I have recently encountered this question in the text book:
I am suppose to write a method to check if a string have:
at least ten characters
only letters and digits
at least three digits
I am trying to solve it by Regx, rather than iterating through every character; this is what I got so far:
String regx = "[a-z0-9]{10,}";
But this only matches the first two conditions. How should I go about the 3rd condition?
You could use a positive lookahead for 3rd condition, like this:
^(?=(?:.*\d){3,})[a-z0-9]{10,}$
^ indicates start of string.
(?= ... ) is the positive lookahead, which will search the whole string to match whatever is between (?= and ).
(?:.*\d){3,} matches at least 3 digits anywhere in the string.
.*\d matches a digit preceded by any (or none) character (if omitted then only consecutive digits would match).
{3,} matches three or more of .*\d.
(?: ... ) is a non-capturing group.
$ indicates end of string.
Related
Using the following regex
^(\d)(?!\1+$)\d{3}-\d{1}$
It works for the pattern but I need to validate that all numbers are not the same even after /separated by the hyphen (-).
Example:
0000-0 not allowed (because of all are same digits)
0000-1 allowed
1111-1 not allowed (because of all are same digits)
1234-2 allowed
TheFourthBird's answer surely works that uses a negative lookahead. Here is another variant of this regex that might be slightly faster:
^(\d)(?!\1{3}-\1$)\d{3}-\d$
RegEx Demo
Explanation:
^(\d) matches and captures first digit after start in group #1
(?!\1{3}-\1$) is a negative lookahead that will fail the match if we have 3 repetitions and a hyphen and another repeat of 1st digit.
You could exclude only - or the same digit only to the right till the end of the string:
^(\d)(?!(?:\1|-)*$)\d{3}-\d$
^ Start of string
(\d) Capture group 1, match a digit
(?! Negative lookahead, assert what is to the right is not
(?:\1|-)*$ Optionally repeat either the backrefernce to what is already captured or - till the end of the string
) Close the non capture group
\d{3}-\d Match 3 digits - and a digit
$ End of string
Regex demo
If you don't want to match double -- or an - at the end of the string and match optional repetitions:
^(\d)(?!(?:\1|-)*$)\d*(?:-\d+)*$
Explanation
^ Start of string
(\d) Capture a single digits in group 1
(?!(?:\1|-)*$) Negative lookahead, assert not only - and the same digit till the end of the string
\d* Match optional digits
(?:-\d+)* Optionally repeat matching - and 1+ digits
$ End of string
Regex demo
You'll need a back reference, for example:
^(\d){4}-\1$
I'm new to Regex in Java and I wanted to know how can I build one that only takes a string that consists of one or two comma-separated lists of uppercase letters, separated by a single whitespace.
I would need to filter out strings that start with a comma, that end with a comma or strings that have multiple consecutive commas.
All these would be invalid:
"D,, D"
"D D,,"
"D, ,D"
"D, ,,D"
"D,, ,D"
"D,,"
",,A"
",A"
"A,"
All these would be valid:
"D,D T,F"
"D,D T"
"A,A"
"A"
I used (\s?("[\w\s]*"|\d*)\s?(,,|$)) for consecutive commas but it doesn't do the trick when the comma is at the end or beggining of one of the whitespace separated substring like "D, ,D"
Should I aim to split by whitespace and look for a simpler regex for each of the substrings?
That would be something like this:
^[A-Z](,[A-Z])*( [A-Z](,[A-Z])*)*$
What happens here, is the following:
We expect a letter, optionally followed by one or more times a comma-immediately-followed-by-another-letter.
Then we optionally accept a space, and then the abovementioned pattern. And this is repeated.
Test: https://regex101.com/r/kzLhtw/1
You could, of course, slightly optimize the regex by making all capturing groups non-capturing: just put ?: immediately behind the (, that is, (?:.
