I am new to regex going through the tutorial I found the regex [...] says Matches any single character in brackets.. So I tried
System.out.println(Pattern.matches("[...]","[l]"));
I also tried escaping brackets
System.out.println(Pattern.matches("[...]","\\[l\\]"));
But it gives me false I expected true because l is inside brackets.
It would be helpful if anybody clear my doubts.
Characters that are inside [ and ] (called a character class) are treated as a set of characters to choose from, except leading ^ which negates the result and - which means range (if it's between two characters). Examples:
[-123] matches -, 1, 2 or 3
[1-3] matches a single digit in the range 1 to 3
[^1-3] matches any character except any of the digits in the range 1 to 3
. matches any character
[.] matches the dot .
If you want to match the string [l] you should change your regex to:
System.out.println(Pattern.matches("...", "[l]"));
Now it prints true.
The regex [...] is equivalent to the regexes \. and [.].
The tutorial is a little misleading, it says:
[...] Matches any single character in brackets.
However what it means is that the regex will match a single character against any of the characters inside the brackets. The ... means "insert characters you want to match here". So you need replace the ... with the characters that you want to match against.
For example, [AP]M will match against "AM" and "PM".
If your regex is literally [...] then it will match against a literal dot. Note there is no point repeating characters inside the brackets.
The tutorial is saying:
Matches any single character in brackets.
It means you replace ... with a single character, for example [l]
These will print true:
System.out.println(Pattern.matches("[l]","l"));
System.out.println(Pattern.matches("[.]","."));
System.out.println(Pattern.matches("[.]*","."));
System.out.println(Pattern.matches("[.]*","......"));
System.out.println(Pattern.matches("[.]+","......"));
Related
For example I have text like below :
case1:
(1) Hello, how are you?
case2:
Hi. (1) How're you doing?
Now I want to match the text which starts with (\d+).
I have tried the following regex but nothing is working.
^[\(\d+\)], ^\(\d+\).
[] are used to match any of the things you specify inside the brackets, and are to be followed by a quantifier.
The second regexp will work: ^\(\d+\), so check your code.
Check also so there's no space in front of the first parenthesis, or add \s* in front.
EDIT: Also, java can be tricky with escapes depending on if the regexp you type is directly translated to a regexp or is first a string literal. You may need to double escape your escapes.
In Java you have to escape parenthesis, so "\\(\\d+\\)" should match (1) in case one and two. Adding ^ as you did "^\\(\\d+\\)" will match only case1.
You have to use double back slashes within java string. Consider this
"\n" give you [line break]
"\\n" give you [backslash][n]
If you are going to downvote my post, at least comment to tell me WHY it's not useful.
I believe Java's Regex Engine supports Positive Lookbehind, in which case you can use the following regex:
(?<=[(][0-9]{1,9999}[)]\s?)\b.*$
Which matches:
The literal text (
Any digit [0-9], between 1 and 9999 times {1,9999}
The literal text )
A space, between 0 and 1 times \s?
A word boundary \b
Any character, between 0 and unlimited times .*
The end of a string $
I want to be able to write a regular expression in java that will ensure the following pattern is matched.
<D-05-hello-87->
For the letter D, this can either my 'D' or 'E' in capital letters and only either of these letters once.
The two numbers you see must always be a 2 digit decimal number, not 1 or 3 numbers.
The string must start and end with '<' and '>' and contain '-' to seperate parts within.
The message in the middle 'hello' can be any character but must not be more than 99 characters in length. It can contain white spaces.
Also this pattern will be repeated, so the expression needs to recognise the different individual patterns within a logn string of these pattersn and ensure they follow this pattern structure. E.g
So far I have tried this:
([<](D|E)[-]([0-9]{2})[-](.*)[-]([0-9]{2})[>]\z)+
But the problem is (.*) which sees anything after it as part of any character match and ignores the rest of the pattern.
How might this be done? (Using Java reg ex syntax)
Try making it non-greedy or negation:
(<([DE])-([0-9]{2})-(.*?)-([0-9]{2})>)
Live Demo: http://ideone.com/nOi9V3
Update: tested and working
<([DE])-(\d{2})-(.{1,99}?)-(\d{2})>
See it working: http://rubular.com/r/6Ozf0SR8Cd
You should not wrap -, < and > in [ ]
Assuming that you want to stop at the first dash, you could use [^-]* instead of .*. This will match all non-dash characters.
Any Regex masters out there? I need a regular expression in Java that matches:
"RANDOMSTUFF SPECIFICWORD"
Including the quotation marks.
Thus I need
to match the first quote,
RANDOMSTUFF (any number of words with spaces between preceding SPECIFICWORD)
SPECIFICWORD (a specific word which I won't specify here.)
and the ending quote.
I don't want to match things such as:
RANDOMSTUFF SPECIFICWORD
"RANDOMSTUFF NOTTHESPECIFICWORD"
"RANDOMSTUFF SPECIFICWORD MORERANDOMSTUFF"
\".*\sSPECIFICWORD\"
If you don't want to allow quotes in between, use \"[^"]*\sSPECIFICWORD\"
. matches any character
* says 0 or more of the preceding character (in this case, 0 or more of any characters)
\s matches any whitespace character
SPECIFICWORD will be treated as a string literal, assuming there are no special characters (escape them if there are)
\" matches the quote
[^"] means any character except a quote (the ^ is what makes it 'except')
Also, this link could be useful. Regex's are powerful expressions and are applicable across virtually any language, so it would be a good thing to become comfortable with using them.
