I am using vaadins builtin RegexpValidator to check for valid form fields.I have a description field that can have any charater as long as it's not empty.Initialy i was using ".+" which pretty much worked but when i converted that field from TextField to TextArea it didn't match my strings anymore as ".+" doesn't check for newlines or blank spaces.
I have tried doing "(.|\n|\r)+" but that includes writing a blank space or a newline aswell.
I need only to make sure that i have entered atleast one character it doesn't matter what.
Normally with regex you can check for blanks with "^\s*$" but vaadins RegexpValidator result must match your string so what i am looking is basicaly the opposite of "^\s*$" ? but including atleast one character? RegexpValidator is really confusing me
If you want to allow a pure whitespace text you can use
[\s\S]+
\s a whitespace character
\S a non whitespace character
that would match at least one character and would also match newlines, because they are included in \s
If you want to have at least one Non-whitespace character you can use
^\s*\S
That would check for 0 or more whitespaces at the start of the string (this would cover leading newlines) and it would be successful when it finds the first non whitespace.
Related
Want to match the character at position 7 to either be - or an Uppercase letter
This is what I have ^.{6}[-(A-Z)]
Though this matches the first 7 characters, it doesn't match the whole string. Any help appreciated.
I am using Java and wanting .matches() to return true for this String
Though this matches the first 7 characters, it doesn't match the whole string.
That's the right explanation of what is going on. You can skip over the rest of the string by adding .* at the end. Additionally, the ^ anchor at the front of the expression is implied, so you can drop it for a pattern of
.{6}[A-Z-].*
As mentioned You can use .* to match anything after your specific character so use
^.{6}[-A-Z].*
and also no need of () if you don't want to capture that specific character
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.
am using regex expression to check if a string contains white space.
my regex is : ^\\s+$
for example if my string is my name then regex matches should return true.
but it is returning true only if my string contains only spaces no other character.
How to check if a string contains a whitespace or tab or carriage return characters in between/start/end of some string.
^(.*\s+.*)+$ seems to work for me. Accepts anything as long as there is at least one space in the string. This will match the entire string.
If you only want to check for the presence of a space, you can just use \s without any begin or end markers in the string. The difference is that this will only match the individual spaces.
Your regex is not correct.
That's a string representing a regular expression. (as tchrist pointed out correctly)
The corresponding pattern that you get when using Pattern.compile() matches only strings containing one or more whitespace characters, starting from the beginning until the end. Thus, the matching string only consists of whitespace characters.
Try this string instead for Pattern.compile():
"\\s+"
The difference is that without the anchors "^" and "$" there may be other characters around the whitespace character. The whitespace character(s) may be everywhere in the string.
Using this pattern-string the whitespace character(s) must be at the beginning:
"^\\s+"
And here the sequence of whitespace characters has to be at the end:
"\\s+$"
Use org.apache.commons.lang.StringUtils.containsAny(). See http://commons.apache.org/lang/api-3.1/org/apache/commons/lang3/StringUtils.html.
I'm constructing a regex which will accept at least 1 alpha numerical character and any number of spaces.
Right now I've got...[A-Za-z0-9]+[ \t\r\n]* which I understand to be at least 1 alphanumeric OR at least 1 space. How would I fix this?
EDIT: To answer the comments below I want it to accept strings which contain ATLEAST 1 alphanumeric AND any number of (including no) spaces. Right now it will accept JUST a whitespace.
EDIT2: To clarify, I don't want the any number of whitespace (including 0) to be accepted unless there is at least 1 alphanumeric character
\s*\p{Alnum}[\p{Alnum}\s]*
Your regex, [A-Za-z0-9]+[ \t\r\n]*, requires the string to start with a letter or digit (or, more accurately, it doesn't start matching until it sees one). Adding \s* allows the match to start with whitespace, but you still won't match any alphanumerics after the first whitespace character that follows an alphanumeric (for example, it won't match the xyz in abc xyz. Changing the trailing \s* to [\p{Alnum}\s]* fixes that problem.
On a side note, \p{Alnum} is exactly equivalent to [A-Za-z0-9] in Java, which is not the case in all regex flavors. I used \p{Alnum}, not just because it's shorter, but because it gives more protection from typos like [A-z] (which is syntactically valid, but almost certainly not what the author really meant).
EDIT: Performance should be considered, too. I originally included a + after the first \p{Alnum}, but I realized that wasn't a good idea. If this were part of a longer regex, and the regex didn't match right away, it could end up wasting a lot of time trying to match the same groups of characters with \p{Alnum}+ or [\p{Alnum}\s]*. The leading \s* is okay, though, because \s doesn't match any of the characters that \p{Alnum} matches.
Any one or more word char zero or more whitespace
\w+\s*
Hey try this ([^\s]+\s*) [^\s] means catch everything that is not white space, while \s* means that an white space is optional (if you really want at least one white space put + instead of )
Edit: sory mine catch everithing not only alphanumeric (put ([a-zA-Z0-9]+\s) for alphanumeric)
This should do the trick:
\s*\p{Alnum}+\s*
\p{Alnum} is an alphanumeric character: [\p{Alpha}\p{Digit}]
* says "zero or more times"
+ says "at least one" (not "or" as you seem to believe, or is written |)
| means "or"
\s is a whitespace character: [ \t\n\x0B\f\r]
EDIT: To answer the comments below I want it to accept strings which contain AT LEAST 1 alphanumeric AND any number of (including no) spaces.
The pattern I suggested requires at least one alpha numeric character.
EDIT2: To clarify, I don't want the any number of whitespace (including 0) to be accepted unless there is at least 1 alphanumeric character
The pattern I suggested will not accept only white space characters only.
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.