java – How to count the number of occurrences of a character in a string?

Question:

There is a need to implement a check for multiple commas in a string. Those. only one comma can be used.

But in general, it is enough to know the number of occurrences of a character in a string. The rest will not be a problem to add.

What are the ways to do this (count the number of occurrences of a character)?

Answer:

1. Регулярка

2. Использовать StringUtils.countMatches из org.apache.commons.lang3.StringUtils

   int occurrence = StringUtils.countMatches("a,b,c,d", ",");
   System.out.println(occurrence);
   

3. Использовать string.split

   String string = "a,b,c,d";
   String []splitArray = string.split("\\,");
   System.out.println("Запятых тут : " + (splitArray.length - 1) + " штук.");
   

4. String testString =  "a,b,c,d";
   int occurrencesCount = testString.length() - testString.replace(",", "").length();
   System.out.println(occurrencesCount);
   

5. Java8

   String testString =  "a,b,c,d";
   long occurrencesCount = testString.chars().filter(ch -> ch == ',').count();
   System.out.println(occurrencesCount);
   

6. Еще какой-нибудь способ

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UPD found a similar thread on enSO: How do I count the number of occurrences of a char in a String? , a few examples from there:

• Spring Framework

   int spring = org.springframework.util.StringUtils.countOccurrencesOf(testString, ","); System.out.println("spring = " + spring);

• replaceAll

   int replaceAll = testString.replaceAll("[^,]", "").length(); System.out.println("replaceAll = " + replaceAll);

• Java8

   long java8 = testString.codePoints().filter(ch -> ch == ',').count(); System.out.println("java8 = " + java8);

• StringTokenizer

   int stringTokenizer = new StringTokenizer(" " +testString + " ", ",").countTokens()-1; System.out.println("stringTokenizer = " + stringTokenizer);

  But in this case, you need to be careful, because, for example, for the string abcd this example will work, but for the string a … bc … d or … abcd or a …. b ….. .c ….. d … etc. – will not be. As a result, the sequence of dots between each letter is counted as only one character.