Develop Reference

Regex to check if String is one word in Java

I need regex to check if String has only one word (e.g. "This", "Country", "Boston ", " Programming ").
So far I used an alternative way of doing it which is to check if String contains spaces. However, I am sure that this can be done using regex.
One possible way in my opinion is "^\w{2,}\s". Does this work properly? Are there any other possible answers?

The pattern ^\w{2,}\s matches 2 or more word characters from the start of the string, followed by a mandatory whitespace char (that can also match a newline)
As the pattern is also unanchored, it can also match Boston in Boston test
If you want to match a single word with as least 2 characters surrounded by optional horizontal whitespace characters using \h* and add an anchor $ to assert the end of the string.
^\h*\w{2,}\h*$
Regex demo
In Java
String regex = "^\\h*\\w{2,}\\h*$";

Regex to exactly match against a given set of keywords in between delimiters?

Im trying a regex to exactly match against a given set of keywords in between delimiters?
For example:
Keywords: keyone, keytwo, keythree
Start delimiter: ;
End delimiter: ;
Text under test: some text ;keyone; other text ;keytwo; some text ;keythreeeee;
Regex i tried : ;([keyonekeytwokeythree]+);
Problem with this regex is, this matching with keythreeeee also. My expectation is it should not match keythreeeee because this is not exact match.

You should read up on regular expression syntax.
([keyonekeytwokeythree]+)
The square bracket syntax tells the regexp matcher to match 'any number of characters from the set keyonekeytwokeythree'. It will thus also match yekenoeerth.
You're looking for something like:
;(keyone|keytwo|keythree);

You should use a regex like this:
;(keyone|keytwo|keythree);

I first take all the text inside delimiters.
(delmiterSart)(.)*(delimiterEnd)
and then on this selected text i try to search you word
(key1|key2|keyn)+

capture all characters between match character (single or repeated) on string

I'm trying to extract the string preceding a specific character (even when character is repeated, like this (ie: underscore '_'):
this_is_my_example_line_0
this_is_my_example_line_1_
this_is_my_example_line_2___
_this_is_my_ _example_line_3_
__this_is_my___example_line_4__
and after running my regex I should get this (the regex should ignore the any instances of the matching character in the middle of the string):
this_is_my_example_line_0
this_is_my_example_line_1
this_is_my_example_line_2
this_is_my_ _example_line_3
this_is_my___example_line_4
In other words I'm trying to 'trim' the matched character(s) at the beginning and end of string.
I'm trying to use a Regex in Java to accomplish this, my idea is to capture the group of characters between the special character(s) at the end or beginning of the line.
So far I can only do this successfully for example 3 with this regexp:
/[^_]+|_+(.*)[_$]+|_$+/
[^_]+ not 'underscore' once or more
| OR
_+ underscore once or more
(.*) capture all characters
[_$]+ not 'underscore' once or more followed by end of line
|_$+ OR 'underscore' once or more followed by end of line
I just realized that this excludes the first word of the message on example 0,1,2 since the string doesn't start with underscore and it only starts matching after finding a underscore..
Is there an easier way not involving regex?
I don't really care about the first character (although it would be nice) I only need to ignore the repeating character at the end.. it looks that (by this regex tester) just doing this, would work? /()_+$/ the empty parenthesis matches anything before a single or repeting matches at the end of the line.. would that be correct?
Thank you!

There are a couple of options here, you could either replace matches of ^_+|_+$ with an empty string, or extract the contents of the first capture group from the match of ^_*(.*?)_*$. Note that if your strings may be multiple lines and you want to perform the replacement on each line then you will need to use the Pattern.MULTILINE flag for either approach. If your strings may be multiple lines and you only want to replacement to occur at the very beginning and end, don't use Pattern.MULTILINE but use Pattern.DOTALL for the second approach.
For example: http://regexr.com?355ff

How about [^_\n\r](.*[^_\n\r])??
Demo
String data=
"this_is_my_example_line_0\n" +
"this_is_my_example_line_1_\n" +
"this_is_my_example_line_2___\n" +
"_this_is_my_ _example_line_3_\n" +
"__this_is_my___example_line_4__";
Pattern p=Pattern.compile("[^_\n\r](.*[^_\n\r])?");
Matcher m=p.matcher(data);
while(m.find()){
System.out.println(m.group());
}
output:
this_is_my_example_line_0
this_is_my_example_line_1
this_is_my_example_line_2
this_is_my_ _example_line_3
this_is_my___example_line_4

Check string contains whitespace along with some other char sequence using regex in java

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.

Java String validation only one alphanumeric with Regex

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]+)