I need to get a regex where a phone number must begin with a +. There can be a comma seperated list eg
List:
tel1: +E1234498912345678#fake.com, tel2: +498912345678, tel1: +E123449D1238912345678#fake.com
is a valid list. E is a valid special case
My regex is this:
^(tel1:)|(tel2:)( )(\+.)$
but it accepts numbers without a + as being valid which is not what I want. The number MUST be preceded by a + otherwise it's invalid. Any hints?
You can try the following:
(tel[12]:\s*\+[eE]?\w+(#\w+(\.\w+)+)?(,\s*)?)+
This should match telephone numbers separated by a comma, in a single line.
It also oversees the use of a special character E or e.
Also, domains may not only end in .com. .net or .com.uk should also be valid.
Related
I am trying to use Regex to dynamically capture all numbers in a string such as 1234-12-1234 or 1234-123-1234 without knowing the number of characters that will occur in each string segment. I have been able to capture this using positive look ahead via the following expression: [0-9]*(?=-). However, when I try to replace the numbers to Xs such that each number that occurs before the last dash is replaced by an X, the Regex does not return X's for numbers 1:1. Instead, each section returns exactly two X's. How can I get the regex to return the following:
1234-123-1234 -> XXXX-XXX-1234
1234-12-1234 -> XXXX-XX-1234
instead of the current
1234-123-1234 -> XX-XX-1234
?
Link to demo
The problem is that by placing the * directly after the digit match, more than one digit would get replaced with a single X. And then zero digits would get replaced with a single X. Therefore any number of digits would be effectively replaced as two X's.
Use this instead:
[0-9](?=.*-)
I am trying to extract words after the first space using
species<-gsub(".* ([A-Za-z]+)", "\1", x=genus)
This works fine for the other rows that have two words, however row [9] "Eulamprus tympanum marnieae" has 3 words and my code is only returning the last word in the string "marnieae". How can I extract the words after the first space so I can retrieve "tympanum marnieae" instead of "marnieae" but have the answers stored in one variable called >species.
genus
[9] "Eulamprus tympanum marnieae"
Your original pattern didn't work because the subpattern [A-Za-z]+ doesn't match spaces, and therefore will only match a single word.
You can use the following pattern to match any number of words (other than 0) after the first, within double quotes:
"[A-Za-z]+ ([A-Za-z ]+)" https://regex101.com/r/p6ET3I/1
https://regex101.com/r/p6ET3I/2
This is a relatively simple, but imperfect, solution. It will also match trailing spaces, or just one or more spaces after the first word even if a second word doesn't exist. "Eulamprus " for example will successfully match the pattern, and return 5 spaces. You should only use this pattern if you trust your data to be properly formatted.
A more reliable approach would be the following:
"[A-Za-z]+ ([A-Za-z]+(?: [A-Za-z]+)*)"
https://regex101.com/r/p6ET3I/3
This pattern will capture one word (following the first), followed by any number of addition words (including 0), separated by spaces.
However, from what I remember from biology class, species are only ever comprised of one or two names, and never capitalized. The following pattern will reflect this format:
"[A-Za-z]+ ([a-z]+(?: [a-z]+)?)"
https://regex101.com/r/p6ET3I/4
I'm required to write a regular expression that has the following rules:
Digits between 1 to 4
hyphen (only one and can occur at any position)
Length of Text must be less than or equal to 6 (including the potential hyphen)
May end with a letter or a number, but not a hyphen.
Some valid examples are:
1-3411
12-413
123-2A
11-1
These examples are invalid:
12--11 ( since it contains two hyphens)
1-2345 ( since it contains number 5)
11-2311 ( since length is more than 6)
The RegEx that I wrote is:
^[1-4]-[1-4]{4}|^[1-4]{2}-[1-4]{3}|^[1-4]{3}-[1-4]{2}|^[1-4]{4}-[1-4]
However, this does not seem to be working, and it doesn't handle the case of a single character being is present in the end.
Can some some please help me determine a way of handling this?
<>
is character occurs in last position then before character we must have a digit not hypen .
i.e 11-a ( must fail)
11-1a (must pass)
^(?!(?:[^-\n]*-){2})(?:[1-4-]{1,5}[1-4]|[1-4-]{1,5}[a-zA-Z])$
You can handle that using a lookahead.See demo.
https://regex101.com/r/tS1hW2/16
If you have such a complex requirement, it is always easy to use lookarrounds to form an and-pattern matching each condition at the same time. Sometimes you need to split up ONE condition into two:
Base-Match: 6 or less digits: ^.{1,6}$
(AND) Only 1-4 and hyphen and letter: ^[1-4a-z\-]+$ (not accurate, requires next line)
(AND) First 1...5 elements NO Letter: ^[1-4\-]{1,5}[1-4a-z]$
(AND) No double hypen and not at the end: ^[^-]*-[^-]+$
Putting all together leads to:
(?=^[1-4\-]{1,5}[1-4a-z]$)(?=^[^-]*-[^-]*$)(?=^[1-4a-z\-]+$)^.{1,6}$
Debuggex Demo
I have been searching for regular expression which accepts at least two digits and one special character and minimum password length is 8. So far I have done the following: [0-9a-zA-Z!##$%0-9]*[!##$%0-9]+[0-9a-zA-Z!##$%0-9]*
Something like this should do the trick.