You might use
^[A-Z](?: [A-Z])*(?:,[A-Z](?: [A-Z])*){0,2}$
^ Start of string
[A-Z] Match a single char A-Z
(?: [A-Z])* Optionally repeat a space and and a single char A-Z
(?: Non capture group
,[A-Z](?: [A-Z])* Match a comma, char A-Z followed by optionally repeat matching a space and a char A-Z
){0,2} Close the group and repeat 0-2 times
$ End of string
Regex demo
"a string that consists of one or two comma-separated lists of uppercase letters, separated by a single whitespace"
Not sure how to exactly interpretate the above, but my reading is: One or two comma-seperated lists where each list may only consist of uppercase characters. In the case of two lists, the two lists are seperated by a single space.
You could try:
^(?!.* .* )[A-Z](?:[ ,][A-Z])*$
See the online demo
^ - Start string anchor.
(?!.* .* ) - Negative lookahead to prevent two spaces present.
[A-Z] - A single uppercase alpha-char.
(?: - Open non-capture group:
[ ,] - A comma or space.
[A-Z] - A single uppercase alpha-char.
)* - Close non-capture group and match 0+ times upt to;
$ - End string anchor.
I want to extract URL strings from a log which looks like below:
<13>Mar 27 11:22:38 144.0.116.31 AgentDevice=WindowsDNS AgentLogFile=DNS.log PluginVersion=X.X.X.X Date=3/27/2019 Time=11:22:34 AM Thread ID=11BC Context=PACKET Message= Internal packet identifier=0000007A4843E100 UDP/TCP indicator=UDP Send/Receive indicator=Snd Remote IP=X.X.X.X Xid (hex)=9b01 Query/Response=R Opcode=Q Flags (hex)=8081 Flags (char codes)=DR ResponseCode=NOERROR Question Type=A Question Name=outlook.office365.com
I am looking to extract Name text which contains more that 5 digits.
A possible way suggested is (\d.*?){5,} but does not seem to work, kindly suggest another way get the field.
Example of string match:
outlook12.office345.com
outlook.office12345.com
You can look for the following expression:
Name=([^ ]*\d{5,}[^ ]*)
Explanation:
Name= look for anything that starts with "Name=", than capture if:
[^ ]* any number of characters which is not a space
\d{5,} then 5 digits in a row
[^ ]* then again, all digits up to a white space
This regular expression:
(?<=Name=).*\d{5,}.*?(?=\s|$)
would extract strings like outlook.office365666.com (with 5 or more consecutive digits) from your example input.
Demo: https://regex101.com/r/YQ5l2w/1
Try this pattern: (?=\b.*(?:\d[^\d\s]*){5,})\S*
Explanation:
(?=...) - positive lookahead, assures that pattern inside it is matched somewhere ahead :)
\b - word boundary
(?:...) - non-capturing group
\d[^\d\s]* - match digit \d, then match zero or more of any characters other than whitespace \s or digit \d
{5,} - match preceeding pattern 5 or more times
\S* - match zero or more of any characters other than space to match the string if assertion is true, but I think you just need assertion :)
Demo
If you want only consecutive numbers use simplified pattern (?=\b.*\d{5,})\S*.
Another demo
Of course, you have to add positive lookbehind: (?<=Name=) to assert that you have Name= string preceeding
Try this regex
([a-z0-9]{5,}.[a-z0-9]{5,})+.com
https://regex101.com/r/OzsChv/3
It Groups,
outlook.office365.com
outlook12.office345.com
also all url strings
I need to replace all non-digit charaters in the string. For instance:
String: 987sdf09870987=-0\\\`42
Replaced: 987**sdf**09870987**=-**0**\\\`**42
That's all non-digit char-sequence wrapped into ** charaters. How can I do that with String::replaceAll()?
(?![0-9]+$).*
the regex doesn't match what I want. How can I do that?
