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Reference - What does this regex mean?
(1 answer)
Closed 5 years ago.
Does this regular expression mean that at least one of the following that isn't a-z:
(?=.*(?:[a-z]))
It's part of the following expression:
/^(?=[A-Za-z0-9\'\s\d\.]{2,50}$)(?=.*(?:[a-z]))[a-zA-Z0-9]+[A-Za-z0-9\'\s\.]+$/m
No, (?=.*(?:[a-z])) means that there could be whatever but must finish with a lowercase letter.
This regex means:
/^(?=[A-Za-z0-9\'\s\d\.]{2,50}$)(?=.*(?:[a-z]))[a-zA-Z0-9]+[A-Za-z0-9\'\s\.]+$/m
Match the line that starts with 2 to 50 alphanumeric, single quote, spaces or a dot, and then follows with lower case letter, and continues with alphanumerics and must ends followed by alphanumerics, spaces, single quote or dot.
Here you can see a better graphical approach for your regex:
Actually, this can be improved as:
/^(?=[A-Za-z\d'\s.]{2,50}$)(?=.*[a-z])[a-zA-Z\d]+[A-Za-z\d'\s.]+$/m
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 3 years ago.
I need a regular expression to validate strings with the prefix 'CON' followed by an optional space followed by 8 digits.
I've tried various expressions, I got tangled up and now I'm lost.
^(CON+s\?d{8})$
\bCON\b\S?D{8}
Syntax is off a bit
^(CON\s?\d{8})
( starts a capturing group
CON is exactly matched
\s matches any white space character and the ? makes it optional
\d{8} matches 8 digits
) ends the capturing group
You were pretty well off to start, Hope this helps :)
keeping in mind If there is no space, then there shouldn't be 8 more digits
^CON(\ \d{8})?
If the string you are looking for can be part of a larger string (note that in this case it may be preceded or followed by anything, even other digits):
CON\s?\d{8}
If the string must match in full, use ^$ to designate that:
^CON\s?\d{8}$
You can add variations to it, if say you want it to begin/end with a word boundary - use \bto indicate that. If you want it to end in a non-digit, use \D+ at the end, instead of $.
Finally, if you want the string to end with an EOL or a non-digit, you may use an expression like this:
CON\s?\d{8}(\D+|$) or the same with a non-capturing group: CON\s?\d{8}(?:\D+|$)
This question already has an answer here:
Restricting character length in a regular expression
(1 answer)
Closed 4 years ago.
I would like to match 1 or more capital letters, [A-Z]+ followed by 0 or more numbers, [0-9]* but the entire string needs to be less than or equal to 8 characters in total.
No matter what regex I come up with the total length seems to be ignored. Here is what I've tried.
^[A-Z]+[0-9]*{1,8}$ //Range ignored, will not work on regex101.com but will on rubular.com/
^([A-Z]+[0-9]*){1,8}$ //Range ignored
^(([A-Z]+[0-9]*){1,8})$ //Range ignored
Is this not possible in regex? Do I just need to do the range check in the language I'm writing in? That's fine but I thought it would be cleaner to keep in all in regex syntax. Thanks
The behaviour is expected. When you write the following pattern:
^([A-Z]+[0-9]*){1,8}$
The {1,8} quantifier is telling the regex to repeat the previous pattern, therefore the capturing group in this case, between one to eight times. Due to the greedyness of your operators, you will match and capture indefinitely.
You need to use a lookahead to obtain the desired behaviour:
^(?=.{1,8}$)[A-Z]+[0-9]*$
^ Assert beginning of string.
(?=.{1,8}$) Ensure that the string that follows is between one and eight characters in length.
[A-Z]+[0-9]*$ Match any upper case letters, one or more, and any digits, zero or more.
$ Asserts position end of string.
See working demo here.
The regex ^([A-Z]+[0-9]*){1,8}$ would match [A-Z]+[0-9]* 1 - 8 times. That would match for example a repetition of 8 times A1A1A1A1A1A1A1A1 but not a repetition of 9 times A1A1A1A1A1A1A1A1A1
You might use a positive lookahead (?=[A-Z0-9]{1,8}$) to assert the length of the string:
^(?=[A-Z0-9]{1,8}$)[A-Z]+[0-9]*$
That would match
^ From the start of the string
(?=[A-Z0-9]{1,8}$) Positive lookahead to assert that what follows matches any of the characters in the character class [A-Z0-9] 1 - 8 times and assert the end of the string.
[A-Z]+[0-9]*$ Match one or more times an uppercase character followed by zero or more times a digit and assert the end of the string. $
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 3 years ago.
I have a huge file and need to replace some strings, the problem is that they are dynamic but always follow a pattern:
year[4 digits number]/month[2 digits number]/timestamp[8 digits number]/file[random string ending with extension]
Some examples:
2017/07/24204301/a-4.png
2017/07/24204318/a-5-e1501986401369.png
2017/11/24211223/questao10branca-172x300.png
I need to remove the timestamp on all occurrences, then the above example would become:
2017/07/a-4.png
2017/07/a-5-e1501986401369.png
2017/11/questao10branca-172x300.png
How can I achieve this using Regexp and Visual Studio Code?
Given the examples you presented, there are a couple of regular expressions that will work for you.
See regex in use here
/\d{8}(?=/)
/ Match this literally
\d{8} Match any digit exactly 8 times
(?=/) Positive lookahead ensuring what follows is a literal /
See regex in use here
(?<=^\d{4}/\d{2}/)\d{8}/
(?<=^\d{4}/\d{2}/) Negative lookbehind ensuring what precedes is the following:
^ Assert position at the start of the string
\d{4} Match any digit exactly 4 times
/ Match this literally
\d{2} Match any digit exactly twice
/ Match this literally
\d{8} Match any digit exactly 8 times
/ Match this literally
This question already has answers here:
Regex how to match an optional character
(5 answers)
Closed 6 years ago.
I just want to write a regular expression 4 digits and '.' and 5 digits and optional 'A'
Ex: 1111.2345A where A is optional.
^[0-9]{4}[\.][0-9]{4}$
This reg ex will give 1111.2345, but how to add Optional 'N' at last.
Use ? at the end for characters:
[A-Za-z]?
This will match at most 1 presence of a character (lower or upper case).
You can check for a character zero or one times with this:
'[A]{0,1}'
Put that at the end of your string and it will try and match the character 'A' zero or one times. You may also use the symbol ? to match zero or one times. All about preference.
To get a single, optional A at the end, append A? to your regular expression:
^[0-9]{4}[\.][0-9]{4}A?$
Btw. instead of [0-9] you could use \d which stands for 'digit':
^\d{4}\.\d{4}A?$
This question already has answers here:
Reference - What does this regex mean?
(1 answer)
Using explicitly numbered repetition instead of question mark, star and plus
(4 answers)
Closed 3 years ago.
I've been searching for a long time but didn't find an answer for my question, can tell me what the meaning of
(?:[-\w\d{1-3}]+\.)+
and not
(?:[-\w\d{1,3}]+\.)+
I don't understand the {1-3} part and can't find anywhere what it's mean.
Thank you
Everything between [] are characters to be matched. So it matches each of those characters:
- the literal character -
\w match any word character [a-zA-Z0-9_]
\d match a digit [0-9]
{ the literal character {
1-3 a single character in the range between 1 and 3
} the literal character }
the 1-3 makes no sense there, as well as the \d. Both are included in \w
Even what you would say that is correct {1,3} inside the [] makes no sense.