^(?=(.*\d){2})(?=.*[a-zA-Z])(?=.*[!##$%])[0-9a-zA-Z!##$%]{8,}
(?=(.*\d){2}) - uses lookahead (?=) and says the password must contain at least 2 digits
(?=.*[a-zA-Z]) - uses lookahead and says the password must contain an alpha
(?=.*[!##$%]) - uses lookahead and says the password must contain 1 or more special characters which are defined
[0-9a-zA-Z!##$%] - dictates the allowed characters
{8,} - says the password must be at least 8 characters long
It might need a little tweaking e.g. specifying exactly which special characters you need but it should do the trick.
There is no reason, whatsoever, to implement all rules in a single regex.
Consider doing it like thus:
Pattern[] pwdrules = new Pattern[] {
Pattern.compile("........"), // at least 8 chars
Pattern.compile("\d.*\d"), // 2 digits
Pattern.compile("[-!"ยง$%&/()=?+*~#'_:.,;]") // 1 special char
}
String password = ......;
boolean passed = true;
for (Pattern p : pwdrules) {
Matcher m = p.matcher(password);
if (m.find()) continue;
System.err.println("Rule " + p + " violated.");
passed = false;
}
if (passed) { .. ok case.. }
else { .. not ok case ... }
This has the added benefit that passwort rules can be added, removed or changed without effort. They can even reside in some ressource file.
In addition, it is just more readable.
Try this one:
^(?=.*\d{2,})(?=.*[$-/:-?{-~!"^_`\[\]]{1,})(?=.*\w).{8,}$
Here's how it works shortly:
(?=.*\d{2,}) this part saying except at least 2 digits
(?=.*[$-/:-?{-~!"^_[]]{1,})` these are special characters, at least 1
(?=.*\w) and rest are any letters (equals to [A-Za-z0-9_])
.{8,}$ this one says at least 8 characters including all previous rules.
Below is map for current regexp (made with help of Regexper)
UPD
Regexp should look like this ^(?=(.*\d){2,})(?=.*[$-\/:-?{-~!"^_'\[\]]{1,})(?=.*\w).{8,}$
Check out comments for more details.
Try this regex. It uses lookahead to verified there is a least two digits and one of the special character listed by you.
^(?=.*?[0-9].*?[0-9])(?=.*[!##$%])[0-9a-zA-Z!##$%0-9]{8,}$
EXPLANATION
^ #Match start of line.
(?=.*?[0-9].*?[0-9]) #Look ahead and see if you can find at least two digits. Expression will fail if not.
(?=.*[!##$%]) #Look ahead and see if you can find at least one of the character in bracket []. Expression will fail if not.
[0-9a-zA-Z!##$%0-9]{8,} #Match at least 8 of the characters inside bracket [] to be successful.
$ # Match end of line.
Regular expressions define a structure on the string you're trying to match. Unless you define a spatial structure on your regex (e.g. at least two digits followed by a special char, followed by ...) you cannot use a regex to validate your string.
Try this : ^.*(?=.{8,15})(?=.*\d)(?=.*\d)[a-zA-Z0-9!##$%]+$
Please read below link for making password regular expression policy:-
Regex expression for password rules
I am trying to validate a comma separated list for numbers 1-8.
i.e. 2,4,6,8,1 is valid input.
I tried [0-8,]* but it seems to accept 1234 as valid. It is not requiring a comma and it is letting me type in a number larger than 8. I am not sure why.
[0-8,]* will match zero or more consecutive instances of 0 through 8 or ,, anywhere in your string. You want something more like this:
^[1-8](,[1-8])*$
^ matches the start of the string, and $ matches the end, ensuring that you're examining the entire string. It will match a single digit, plus zero or more instances of a comma followed by a digit after it.
/^\d+(,\d+)*$/
for at least one digit, otherwise you will accept 1,,,,,4
[0-9]+(,[0-9]+)+
This works better for me for comma separated numbers in general, like: 1,234,933
You can try with this Regex:
^[1-8](,[1-8])+$
If you are using python and looking to find out all possible matching strings like
XX,XX,XXX or X,XX,XXX
or 12,000, 1,20,000 using regex
string = "I spent 1,20,000 on new project "
re.findall(r'(\b[1-8]*(,[0-9]*[0-9])+\b)', string, re.IGNORECASE)
Result will be ---> [('1,20,000', ',000')]
You need a number + comma combination that can repeat:
^[1-8](,[1-8])*$
If you don't want remembering parentheses add ?: to the parens, like so:
^[1-8](?:,[1-8])*$