(\\D+)
You can use this and replace by **$1**.See demo.
https://regex101.com/r/fM9lY3/2
You can use a negated character class for a non-digit and use the 0th group back-reference to avoid overhead with capturing groups (it is minimal here, but still is):
String x = "987sdf09870987=-0\\\\\\`42";
x = x.replaceAll("[^0-9]+", "**$0**");
System.out.println(x);
See demo on IDEONE. Output: 987**sdf**09870987**=-**0**\\\`**42.
Also, in Java regex, character classes look neater than multiple escape symbols, that is why I prefer this [^0-9]+ pattern meaning match 1 or more (+) symbols other than (because of ^) digits from 0 to 9 ([0-9]).
A couple of words about your (?![0-9]+$).* regex. It consists of a negative lookahead (?![0-9]+$) that checks if from the current position onward there are no digits only (if there are only digits up to the end of string, the match fails), and .* matching any characters but a newline. You can see example of what it is doing here. I do not think it can help you since you need to actually match non-numbers, not just check if digits are absent.
I'm trying to write a regular expression for Java that matches if there is a semicolon that does not have two (or more) leading '-' characters.
I'm only able to get the opposite working: A semicolon that has at least two leading '-' characters.
([\-]{2,}.*?;.*)
But I need something like
([^([\-]{2,})])*?;.*
I'm somehow not able to express 'not at least two - characters'.
Here are some examples I need to evaluate with the expression:
; -- a : should match
-- a ; : should not match
-- ; : should not match
--; : should not match
-;- : should match
---; : should not match
-- semicolon ; : should not match
bla ; bla : should match
bla : should not match (; is mandatory)
-;--; : should match (the first occuring semicolon must not have two or more consecutive leading '-')
It seems that this regex matches what you want
String regex = "[^-]*(-[^-]+)*-?;.*";
DEMO
Explanation: matches will accept string that:
[^-]* can start with non dash characters
(-[^-]+)*-?; is a bit tricky because before we will match ; we need to make sure that each - do not have another - after it so:
(-[^-]+)* each - have at least one non - character after it
-? or - was placed right before ;
;.* if earlier conditions ware fulfilled we can accept ; and any .* characters after it.
More readable version, but probably little slower
((?!--)[^;])*;.*
Explanation:
To make sure that there is ; in string we can use .*;.* in matches.
But we need to add some conditions to characters before first ;.
So to make sure that matched ; will be first one we can write such regex as
[^;]*;.*
which means:
[^;]* zero or more non semicolon characters
; first semicolon
.* zero or more of any characters (actually . can't match line separators like \n or \r)
So now all we need to do is make sure that character matched by [^;] is not part of --. To do so we can use look-around mechanisms for instance:
(?!--)[^;] before matching [^;] (?!--) checks that next two characters are not --, in other words character matched by [^;] can't be first - in series of two --
[^;](?<!--) checks if after matching [^;] regex engine will not be able to find -- if it will backtrack two positions, in other words [^;] can't be last character in series of --.
How about just splitting the string along -- and if there are two or more sub strings, checking if the last one contains a semicolon?
How about using this regex in Java:
[^;]*;(?<!--[^;]{0,999};).*
Only caveat is that it works with up to 999 character length between -- and ;
Java Regex Demo
I think this is what you're looking for:
^(?:(?!--).)*;.*$
In other words, match from the start of the string (^), zero or more characters (.*) followed by a semicolon. But replacing the dot with (?:(?!--).) causes it to match any character unless it's the beginning of a two-hyphen sequence (--).
If performance is an issue, you can exclude the semicolon as well, so it never has to backtrack:
^(?:(?!--|;).)*;.*$
EDIT: I just noticed your comment that the regex should work with the matches() method, so I padded it out with .*. The anchors aren't really necessary, but they do no harm.
You need a negative lookahead!
This regex will match any string which does not contain your original match pattern:
(?!-{2,}.*?;.*).*?;.*
This Regex matches a string which contains a semicolon, but not one occuring after 2 or more dashes.
Example: