How do a I create a validator, which has these simple rules. An expression is valid if
it must start with a letter
it must end with a letter
it can contain a dash (minus sign), but not at start or end of the expression
^[a-zA-Z]+-?[a-zA-Z]+$
E.g.
def validate(whatever)
reg = /^[a-zA-Z]+-?[a-zA-Z]+$/
return (reg.match(whatever)) ? true : false;
end
/^[A-Za-z]+(-?[A-Za-z]+)?$/
this seems like what you want.
^ = match the start position
^[A-Za-z]+ = start position is followed by any at least one or more letters.
-? = is there zero or one hyphens (use "*" if there can be multiple hyphens in a row).
[A-Za-z]+ = hyphen is followed by one or more letters
(-?[A-Za-z]+)? = for the case that there is a single letter.
$= match the end position in the string.
xmammoth pretty much got it, with one minor problem. My solution is:
^[a-zA-Z]+\-?[a-zA-Z]+$
Note that the original question states, it can contain a dash. The question mark is needed after the dash to make sure that it is optional in the regex.
^[A-Za-z].*[A-Za-z]$
In other words: letter, anything, letter.
Might also want:
^[A-Za-z](.*[A-Za-z])?$
so that a single letter is also matched.
What I meant, to be able to create tags. For example: "Wild-things" or "something-wild" or "into-the-wild" or "in-wilderness" "my-wild-world" etc...
This regular expression matches sequences, that consist of one or more words of letters, that are concatenated by dashes.
^[a-zA-Z]+(?:-[a-zA-Z]+)*$
Well,
[A-Za-z].*[A-Za-z]
According to your rules that will work. It will match anything that:
starts with a letter
ends with a letter
can contain a dash (among everything else) in between.
Related
I am building a Rails 5.2 app.
In this app I got outputs from different suppliers (I am building a webshop).
The name of the shipping provider is in this format:
dhl_freight__233433
It could also be in this format:
postal__US-320202
How can I remove all that is before (and including) the __ so all that remains are the things after the ___ like for example 233433.
Perhaps some sort of RegEx.
A very simple approach would be to use String#split and then pick the second part that is the last part in this example:
"dhl_freight__233433".split('__').last
#=> "233433"
"postal__US-320202".split('__').last
#=> "US-320202"
You can use a very simple Regexp and a ask the resulting MatchData for the post_match part:
p "dhl_freight__233433".match(/__/).post_match
# another (magic) way to acces the post_match part:
p $'
Postscript: Learnt something from this question myself: you don't even have to use a RegExp for this to work. Just "asddfg__qwer".match("__").post_match does the trick (it does the conversion to regexp for you)
r = /[^_]+\z/
"dhl_freight__233433"[r] #=> "233433"
"postal__US-320202"[r] #=> "US-320202"
The regular expression matches one or more characters other than an underscore, followed by the end of the string (\z). The ^ at the beginning of the character class reads, "other than any of the characters that follow".
See String#[].
This assumes that the last underscore is preceded by an underscore. If the last underscore is not preceded by an underscore, in which case there should be no match, add a positive lookbehind:
r = /(?<=__[^_]+\z/
This requires the match to be preceded by two underscores.
There are many ruby ways to extract numbers from string. I hope you're trying to fetch numbers out of a string. Here are some of the ways to do so.
Ref- http://www.ruby-forum.com/topic/125709
line.delete("^0-9")
line.scan(/\d/).join('')
line.tr("^0-9", '')
In the above delete is the fastest to trim numbers out of strings.
All of above extracts numbers from string and joins them. If a string is like this "String-with-67829___numbers-09764" outut would be like this "6782909764"
In case if you want the numbers split like this ["67829", "09764"]
line.split(/[^\d]/).reject { |c| c.empty? }
Hope these answers help you! Happy coding :-)
I have one path in form of string like this Folder1/File.png
But in this string sometimes if file is hidden or folder is hidden I don't want it to be matched by my regex.
regex = %r{([a-zA-Z0-9_ -]*)\/[^.]+$}
input_path = "Folder_1/.file" # This shouldn't be matched.
input_path = "Folder/file.png" # This should be matched.
But my regex works for first input but its not even matching second one.
You are currently looking for \/[^.]+$, that is a / followed by any character except . until the end. Since the filename+extension format has a . character, it fails to match the second case.
Instead of using [^.]+$, check only that the character following / is not ., and match everything after that:
([a-zA-Z0-9_ -]*)\/[^.].*$
While there are some suggestions here that work, my suggestion would be
\/[^.][^\/\n]+$
It finds a slash, followed by anything but a dot, which in turn is followed by one, or more, of anything but a slash or a newline.
To handle the two lines given as an example,
Folder_1/.file
Folder/file.png
it takes 8 steps.
The suggested ones all work, but ([a-zA-Z0-9_ -]*)\/[^.] takes 75 steps, ([a-zA-Z0-9_ -]*)\/[^.]+\.[^.]+\z 78 steps and ([a-zA-Z0-9_ -]*)\/[^.].*$ takes 77 steps.
