public static void getCharacter(String str){

        int count[]= new int[256];

        for(int i=0;i<str.length(); i++){


            count[str.charAt(i)]++;

        }
        System.out.println("The ascii values are:"+ Arrays.toString(count));

        //Now display wht character is repeated how many times

        for (int i = 0; i < count.length; i++) {
            if (count[i] > 0)
               System.out.println("Number of " + (char) i + ": " + count[i]);
        }


    }
}

这是一个稍微不同风格的递归解决方案:

public static int countOccurrences(String haystack, char needle)
{
    return countOccurrences(haystack, needle, 0);
}

private static int countOccurrences(String haystack, char needle, int accumulator)
{
    if (haystack.length() == 0) return accumulator;
    return countOccurrences(haystack.substring(1), needle, haystack.charAt(0) == needle ? accumulator + 1 : accumulator);
}

如果你使用Spring框架，你也可以使用“StringUtils”类。 方法是“countOccurrencesOf”。

String s = "a.b.c.d";
long result = s.chars().filter(ch -> ch == '.').count();

一个简短的例子是

String text = "a.b.c.d";
int count = text.split("\\.",-1).length-1;

总结一下其他的答案，以及我所知道的所有用一句话来做到这一点的方法:

   String testString = "a.b.c.d";

1)使用Apache Commons

int apache = StringUtils.countMatches(testString, ".");
System.out.println("apache = " + apache);

2)使用Spring框架

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

3)使用replace

int replace = testString.length() - testString.replace(".", "").length();
System.out.println("replace = " + replace);

4)使用replaceAll(案例1)

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

5)使用replaceAll(情况2)

int replaceAllCase2 = testString.length() - testString.replaceAll("\\.", "").length();
System.out.println("replaceAll (second case) = " + replaceAllCase2);

6)使用split

int split = testString.split("\\.",-1).length-1;
System.out.println("split = " + split);

7)使用Java8(案例1)

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

8)使用Java8(情况2)，可能比情况1对unicode更好

long java8Case2 = testString.codePoints().filter(ch -> ch =='.').count();
System.out.println("java8 (second case) = " + java8Case2);

9)使用StringTokenizer

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

来自评论:要小心StringTokenizer，对于a.b.c.d它将工作，但对于a…b.c....D或a.b.c。D或a....b......c..... D…等等，它不会工作。这只是有意义的。字符之间只有一次

更多信息在github

性能测试(使用JMH，模式= AverageTime，得分0.010优于0.351):

Benchmark              Mode  Cnt  Score    Error  Units
1. countMatches        avgt    5  0.010 ±  0.001  us/op
2. countOccurrencesOf  avgt    5  0.010 ±  0.001  us/op
3. stringTokenizer     avgt    5  0.028 ±  0.002  us/op
4. java8_1             avgt    5  0.077 ±  0.005  us/op
5. java8_2             avgt    5  0.078 ±  0.003  us/op
6. split               avgt    5  0.137 ±  0.009  us/op
7. replaceAll_2        avgt    5  0.302 ±  0.047  us/op
8. replace             avgt    5  0.303 ±  0.034  us/op
9. replaceAll_1        avgt    5  0.351 ±  0.045  us/op