Question

我在 Java 中有如下字符串。字符串的长度未知，作为示例，它将类似于以下内容。

String str = "I love programming. I'm currently working with Java and C++."

对于某些要求，我想获得前 15 个字符。然后是 30、45、70 个下一个字符。一旦字符串被拆分，如果名称没有意义，那么它应该从最近的空间拆分。对于上面的示例输出如下。

String strSpli1 = "I love "; //Since 'programming' is splitting it was sent to next split
String strSpli2 = "programming. I'm currently ";//Since 'working'  is splitting it was sent to next split
String strSpli3 = "working with Java and C++.";

请帮助我实现这一目标。

为有此类要求的任何人更新了答案。

 String str = "I love programming. I'm currently working with Java and C++.";
                String strSpli1 = "";
                String strSpli2 = "";
                String strSpli3 = "";

    try {
                strSpli1 = str.substring(15);
                int pos = str.lastIndexOf(" ", 16);
                if (pos == -1) {
                    pos = 15;
                }
                strSpli1 = str.substring(0, pos);
                str = str.substring(pos);
    
                try {
                    strSpli2 = str.substring(45);
                    int pos1 = str.lastIndexOf(" ", 46);
                    if (pos1 == -1) {
                        pos1 = 45;
                    }
                    strSpli2 = str.substring(0, pos1);
                    str = str.substring(pos1);
    
                    try {
                        strSpli3 = str.substring(70);
                        int pos2 = str.lastIndexOf(" ", 71);
                        if (pos2 == -1) {
                            pos2 = 45;
                        }
                        strSpli3 = str.substring(0, pos2);
                        str = str.substring(pos2);
    
                    } catch (Exception ex) {
                        strSpli3 = str;
                    }
    
                } catch (Exception ex) {
                    strSpli2 = str;
                }
            } catch (Exception ex) {
                strSpli1 = str;
            }

谢谢

Answer 1

使用 lastIndexOf() 的 2 参数版本从给定位置开始向后搜索空间。前 15 个字符的示例：

int pos = str.lastIndexOf(" ", 16);
if (pos == -1) {
    pos = 15;
}
String found = str.substring(0, pos);
str = str.substring(pos+1);

缺少检查，例如确保字符串以至少 15 个字符开头，或者 pos+1 对给定长度有效

建议看看java.text.BreakIterator

Answer 2

你为什么使用这么多 try catch ？试试这个。

public class MyClass {
public static void main(String args[]) {
    String str = "I love programming. I'm currently working with Java and C++.";
    String strSpli1 = "";
    String strSpli2 = "";
    String strSpli3 = "";
    strSpli1 = str.substring(0, 7);
    strSpli2 = str.substring(7, 33);
    strSpli3 = str.substring(34, str.length());
   

     System.out.println(strSpli1+"\n");
     System.out.println(strSpli2+"\n");
     System.out.println(strSpli3+"\n");

}

使用子字符串（开始索引，结束索引）。

如何使用Java根据长度和空间分割字符串

2 个答案: