I don't have an experience on Regular Expressions. I need to a regular expression which doesn't allow to repeat of special characters (+-*/& etc.)
The string can contain digits, alphanumerics, and special characters.
This should be valid : abc,df
This should be invalid : abc-,df
i will be really appreciated if you can help me ! Thanks for advance.
Two solutions presented so far match a string that is not allowed.
But the tilte is How to prevent..., so I assume that the regex
should match the allowed string. It means that the regex should:
match the whole string if it does not contain 2
consecutive special characters,
not match otherwise.
You can achieve this putting together the following parts:
^ - start of string anchor,
(?!.*[...]{2}) - a negative lookahead for 2 consecutive special
characters (marked here as ...), in any place,
a regex matching the whole (non-empty) string,
$ - end of string anchor.
So the whole regex should be:
^(?!.*[!##$%^&*()\-_+={}[\]|\\;:'",<.>\/?]{2}).+$
Note that within a char class (between [ and ]) a backslash
escaping the following char should be placed before - (if in
the middle of the sequence), closing square bracket,
a backslash itself and / (regex terminator).
Or if you want to apply the regex to individual words (not the whole
string), then the regex should be:
\b(?!\S*[!##$%^&*()\-_+={}[\]|\\;:'",<.>\/?]{2})\S+
[\,\+\-\*\/\&]{2,} Add more characters in the square bracket if you want.
Demo https://regex101.com/r/CBrldL/2
Use the following regex to match the invalid string.
[^A-Za-z0-9]{2,}
[^\w!\s]{2,} This would be a shortest version to match any two consecutive special characters (ignoring space)
If you want to consider space, please use [^\w]{2,}
Related
I am just writing some piece of java code where I need to validate groupId (maven) passed by user.
For example - com.fb.test1.
I have written regex which says string should not start and end with '.' and can have alphanumeric characters delimited by '.'
[^\.][[a-zA-Z0-9]+\\.{0,1}]*[a-zA-Z0-9]$
But this regex not able to find out consecutive '.' For example - com..fb.test. I have added {0,1} followed by decimal to restrict it limitation to 1 but it didnt work.
Any leads would be highly appreciated.
The quantifier {0,1} and the dot should not be in the character class, because you are repeating the whole character class allowing for 0 or more dots, including { , } chars.
You can also exclude a dot to the left using a negative lookbehind instead of matching an actual character that is not a dot.
In Java you could write the pattern as
(?<!\\.)[a-zA-Z0-9]+(?:\\.[a-zA-Z0-9]+)+[a-zA-Z0-9]$
Note that the $ makes sure that that match is at the end of the string.
Regex demo
Forgive me. I am not familiarized much with Regex patterns.
I have created a regex pattern as below.
String regex = Pattern.quote(value) + ", [NnoneOoff0-9\\-\\+\\/]+|[NnoneOoff0-9\\-\\+\\/]+, "
+ Pattern.quote(value);
This regex pattern is failing with 2 different set of strings.
value = "207e/160";
Use Case 1 -
When channelStr = "207e/160, 149/80"
Then channelStr.matches(regex), returns "true".
Use Case 2 -
When channelStr = "207e/160, 149/80, 11"
Then channelStr.matches(regex), returns "false".
Not able to figure out why? As far I can understand it may be because of the multiple spaces involved when more than 2 strings are present with separated by comma.
Not sure what should be correct pattern I should write for more than 2 strings.
Any help will be appreciated.
If you print your pattern, it is:
\Q207e/160\E, [NnoneOoff0-9\-\+\/]+|[NnoneOoff0-9\-\+\/]+, \Q207e/160\E
It consists of an alternation | matching a mandatory comma as well on the left as on the right side.
Using matches(), should match the whole string and that is the case for 207e/160, 149/80 so that is a match.
Only for this string 207e/160, 149/80, 11 there are 2 comma's, so you do get a partial match for the first part of the string, but you don't match the whole string so matches() returns false.
