Basically my question is this, why is:
String word = "unauthenticated";
word.matches("[a-z]");
returning false? (Developed in java1.6)
Basically I want to see if a string passed to me has alpha chars in it.
The String.matches() function matches your regular expression against the whole string (as if your regex had ^ at the start and $ at the end). If you want to search for a regular expression somewhere within a string, use Matcher.find().
The correct method depends on what you want to do:
Check to see whether your input string consists entirely of alphabetic characters (String.matches() with [a-z]+)
Check to see whether your input string contains any alphabetic character (and perhaps some others) (Matcher.find() with [a-z])
Your code is checking to see if the word matches one character. What you want to check is if the word matches any number of alphabetic characters like the following:
word.matches("[a-z]+");
with [a-z] you math for ONE character.
What you’re probably looking for is [a-z]*
Related
String replace change more that i want.
For example
String input = "The blue house Theatres";
input = input.replace("the", "AAA");
output it will be:
"AAA blue house AAAatres"
I don't whant to change when is a subpart of a word.
First you should try to use replaceAll(regex, replacement) instead of replace(literal, replacement) since the latter works on literals only, i.e. you can't use expressions, while the former uses regular expressions to find matches.
Next your regular expression should use word boundaries, e.g. \bthe\b where \b marks a word boundary.
Finally if you want to do a case-insensitive replacement you'll need to either handle the possible cases in the epxression (e.g. \b[tT]he\b) or switch the expression to case-insensitive mode by prepending it with (?i), i.e. (?i)\bthe\b. Note that the expression [tT]he would not match THE while the case-insensitive expression would, so depending on your requirements you'd need to choose one or the other.
Using all that you'd get input = input.replaceAll("(?i)\\bthe\\b", "AAA");.
Edit:
According to your comment on the question you don't want to use word boundaries but only look for characters before and after. You can achieve that with negative look-around expressions, e.g. (?i)(?<![a-z])the(?![a-z]). Note that I used the quite simple character class [a-z] here, if you need to exclude more characters you'd need to expand it.
The above expression would match !The, the, THE? etc. but not Theatre or aether etc. since if requires the match to not be preceded by a character ((?<![a-z])) and not be followed by one ((?![a-z])).
Use a regex with word boundaries \b:
String input = "The blue house Theatres";
input.replaceAll("\\bThe\\b", "AAA");
Imagine, that I have the string 12.34some_text.
How can I match the substring following after the second character (4 in my case) after the . character. In that particular case the string I want to match is some_text.
For the string 56.78another_text it will be another_text and so on.
All accepted strings have the pattern \d\d\.\d\d\w*
If you wish to match everything from the second character after a specific one (i.e. the dot) you can use a lookbehind, like this:
(?<=[.]\d{2})(\w*)
demo
(?<=[.]\d{2}) is a positive lookbehind that matches a dot [.] followed by two digits \d{2}.
Since you are using java and the given pattern is \d\d\.\d\d\w* you will get some_text from 12.34some_textby using
String s="12.34some_text";
s.substring(5,s.length());
and you can compare the substring!
What would be a regular expression that would evaluate to true if the string has one or more letters anywhere in it.
For example:
1222a3999 would be true
a222aZaa would be true
aaaAaaaa would be true
but:
1111112())-- would be false
I tried: ^[a-zA-Z]+$ and [a-zA-Z]+ but neither work when there are any numbers and other characters in the string.
.*[a-zA-Z].*
The above means one letter, and before/after it - anything is fine.
In java:
String regex = ".*[a-zA-Z].*";
System.out.println("1222a3999".matches(regex));
System.out.println("a222aZaa ".matches(regex));
System.out.println("aaaAaaaa ".matches(regex));
System.out.println("1111112())-- ".matches(regex));
Will provide:
true
true
true
false
as expected
^.*[a-zA-Z].*$
Depending on the implementation, match() functions check if the entire string matches (which is probably why your [a-zA-Z] or [a-zA-Z]+ patterns didn't work).
Either use match() with the above pattern or use some sort of search() method instead.
This regexp should do it:
[a-zA-Z]
It matches as long as there's a single letter anywhere in the string, it doesn't care about any of the other characters.
[a-zA-Z]+
should have worked as well, I don't know why it didn't for you.
.*[a-zA-Z]?.*
Should get you the result you want.
The period matches any character except new line, the asterisk says this should exist zero or more times. Then the pattern [a-zA-Z]? says give me at least one character that is in the brackets because of the use of the question mark. Finally the ending .* says that the alphabet characters can be followed by zero or more characters of any type.
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]+)
I am taking user input through UI, and I have to validate it. Input text should obey the following ondition
It should either end with one or more
white space characters OR with just
single '='
I can use
".*[\s=]+"
but it matches multiple '=' also which I don't want to.
Please help.
You can use alternation:
(\s+|=)$
This expression means match one or more whitespace character or one equals, at the end of the string. The $ is an anchor which matches the end of the string (as you mentioned you're looking for characters at the end of the string).
(As tchrist correctly pointed out in the comments, $ matches the end of line instead of end of string when in multiline mode. If this is true in your case, and you are indeed looking for the end of the string instead of the end of the line, you can use \Z instead, which matches the end of the string regardless of multiline mode.)
If you want to ensure that there is only one = at the end, you can use a lookaround (in this case, a negative lookbehind, specifically). A lookaround is a zero-width assertion which tells the regex engine that the assertion must pass for the pattern to match, but it does not consume any characters.
(\s+|(?<!=)=)$
In this case, (?<!=) tells the regex engine, the character before the current position cannot be an =. When put into the expression, (?<!=)= means that the = will only match if the previous character is not also a =.
Begin string
Anything not "=" ( to avoid the double "==")
One or more blank spaces OR one "="
End of string
^([^=]*[\s+|=])$
Should work :-)
Try this expression:
".*(\\s+|=)"