CODEWITHSUNDEEP
javastring-comparisonpunctuation
sephtian
sephtian

Reputation: 455

Check if string is a punctuation character

Let's say I have a String array that contains some letters and punctuation

String letter[] = {"a","b","c",".","a"};

In letter[3] we have "."

How can I check if a string is a punctuation character? We know that there are many possible punctuation characters (,.?! etc.)

My progress so far:

for (int a = 0; a < letter.length; a++) {
    if (letter[a].equals(".")) { //===>> i'm confused in this line
        System.out.println ("it's punctuation");
    } else {
        System.out.println ("just letter");
    }
}

Upvotes: 22

Views: 84438

Answers (7)

Dastin
Dastin

Reputation: 4507

I have tried regex: "[\\p{Punct}]" or "[\\p{IsPunctuation}]" or withouth [], it doesn't work as expected.

My string is: How are you?, or even "How are you?", he asked.

Then call: my_string.matches(regex); but it seems like it only recognises if the string is only one punctuation (e.g "?", ".",...).

Only this regex works for me: "(.*)[\\p{P}](.*)", it will include all preceding and proceeding characters of the punctuation because the matches() requires to match all the sentence.

Upvotes: 1

jfranzoy
jfranzoy

Reputation: 35

import String ... if(string.punctuation.contains(letter[a]))

Upvotes: -3

Karthik T
Karthik T

Reputation: 31952

Do you want to check more punctuations other than just .?

If so you can do this.

String punctuations = ".,:;";//add all the punctuation marks you want.
...
if(punctuations.contains(letter[a]))

Upvotes: 21

Hans Brende
Hans Brende

Reputation: 8567

Depending on your needs, you could use either

Pattern.matches("\\p{Punct}", str)

or

Pattern.matches("\\p{IsPunctuation}", str)

The first pattern matches the following 32 characters: !"#$%&'()*+,-./:;<=>?@[\]^_`{|}~

The second pattern matches a whopping 632 unicode characters, including, for example: «, », ¿, ¡, §, , ‘, ’, “, ”, and ‽.

Interestingly, not all of the 32 characters matched by the first pattern are matched by the second. The second pattern does not match the following 9 characters: $, +, <, =, >, ^, `, |, and ~ (which the first pattern does match).

If you want to match for any character from either character set, you could do:

Pattern.matches("[\\p{Punct}\\p{IsPunctuation}]", str)

Upvotes: 35

Rob Korobkin
Rob Korobkin

Reputation: 19

function has_punctuation(str) {

  var p_found = false;
  var punctuations = '`~!@#$%^&*()_+{}|:"<>?-=[]\;\'.\/,';
  $.each(punctuations.split(''), function(i, p) {
    if (str.indexOf(p) != -1) p_found = true;
  });

  return p_found;

}

Upvotes: -5

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726589

Here is one way to do it with regular expressions:

if (Pattern.matches("\\p{Punct}", str)) {
    ...
}

The \p{Punct} regular expression is a POSIX pattern representing a single punctuation character.

Upvotes: 66

Jj Tuibeo
Jj Tuibeo

Reputation: 773

Try this method: Character.isLetter(). It returns true if the character is a letter (a-z, uppercase or lowercase), returns false if character is numeric or symbol.

e.g. boolean answer = Character.isLetter('!');

answer will be equal to false.

Upvotes: 3

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