See the matches in this regex demo.
To match all the values, you can use a repeating pattern:
^[NnoeOf0-9+/-]+(?:,\h*[NnoeOf0-90+/-]+)*$
^ Start of string
[NnoeOf0-9\\+/-]+
(?: Non capture group
,\h* Match a comma and optional horizontal whitespace chars
[NnoeOf0-90-9\\+/-]+ Match 1+ any of the listed in the character class
)* Close the non capture group and optionally repeat it (if there should be at least 1 comma, then the quantifier can be + instead of *)
$ End of string
Regex demo
Example using matches():
String channelStr1 = "207e/160, 149/80";
String channelStr2 = "207e/160, 149/80, 11";
String regex = "^[NnoeOf0-9+/-]+(?:,\\h*[NnoeOf0-90+/-]+)*$";
System.out.println(channelStr1.matches(regex));
System.out.println(channelStr2.matches(regex));
Output
true
true
Note that in the character class you can put - at the end not having to escape it, and the + and / also does not have to be escaped.
You can use regex101 to test your RegEx. it has a description of everything that's going on to help with debugging. They have a quick reference section bottom right that you can use to figure out what you can do with examples and stuff.
A few things, you can add literals with \, so \" for a literal double quote.
If you want the pattern to be one or more of something, you would use +. These are called quantifiers and can be applied to groups, tokens, etc. The token for a whitespace character is \s. So, one or more whitespace characters would be \s+.
It's difficult to tell exactly what you're trying to do, but hopefully pointing you to regex101 will help. If you want to provide examples of the current RegEx you have, what you want to match and then the strings you're using to test it I'll be happy to provide you with an example.
^(?:[NnoneOoff0-9\\-\\+\\/]+ *(?:, *(?!$)|$))+$
^ Start
(?: ... ) Non-capturing group that defines an item and its separator. After each item, except the last, the separator (,) must appear. Spaces (one, several, or none) can appear before and after the comma, which is specified with *. This group can appear one or more times to the end of the string, as specified by the + quantifier after the group's closing parenthesis.
Regex101 Test
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 have a state machine which is capable of matching the comments. So it can handle :
/* /* */ */
But I bogged down of skipping the contents that are inside the comment lines. Currently my comments-word regex looks something strange :
[0-9A-Za-zA-Z0-9\*\(\*\*\)\.\{\}\_\;\,\-\:" "\#]*
Are there any simple regex ( in java ) which matches all the characters? Alphabets along with special characters?
Thanks for the help.
use . (dot) if you want to match any character.
See here: Dot
. matches anything once. .* will match 0 or more of anything, while .+ will match one or more, depending on your needs.
. is the character that matches all other characters, with the possible exception of newlines (depending on whether DOTALL is enabled).
If you want to match everything EXCEPT a certain character or two, use [^...] syntax (such as [^0-9a-fA-F] to avoid matching every hexadecimal digit).
It is often useful to add a trailing ? to expressions with a dot, to match the fewest characters as possible (such as .*? or .+?). Otherwise, an unterminated dot expression may match the rest of the string.
I want to do validation for a String which can only contains alphanumeric and only one special character. I tried with (\\W).{1,1}(\\w+).
But it is true only when I start with a special character. But I can have one special character at any place in String.
Use the ^ and $ anchors to instruct the regex engine to start matching from the beginning of the string and stop matching at the end of the string, so taking your regex:
^(\\W).{1,1}(\\w+)$
Please take a look at this Oracle (Java) tutorial on regular expressions.
Try this regexp: \w*\W?\w* (Java string: "\\w*\\W?\\w*")
This expression has a drawback of matching zero-length strings. If your input must have exactly one special character, remove the question mark ? from the expression.
use matcher.find() and not matcher.match() and search for \\w and remove plus (+) because it will match all alphanumeric characters sequence in your string.If your string contains only them, your regex will match whole string.
if I understand your regex correctly, this could solve your problem:
([\w]+)([^\w])([\w]+)