I have to validate a String with only white spaces. In my String white spaces between a String is allowed but only white spaces not allowed. For example "conditions apply","conditions" etc is allowed but not " ". ie, only white space not allowed
I want a regular expression in JavaScript for this
Try this regular expression
".*\\S+.*"
Do you really need to use a regex?
if (str.trim().length() == 0)
return false;
else
return true;
As mentionned in the comments, this could be simplified to a one-liner
return str.trim().length() > 0;
or, since Java 6
return !str.trim().isEmpty();
you can do it like this:
// This does replace all whitespaces at the end of the string
String s = " ".trim();
if(s.equals(""))
System.out.println(true);
else
System.out.println(s);
what about checking if the string does not match "\\s+"?
The regular expression is ^\\s*$ is used to match a whitespace only string, you can validate against this.
^ # Match the start of the string
\\s* # Match zero of more whitespace characters
$ # Match the end of the string
Anchoring to the start and the end of the string is important here.
Related
I have set of inputs ++++,----,+-+-.Out of these inputs I want the string containing only + symbols.
If you want to see if a String contains nothing but + characters, write a loop to check it:
private static boolean containsOnly(String input, char ch) {
if (input.isEmpty())
return false;
for (int i = 0; i < input.length(); i++)
if (input.charAt(i) != ch)
return false;
return true;
}
Then call it to check:
System.out.println(containsOnly("++++", '+')); // prints: true
System.out.println(containsOnly("----", '+')); // prints: false
System.out.println(containsOnly("+-+-", '+')); // prints: false
UPDATE
If you must do it using regex (worse performance), then you can do any of these:
// escape special character '+'
input.matches("\\++")
// '+' not special in a character class
input.matches("[+]+")
// if "+" is dynamic value at runtime, use quote() to escape for you,
// then use a repeating non-capturing group around that
input.matches("(?:" + Pattern.quote("+") + ")+")
Replace final + with * in each of these, if an empty string should return true.
The regular expression for checking if a string is composed of only one repeated symbol is
^(.)\1*$
If you only want lines composed by '+', then it's
^\++$, or ^++*$ if your regex implementation does not support +(meaning "one or more").
For a sequence of the same symbol, use
(.)\1+
as the regular expression. For example, this will match +++, and --- but not +--.
Regex pattern: ^[^\+]*?\+[^\+]*$
This will only permit one plus sign per string.
Demo Link
Explanation:
^ #From start of string
[^\+]* #Match 0 or more non plus characters
\+ #Match 1 plus character
[^\+]* #Match 0 or more non plus characters
$ #End of string
edit, I just read the comments under the question, I didn't actually steal the commented regex (it just happens to be intellectual convergence):
Whoops, when using matches disregard ^ and $ anchors.
input.matches("[^\\+]*?\+[^\\+]*")
I need to add spaces between all punctuation in a string.
\\ "Hello: World." -> "Hello : World ."
\\ "It's 9:00?" -> "It ' s 9 : 00 ?"
\\ "1.B,3.D!" -> "1 . B , 3 . D !"
I think a regex is the way to go, matching all non-punctuation [a-ZA-Z\\d]+, adding a space before and/or after, then extracting the remainder matching all punctuation [^a-ZA-Z\\d]+.
But I don't know how to (recursively?) call this regex. Looking at the first example, the regex will only match the "Hello". I was thinking of just building a new string by continuously removing and appending the first instance of the matched regex, while the original string is not empty.
private String addSpacesBeforePunctuation(String s) {
StringBuilder builder = new StringBuilder();
final String nonpunctuation = "[a-zA-Z\\d]+";
final String punctuation = "[^a-zA-Z\\d]+";
String found;
while (!s.isEmpty()) {
// regex stuff goes here
found = ???; // found group from respective regex goes here
builder.append(found);
builder.append(" ");
s = s.replaceFirst(found, "");
}
return builder.toString().trim();
}
However this doesn't feel like the right way to go... I think I'm over complicating things...
You can use lookarounds based regex using punctuation property \p{Punct} in Java:
str = str.replaceAll("(?<=\\S)(?:(?<=\\p{Punct})|(?=\\p{Punct}))(?=\\S)", " ");
(?<=\\S) Asserts if prev char is not a white-space
(?<=\\p{Punct}) asserts a position if previous char is a punctuation char
(?=\\p{Punct}) asserts a position if next char is a punctuation char
(?=\\S) Asserts if next char is not a white-space
IdeOne Demo
When you see a punctuation mark, you have four possibilities:
Punctuation is surrounded by spaces
Punctuation is preceded by a space
Punctuation is followed by a space
Punctuation is neither preceded nor followed by a space.