EDIT:
As several other posters have pointed out, adding ^ to the beginning and $ to the end will only match if the entire line matches.
^ matches the beginning of the line
$ matches the end of the line
^.*\s+SPECIFICWORD"$
'^' matches 'from the start of the line'
.* matches anything
\s+ matches 'any amount of whitespace, but at least some'
SPECIFICWORD" is a string literal
$ means 'this is the end of the line'
Note that ^ and $ are not always 'line'-based; most languages allow you to specify a 'multiline' mode that would cause them to match 'start of the string/end of the string' instead of one line at a time.
Will this string be matched as a line by line basis or will it be found within the text? If so, you can add anchors to ensure that it matches the string.
^(\".*\sSPECIFICWPRD\")$
Saying, at the start of the line, look for a double quote followed by zero or more random characters followed by a single whitespace, followed by the specific word, followed by a double quote at the end of the string.
Optionally, there are excellent tools for designing regex patterns and seeing what they match in real time.
Here are a couple of examples:
http://gskinner.com/RegExr/
http://regex101.com/r/zC3fM1
Try:
\"[\w\s]*SPECIFICWORD\"
Works like this:
\" matches opening quote
[\w\s]* matches zero or more of the characters from the following sets:
[a-zA-Z_0-9] (\w part)
[ \t\n\x0B\f\r] (\s part)
SPECIFICWORD matches the SPECIFICWORD
\" matches closing quote
I'm trying to compare following strings with regex:
#[xyz="1","2"'"4"] ------- valid
#[xyz] ------------- valid
#[xyz="a5","4r"'"8dsa"] -- valid
#[xyz="asd"] -- invalid
#[xyz"asd"] --- invalid
#[xyz="8s"'"4"] - invalid
The valid pattern should be:
#[xyz then = sign then some chars then , then some chars then ' then some chars and finally ]. This means if there is characters after xyz then they must be in format ="XXX","XXX"'"XXX".
Or only #[xyz]. No character after xyz.
I have tried following regex, but it did not worked:
String regex = "#[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"]";
Here the quotations (in part after xyz) are optional and number of characters between quotes are also not fixed and there could also be some characters before and after this pattern like asdadad #[xyz] adadad.
You can use the regex:
#\[xyz(?:="[a-zA-z0-9]+","[a-zA-z0-9]+"'"[a-zA-z0-9]+")?\]
See it
Expressed as Java string it'll be:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
What was wrong with your regex?
[...] defines a character class. When you want to match literal [ and ] you need to escape it by preceding with a \.
[a-zA-z][0-9] match a single letter followed by a single digit. But you want one or more alphanumeric characters. So you need [a-zA-Z0-9]+
Use this:
String regex = "#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\]";
When you write [a-zA-z][0-9] it expects a letter character and a digit after it. And you also have to escape first and last square braces because square braces have special meaning in regexes.
Explanation:
[a-zA-z0-9]+ means alphanumeric character (but not an underline) one or more times.
(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? means that expression in parentheses can be one time or not at all.
Since square brackets have a special meaning in regex, you used it by yourself, they define character classes, you need to escape them if you want to match them literally.
String regex = "#\\[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"\\]";
The next problem is with '"[a-zA-z][0-9]' you define "first a letter, second a digit", you need to join those classes and add a quantifier:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
See it here on Regexr
there could also be some characters before and after this pattern like
asdadad #[xyz] adadad.
Regex should be:
String regex = "(.)*#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\](.)*";
The First and last (.)* will allow any string before the pattern as you have mentioned in your edit. As said by #ademiban this (=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? will come one time or not at all. Other mistakes are also very well explained by Others +1 to all other.
I expect: \b([a-zA-Z]+\.?)\b or \b([a-zA-Z]+\.{0,1})\b to work as at least one letter and at most one dot.
But the matcher finds "ab" with an input of "ab" "ab." and "ab.." and I'm expecting it to do the following:
"ab" is found for input "ab"
"ab." is found for input "ab."
nothing is found for input "ab.."
If I replace the regex to work with 0 instead of a dot e.g. \b([a-zA-Z]+0?)\b than it works as expected:
"ab" is found for input "ab"
"ab0" is found for input "ab0"
nothing is found for input "ab00"
So, how do I get my regex to work?
The issue is that \b matches between word characters and non-word characters, not between whitespace and non-whitespace as you seem to be trying. The difference between a . and a 0 is that 0 is considered a "word" character, but . isn't.
So what's happening in your examples is this:
Let's take that last string ab.. and see where \b could match:
a b . .
^ x ^ x x
Remember, \b matches between characters. I've shown where \b could match with a ^, and where it can't with an x. Since \b can only match in front of a or right after b, we're limited to just matching ab so long as you have those \b bits in there.
I think you want something like \bab\.?(?!\S). That says "word boundary, then a then b then maybe a single dot where there is NOT a non-space character immediately after."
If I've misunderstood your question, and you do want the expression to find ab. in the string ab.c or find ab in abc you can do \bab\.?(?!\.)
\b([a-zA-Z]+\.+)\b is "at least one letter followed by at least one dot
\b([a-zA-Z]+\.{0,1})\b is "at least one letter followed by zero or one dot