3
Example I have some characters 123456789.123456789.
How to do to count up to the . after the . count again remembering that the point should not be counted.
3 answers
5
Another alternative by using indexOf():
String str = "123456789.123456789";
// Como o índice da string começa em 0, indexOf retornará exatamente o tamanho da primeira parte.
int t1 = str.indexOf('.');
int t2 = str.length() - t1 - 1; // - 1 para subtrair o "ponto" do total de caracteres da string
See working on Ideone
2
String str = "123456789.123456789";
//divide a string usando o ponto como divisor
String[] partes= str.split(".");
//antes do ponto
int n1 = partes[0].length();
//depois do ponto
int n1 = partes[1].length();
2
Another alternative that can be used is the class StringTokenizer package java.util.
public static void main (String[] args) throws java.lang.Exception{
StringTokenizer st = new StringTokenizer("12.345.6789", ".");
List<String> numeros = new ArrayList<String>();
while(st.hasMoreTokens()){
numeros.add(st.nextToken());
}
// Vai retornar o tamanho da 1ª sequência antes do ponto
System.out.println(numeros.get(0).length());
}
Functional example in Ideone.
In terms of performance, the StringTokenizer is faster than the String#split(), however, StringTokenizer is slow compared to String#indexOf().
In the link below is shown the comparison of the performance of each alternative cited here, among others.
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string.split(".")wouldn’t it work?lenght()to see the size.– Gustavo Cinque
@Gustavocinque Good suggestion. Why not create an answer?
– André Ribeiro
I didn’t have time @André. But now they have, so no problem.
– Gustavo Cinque