Here is code that does the replacement properly:
String ss = s
.replaceAll("(?<=\\S)\\p{Punct}", " $0")
.replaceAll("\\p{Punct}(?=\\S)", "$0 ");
It uses two expressions - one matching the number 2, and one matching the number 3. Since the expressions are applied on top of each other, they take care of the number 4 as well. The number 1 requires no change.
Demo.
i have seen to replace "," to "." by using ".$"|",$", but this logic is not working with alphabets.
i need to replace last letter of a word to another letter for all word in string containing EXAMPLE_TEST using java
this is my code
Pattern replace = Pattern.compile("n$");//here got the real problem
matcher2 = replace.matcher(EXAMPLE_TEST);
EXAMPLE_TEST=matcher2.replaceAll("k");
i also tried "//n$" ,"\n$" etc
Please help me to get the solution
input text=>njan ayman
output text=> njak aymak
Instead of the end of string $ anchor, use a word boundary \b
String s = "njan ayman";
s = s.replaceAll("n\\b", "k");
System.out.println(s); //=> "njak aymak"
You can use lookahead and group matching:
String EXAMPLE_TEST = "njan ayman";
s = EXAMPLE_TEST.replaceAll("(n)(?=\\s|$)", "k");
System.out.println("s = " + s); // prints: s = njak aymak
Explanation:
(n) - the matched word character
(?=\\s|$) - which is followed by a space or at the end of the line (lookahead)
The above is only an example! if you want to switch every comma with a period the middle line should be changed to:
s = s.replaceAll("(,)(?=\\s|$)", "\\.");
Here's how I would set it up:
(?=.\b)\w
Which in Java would need to be escaped as following:
(?=.\\b)\\w
It translates to something like "a character (\w) after (?=) any single character (.) at the end of a word (\b)".
String s = "njan ayman aowkdwo wdonwan. wadawd,.. wadwdawd;";
s = s.replaceAll("(?=.\\b)\\w", "");
System.out.println(s); //nja ayma aowkdw wdonwa. wadaw,.. wadwdaw;
This removes the last character of all words, but leaves following non-alphanumeric characters. You can specify only specific characters to remove/replace by changing the . to something else.
However, the other answers are perfectly good and might achieve exactly what you are looking for.
if (word.endsWith("char oldletter")) {
name = name.substring(0, name.length() - 1 "char newletter");
}
What could be the regular expression to match below String
String str = "<Element>\r\n <Sub>regular</Sub></Element>";
There is a
carriage return "\r", new line character "\n" and a space after <Element>.
My code is as below
if(str.matches("<Element>([\\s])<Sub>(.*)"))
{
System.out.println("Matches");
}
Use the "dot matches newline" switch:
if (str.matches("(?s)<Element>\\s*<Sub>(.*)"))
With the switch turned on, \s will match newline characters.
I slso fixed your regex, removing two sets of redundant brackets, and adding the crucial * after the whitespace regex.
Say I have a string like this in java:
"this is {my string: } ok"
Note, there can be any number of white spaces in between the various characters. How do I check the above string to see if it contains just the substring:
"{my string: }"
Many thanks!
If you are looking to see if a String contains another specific sequence of characters then you could do something like this :
String stringToTest = "blah blah blah";
if(stringToTest.contains("blah")){
return true;
}
You could also use matches. For a decent explanation on matching Strings I would advise you check out the Java Oracle tutorials for Regular Expressions at :
http://docs.oracle.com/javase/tutorial/essential/regex/index.html
Cheers,
Jamie
If you have any number of white space between each character of your matching string, I think you are better off removing all white spaces from the string you are trying to match before the search. I.e. :
String searchedString = "this is {my string: } ok";
String stringToMatch = "{my string: }";
boolean foundMatch = searchedString.replaceAll(" ", "").contains(stringToMatch.replaceAll(" ",""));
Put it all into a string variable, say s, then do s.contains("{my string: }); this will return true if {my string: } is in s.
For this purpose you need to use String#contains(CharSequence).
Note, there can be any number of white spaces in between the various
characters.
For this purpose String#trim() method is used to returns a copy of the string, with leading and trailing whitespace omitted.
For e.g.:
String myStr = "this is {my string: } ok";
if (myStr.trim().contains("{my string: }")) {
//Do something.
}
The easiest thing to do is to strip all the spaces from both strings.
return stringToSearch.replaceAll("\s", "").contains(
stringToFind.replaceAll("\s", ""));
Look for the regex
\{\s*my\s+string:\s*\}
This matches any sequence that contains
A left brace
Zero or more spaces
'my'
One or more spaces
'string:'
Zero or more spaces
A right brace
Where 'space' here means any whitespace (tab, space, newline, cr)