Trying to create Regex code that starts with # and then matches any alphanumeric character.
How do I make a code for Regex to match a line that starts with # and then followed by \w ?
I believe you want the following:
^#\w+
The ^ makes sure the string actually starts with the expression that follows.
We only check for one # character so we just place that there without anything else.
The \w includes all alphanumeric characters and underscores.
The + ensures there is at least one of the preceding token (at least one alphanumeric character in your case).
You should really try to attempt to learn expressions on your own though and try before you ask. I recommend this website to learn from: https://regexr.com/
Related
Given a file name of 22-PLUMB-CLR-RECTANGULAR.0001.rfa I need a RegEx to match it. Basically it's any possible characters, then . and 4 digits and one of four possible file extensions.
I tried ^.?\.\d{4}\.(rvt|rfa|rte|rft)$ , which I would have thought would be correct, but I guess my RegEx understanding has not progressed as much as I thought/hoped. Now, .?\.\d{4}\.(rvt|rfa|rte|rft)$ does work and the ONLY difference is that I am not specifying the start of the string with ^. In a different situation where the file name is always in the form journal.####.txt I used ^journal\.\d{4}\.txt$ and it matched journal.0001.txt like a champ. Obviously when I am specifying a specific string, not any characters with .? is the difference, but I don't understand the WHY.
That never matches the mentioned string since ^.? means match beginning of input string then one optional single character. Then it looks for a sequence of dots and digits and nothing's there. Because we didn't yet pass the first character.
Why does it work without ^? Because without ^ it is allowed to go through all characters to find a match and it stops right before R and continues matching up to the end.
That's fine but with first approach it should be ^.*. Kleene star matches every thing greedily then backtracks but ? is the operator here which makes preceding pattern optional. That means one character, not many characters.
So I've got /.+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]\#[\w+-?]+(.{1})\w{2,}/ pattern I want to use for email validation on client-side, which doesn't work as expected.
I know that my pattern is simple and doesn't cover every standard possibility, but it's part of my regex training.
Local part of address should be valid only when it has at least one digit [0-9] or letter [a-zA-Z] and can be mixed with comma or plus sign or underscore (or all at once) and then # sign, then domain part, but no IP address literals, only domain names with at least one letter or digit, followed by one dot and at least two letters or two digits.
In test string form it doesn't validate a#b.com and does validate baz_bar.test+private#e-mail-testing-service..com, which is wrong - it should be vice versa - validate a#b.com and not validate baz_bar.test+private#e-mail-testing-service..com
What specific error I've got there and where?
I can't locate this, sorry..
You need to change your regex
From: .+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]\#[\w+-?]+(\.{1})\w{2,}
To: .+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]?\#[\w+-]+(\.{1})\w{2,}
Notice that I added a ? before the # sign and removed the ? from the first "group" after the # sign. Adding that ? will make your regex to know that hole "group" is not mandatory.
See it working here: https://regex101.com/r/iX5zB5/2
You're requiring the local part (before #) to be at least two characters with the .+ followed by the character class [^...]. It's looking for any character followed by another character not in the list of exclusions you specify. That explains why "a#b.com" doesn't match.
The second problem is partly caused by the character class range +-? which includes the . character. I think you wanted [-\w+?]+. (Do you really want question marks?) And then later I think you wanted to look for a literal . character but it really ends up matching the first character that didn't match the previous block.
Between the regex provided and the explanatory text I'm not sure what rules you intend to implement though. And since this is an exercise it's probably better to just give hints anyway.
You will also want to use the ^ and $ anchors to makes sure the entire string matches.
Quite a simple one in theory but can't quite get it!
I want a regex in ant which matches anything as long as it has a slash on the end.
Below is what I expect to work
<regexp id="slash.end.pattern" pattern="*/"/>
However this throws back
java.util.regex.PatternSyntaxException: Dangling meta character '*' near index 0
*/
^
I have also tried escaping this to \*, but that matches a literal *.
Any help appreciated!
Your original regex pattern didn't work because * is a special character in regex that is only used to quantify other characters.
The pattern (.)*/$, which you mentioned in your comment, will match any string of characters not containing newlines, however it uses a possibly unnecessary capturing group. .*/$ should work just as well.
If you need to match newline characters, the dot . won't be enough. You could try something like [\s\S]*/$
On that note, it should be mentioned that you might not want to use $ in this pattern. Suppose you have the following string:
abc/def/
Should this be evaluated as two matches, abc/ and def/? Or is it a single match containing the whole thing? Your current approach creates a single match. If instead you would like to search for strings of characters and then stop the match as soon as a / is found, you could use something like this: [\s\S]*?/.
I have a regex for matching letters, numbers and some special characters as follows: ^[A-za-z0-9 .#&,’()+/:]*$
I need to add a single hyphen to this list, not allowing multiple hyphens, but I'm not quite sure how to do it. I saw something along the lines of -{1} but I don't know how to add that to the existing rexex.
I'm using C++ and Qt5.
How about:
^[A-za-z0-9 .#&,’()+/:]*-?[A-za-z0-9 .#&,’()+/:]*$
that could be reduce to:
^[\w .#&,’()+/:]*-?[\w .#&,’()+/:]*$
I don't know if C++ support it, but it could be reduced to:
^([\w .#&,’()+/:])*-?(?1)*$
^[A-za-z0-9.#&,’()+/:]*-[A-za-z0-9.#&,’()+/:]*$ allows a single hyphen anywhere in the string.
Note that the hyphen may come at any part (at the beginning or end of the string also) and it is mandatory also.
To make the hyphen optional, use ^[A-za-z0-9.#&,’()+/:]*-?[A-za-z0-9.#&,’()+/:]*$
I am trying to get one regular expression that does the following:
makes sure there are no white-space characters
minimum length of 8
makes sure there is at least:
one non-alpha character
one upper case character
one lower case character
I found this regular expression:
((?=.*[^a-zA-Z])(?=.*[a-z])(?=.*[A-Z])(?!\s).{8,})
which takes care of points 2 and 3 above, but how do I add the first requirement to the above regex expression?
I know I can do two expressions the one above and then
\s
but I'd like to have it all in one, I tried doing something like ?!\s but I couldn't get it to work. Any ideas?
^(?=.*[^a-zA-Z])(?=.*[a-z])(?=.*[A-Z])\S{8,}$
should do. Be aware, though, that you're only validating ASCII letters. Is Ä not a letter for your requirements?
\S means "any character except whitespace", so by using this instead of the dot, and by anchoring the regex at the start and end of the string, we make sure that the string doesn't contain any whitespace.
I also removed the unnecessary parentheses around the entire expression.
Tim's answer works well, and is a good reminder that there are many ways to solve the same problem with regexes, but you were on the right track to finding a solution yourself. If you had changed (?!\s) to (?!.*\s) and added the ^ and $ anchors to the end, it would work.
^((?=.*[^a-zA-Z])(?=.*[a-z])(?=.*[A-Z])(?!.*\s).{8,})$