This may be totally irrelevant and I may have missed some angle, but I wanted to mention it ;)
Se it here at regex101.
regex = %r{([a-zA-Z0-9_ -]*)\/[^.]}
I can't figure out what does this regex match:
A: "\\/\\/c\\/(\\d*)"
B: "\\/\\/(\\d*)"
I suppose they are matching some kind of number sequence since \d matches any digit but I'd like to know an example of a string that would be a match for this regex.
The pattern syntax is that specified by ICU. Expressions are created with NSRegularExpression in an iOS app and are correct.
The first matches //c/ + 0 or more digits. The second matches // + 0 or more digits. In both the digits are captured.
An example of a match for A) is //c/123
An example of a match for B) is //12345
When I use Cygwin which emulates Bash on Windows, I sometimes run into situations where I have to escape my escape characters which is what I think is making this expression look so weird. For instance, when I use sed to look for a single '\' I sometimes have to write it as '\\\\'. (Funny, StackOverflow proved my point. If you write 4 backslashes in the comment, it only shows two. So if you process it again, they might all disappear depending on your situation).
Considering this, it might be helpful to think of pairs of backslashes as representing only one if you're coming from a similar situation. My guess would be you are. Because of this I would say Erik Duymelinck is probably spot on. This will capture a sequence of digits that may or may not follow a couple slashes and a c:
//c/000
//00000
This regex matches an odd sequence of characters, which, at first glance, almost seem like a regex, since \d is a digit, and followed by an asterisk (\d*) would mean zero-or-more digits. But it's not a digit, because the escape-slash is escaped.
\\/\\/c\\/(\\d*)
So, for instance, this one matches the following text:
\/\/c\/\
\/\/c\/\d
\/\/c\/\dd
\/\/c\/\ddd
\/\/c\/\dddd
\/\/c\/\ddddd
\/\/c\/\dddddd
...
This one is almost the same
\\/\\/(\\d*)
except you just delete the c\/ from the above results:
\/\/\
\/\/\d
\/\/\dd
\/\/\ddd
\/\/\dddd
\/\/\ddddd
\/\/\dddddd
...
In both cases, the final \ and optional d is [capture group][1] one.
My first impression was that these regexes were intended for escaping in Java strings, meaning they would be completely invalid. If the were escaped for Java strings, such as
Pattern p = Pattern.compile("\\/\\/c\\/(\\d*)");
It would be invalid, because after un-escaping, it would result in this invalid regex:
\/\/c\/(\d*)
The single escape-slashes (\) are invalid. But the \d is valid, as it would mean any digit.
But again, I don't think they're invalid, and they're not escaped for a Java string. They're just odd.
This is the regular expression which i have, i need to make sure that string does not start or end with underscore , underscore may appear in between.
/^[a-zA-Z0-9_.-]+$/
I have tried
(?!_)
But doesn't seem to work
Allowed strings:
abcd
abcd_123
Not allowed strings:
abcd_
_abcd_123
Not too hard!
/^[^_].*[^_]$/
"Any character except an underscore at the start of the line (^[^_]), then any characters (.*), then any character except an underscore before the end of the line ([^_]$)."
This does require at least two characters to validate the string. If you want to allow one character lines:
/^[^_](.*[^_]|)$/
"Anything except an underscore to start the line, and then either some characters plus a non-underscore character before end-of-line, or just an immediate end-of-line.
You could approach this in the inverse way,
Check all those that do match starting and ending underscores like this:
/^_|_$/
^_ #starts with underscore
| #OR
_$ #ends with underscore
And then eliminate those that match. The above regexp is much more easier to read.
Check : http://www.rubular.com/r/H3Axvol13b
Or you can try the longer regex:
/^[a-zA-Z0-9.-][a-zA-Z0-9_.-]*[a-zA-Z0-9.-]$|^[a-zA-Z0-9.-]+$|^[a-zA-Z0-9.-][a-zA-Z0-9.-]$/
^[a-zA-Z0-9.-] #starts with a-z, or A-Z, or 0-9, or . -
[a-zA-Z0-9_.-]* #anything that can occur and the underscore
[a-zA-Z0-9.-]$ #ends with a-z, or A-Z, or 0-9, or . -
| #OR
^[a-zA-Z0-9.-]$ #for one-letter words
| #OR
^[a-zA-Z0-9.-][a-zA-Z0-9.-]$ #for two letter words
Check: http://www.rubular.com/r/FdtCqW6haG
/^[a-zA-Z0-9.-][a-zA-Z0-9_.-]+[a-zA-Z0-9.-]$/
Try this
Description:
In the first section, [a-zA-Z0-9.-], regex only allows lower and upper case alphabets, digits, dot and hyphen.
In the next section, [a-zA-Z0-9_.-]+, regex looks for a single or more than one characters that are lower or upper case alphabets, digits dot, hyphen or an underscore.
The last part, [a-zA-Z0-9.-], is the same as the first part that restricts the input to end with an underscore.
Try this:
Recently had the same concern and this is how I did it.
// '"^[a-zA-Z0-9_.-]*$"' → Alphanumeric and 「.」「_」「-」
// "^[^_].*[^_]$" → Reject start and end of string if contains 「_」
// (?=) REGEX AND operator
SLUG_REGEX = '"(?=^[a-zA-Z0-9_.-]*$)(?=^[^_].*[^_]$)"';
I used this snippet for my Laravel Validation so you may need to change the code as needed like " to / based on your code sample and other answers' code.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.