at least once.
maybe /(-?\d+(.\d+)?)+/
irb(main):010:0> "910 -6.258000 6.290".scan(/(\-?\d+(\.\d+)?)+/).map{|x| x[0]}
=> ["910", "-6.258000", "6.290"]
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map(&:to_f)
# => [910.0, -6.258, 6.29]
If you don't want integers to be converted to floats, try this:
str = " 910 -6.258000 6.290"
str.scan(/-?\d+\.?\d+/).map do |ns|
ns[/\./] ? ns.to_f : ns.to_i
end
# => [910, -6.258, 6.29]
I have the following string:
"h3. My Title Goes Here"
I basically want to remove the first four characters from the string so that I just get back:
"My Title Goes Here".
The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first four characters blindly.
I checked the docs and the closest thing I could find was chomp, but that only works for the end of a string.
Right now I am doing this:
"h3. My Title Goes Here".reverse.chomp(" .3h").reverse
This gives me my desired output, but there has to be a better way. I don't want to reverse a string twice for no reason. Is there another method that will work?
To alter the original string, use sub!, e.g.:
my_strings = [ "h3. My Title Goes Here", "No h3. at the start of this line" ]
my_strings.each { |s| s.sub!(/^h3\. /, '') }
To not alter the original and only return the result, remove the exclamation point, i.e. use sub. In the general case you may have regular expressions that you can and want to match more than one instance of, in that case use gsub! and gsub—without the g only the first match is replaced (as you want here, and in any case the ^ can only match once to the start of the string).
You can use sub with a regular expression:
s = 'h3. foo'
s.sub!(/^h[0-9]+\. /, '')
puts s
Output:
foo
The regular expression should be understood as follows:
^ Match from the start of the string.
h A literal "h".
[0-9] A digit from 0-9.
+ One or more of the previous (i.e. one or more digits)
\. A literal period.
A space (yes, spaces are significant by default in regular expressions!)
You can modify the regular expression to suit your needs. See a regular expression tutorial or syntax guide, for example here.
A standard approach would be to use regular expressions:
"h3. My Title Goes Here".gsub /^h3\. /, '' #=> "My Title Goes Here"
gsub means globally substitute and it replaces a pattern by a string, in this case an empty string.
The regular expression is enclosed in / and constitutes of:
^ means beginning of the string
h3 is matched literally, so it means h3
\. - a dot normally means any character so we escape it with a backslash
is matched literally