Question

我有一个像这种格式的字符串

数。主题，统一名称，位置2

我想重新订购像这样的格式

Number. Uni Name, Subject, Location 2

我可以通过这种方式进行操作

int index = Order .indexOf(",");
String newStr =Order .substring(index + 1, Order .length());

但是，我认为所有字符串都有不同的名称和长度，所以可能这不起作用。所以，我的问题是如何以有效的方式动态地重新排序字符串格式？

更新问题

Answer 1

    String[] str = Order.split(",");

    System.out.println(str[1] +","+str[0]+","+str[2]);

Answer 2

您需要使用String#split()功能。Here you can find a good example 所以你可以像这样解决你的问题

String[] str = Order.split(",");
String newOrderedString = str[1] +","+str[0]+","+str[2];
//OR
newOrderedString = str[1] +","+str[2]+","+str[0];
//OR
newOrderedString = str[2] +","+str[0]+","+str[1]; // As you want

Answer 3

这样的事情可行：

String s = "Subject, Uni Name, Location 2";
String[] sp = s.split(",");
String finalString = sp[1] + "," + sp[0] + "," + sp[2];

您可以考虑使用StringBuilder来执行finalString组合。

Answer 4

尝试

 String original = "Subject, Uni Name, Location 2";
 String[] split = original.split(",");
 StringBuilder b= new StringBuilder();
 b.append(split[1]).append(",");
 b.append(split[0]).append(",");
 b.append(split[2]);
 System.out.println(b.toString());  // Uni Name,Subject, Location 2

Answer 5

使用：

String s = "Subject, Uni Name, Location 2";
String[] strs = s.split(",");

然后按照你需要的顺序加入结果（我已经使用trim（）去掉额外的空格并将它们添加回需要的地方）：

String t = strs[1].trim() + ", " + strs[0].trim() + ", " + strs[2].trim();

如果你这么做（例如在一个循环中），使用StringBuilder来构建结果字符串 - 它更有效（和惯用）。

Answer 6

一种方法是正则表达式。这可能不是最的有效方式，但它相当干净：

public static void main(String args[]) {
    final Pattern pattern = Pattern.compile("(\\d++)\\.([^,]++),\\s*+([^,]++),\\s*+(.*+)");
    final Matcher matcher = pattern.matcher("");
    //for each input string
    final String input = "Subject, Uni Name, Location 2";
    matcher.reset(input);
    final String output = matcher.replaceAll("$1, $3, $2, $4");
    System.out.println(output);
}

因此，您预先编译Pattern并重新使用Matcher以获得最高速度。然后，您只需对每个输入replaceAll执行String。

Answer 7

我想说不要用这样的原始字符串做这个。使用对象。

class ReportHeader {
    String number;
    String name;
    String subject;
    String location;

    ReportHeader(number,name,subject,location) {
        this.number = number;
        this.name = name;
        this.subject = subject;
        this.location = location;
    }

}

class interface HeaderFormat {
    String toString(ReportHeader header);
}

// this is how Wilson likes his reports
class ProfessorWilsonsHeaderFormat implements HeaderFormat { 
    public String toString(ReportHeader header) {
        return header.number + "," + header.name + "," + header.subject + "," + header.location;
    }
}

// this is how Thomspon likes his reports
class ProfessorThompsonsHeaderFormat implements HeaderFormat { 
    public String toString(ReportHeader header) {
        return header.number + "," + header.subject + "," + header.name + "," + header.location;
    }
}

Java String重新排序

7 个答案: