I am new to regex. I have this regex:
\[(.*[^(\]|\[)].*)\]
Basically it should take this:
[[a][b][[c]]]
And be able to replace with:
[dd[d]]
abc, d are unrelated. Needless to say the regex bit isn't working. it replaces the entire string with "d" in this case.
Any explanation or aid would be great!
EDIT:
I tried another regex,
\[([^\]]{0})\]
This one worked for the case where brackets contain no inner brackets and nothing else inside. But it doesn't work for the described case.
You need to know that . dot is special character which represents "any character beside new line mark" and * is greedy so it will try to find maximal match.
In your regex \[(.*[^(\]|\[)].*)\] first .* will represent maximal set of characters between [ and [^(\]|\[)].*)\]] and this part can be understood as non [ or ] character, optional other characters .* and finally ]. So this regex will match your entire input.
To get rid of that problem remove both .* from your regex. Also you don't need to use | or ( ) inside [^...].
System.out.println("[[a][b][[c]]]".replaceAll("\\[[^\\]\\[]\\]", "d"));
Output: [dd[d]]
\[(\[a\])(\[b\])\[(\[c\])\]\]
If you need to double backslashes in the current context (such as you are placing it in a "" style string):
\\[(\\[a\\])(\\[b\\])\\[(\\[c\\])\\]\\]
An example replacement for a, b and c is [^\]]*, or if you need to escape backslashes [^\\]]*.
Now you can replace capture one, capture two and capture three each with d.
If the string you are replacing in is not exactly of that format, then you want to do a global replacement with
(\[a\])
replacing a,
(\[[^\]]*\])
doubling backslashes,
(\\[[^\\]]*\\])
Try this:
System.out.println("[[a][b][[c]]]".replaceAll("\\[[^]\\[]]", "d"));
if a,b,c are in real world more than one character, use this:
System.out.println("[[a][b][[c]]]".replaceAll("\\[[^]\\[]++]", "d"));
The idea is to use a character class that contains all characters but [ and ]. The class is: [^]\\[] and other square brackets in the pattern are literals.
Note that a literal closing square bracket don't need to be escaped at the first position in a character class and outside a character class.
Related
I want to check if a String contains a } with any character in front of it except \.
As far as I know I can use . as a metacharacter in aString.contains(...) to allow any character at that position but I don’t know how to create something like a blacklist: aString.contains(“.(except ‘\‘)}“Is that possible without creating an own method?
You need regex (well technically you don't need regex, but it's the best way):
if (aString.matches(".*(?<!\\\\)}.*"))
This regex says the string should be made up as follows
.* zero or more of any character
(?<!\\\\) the previous char is not a backslash
} a curly right bracket
.* zero or more of any character
This also works for the edge case of the first char being the curly bracket.
See live demo.
I have a string kind of:
String text = "(plum) some other words, [apple], another words {pear}.";
I have to find and replace the words in brackets, don't replacing the brackets themselves.
If I write:
text = text.replaceAll("(\\(|\\[|\\{).*?(\\)|\\]|\\})", "fruit");
I get:
fruit some other words, fruit, another words fruit.
So the brackets went away with the fruits, but I need to keep them.
Desired output:
(fruit) some other words, [fruit], another words {fruit}.
Here is your regex:
(?<=[({\[\(])[A-Za-z]*(?=[}\]\)])
Test it here:
https://regex101.com/
In order to use it in Java, remember to add second backslashes:
(?<=[({\\[\\(])[A-Za-z]*(?=[}\\]\\)])
It matches 0 or more letters (uppercase or lowercase) preceded by either of these [,{,( and followed by either of these ],},).
If you want to have at least 1 letter between brackets just replace '*' with '+' like this:
(?<=[({\[\(])[A-Za-z]+(?=[}\]\)])
GCP showed how to use look aheads and look behinds to exclude the brackets from the matched part. But you can also match them, and refer to them in your replacement string with capturing groups:
text.replaceAll("([\\(\\[\\{]).*?([\\)\\]\\}])", "$1fruit$2");
Also note that you can replace the | ORs by a character group [].
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,}
I have a String like this
أصبح::ينال::أخذ::حصل (على)::أحضر
And I want to split it on non Arabic characters using java
And here's my code
String s = "أصبح::ينال::أخذ::حصل (على)::أحضر";
String[] arr = s.split("^\\p{InArabic}+");
System.out.println(Arrays.toString(arr));
And the output was
[, ::ينال::أخذ::حصل (على)::أحضر]
But I expect the output to be
[ينال,أخذ,حصل,على,أحضر]
So I don't know what's wrong with this?
You need a negated class, and to do that, you need square brackets [ ... ]. Try to split with this:
"[^\\p{InArabic}]+"
If \\p{InArabic} matches any arabic character, then [^\\p{InArabic}] will match any non-arabic character.
Another option you can consider is an equivalent syntax, using P instead of p to indicate the opposite of the \\p{InArabic} character class like #Pshemo mentioned:
"\\P{InArabic}+"
This works just like \\W is the opposite of \\w.
The only possible advantage you get with the first syntax over the second (again like #Pshemo mentioned), is that if you want to add other characters to the list of characters which shouldn't match, for example, if you want to match all non \\p{InArabic} except periods, the first one is more flexible:
"[^\\p{InArabic}.]+"
^
Otherwise, if you really want to use \\P{InArabic}, you'll need subtraction within classes:
"[\\P{InArabic}&&[^.]]+"
The expression you want is "\\P{InArabic}+"
This means match any (non-zero) number of characters that are not Arabic.
I consider myself pretty good with Regular Expressions, but this one is appearing to be surprisingly tricky: I want to trim all whitespace, except the space character: ' '.
In Java, the RegEx I have tried is: [\s-[ ]], but this one also strips out ' '.
UPDATE:
Here is the particular string that I am attempting to strip spaces from:
project team manage key
Note: it would be the characters between "team" and "manage". They appear as a long space when editing this post but view as a single space in view mode.
Try using this regular expression:
[^\S ]+
It's a bit confusing to read because of the double negative. The regular expression [\S ] matches the characters you want to keep, i.e. either a space or anything that isn't a whitespace. The negated character class [^\S ] therefore must match all the characters you want to remove.
Using a Guava CharMatcher:
String text = ...
String stripped = CharMatcher.WHITESPACE.and(CharMatcher.isNot(' '))
.removeFrom(text);
If you actually just want that trimmed from the start and end of the string (like String.trim()) you'd use trimFrom rather than removeFrom.
There's no subtraction of character classes in Java, otherwise you could use [\s--[ ]], note the double dash. You can always simulate set subtraction using intersection with the complement, so
[\s&&[^ ]]
should work. It's no better than [^\S ]+ from the first answer, but the principle is different and it's good to know both.
I solved it with this:
anyString.replace(/[\f\t\n\v\r]*/g, '');
It is just a collection of all possible white space characters excluding blank (so actually
\s without blanks). It includes tab, carriage return, new line, vertical tab and form feed characters.