Question

我正在尝试将扫描仪复制到字符串中，然后精确返回该字符串，然后在新字符串的末尾添加新行。因此等于This \ nis a \ ntest的扫描仪将返回一个等于This \ nis a \ ntest \ n的字符串。

到目前为止我的代码

public static String scannerToString (Scanner scnr)
{

    String string = "";

    while (scnr.hasNextLine())
    {

        string = scnr.nextLine();

    }
    return string + "\\n";

}

我改用StringBuilder解决了这个问题。这是我的新代码。

StringBuilder result = new StringBuilder();
    while (scnr.hasNextLine())
    {
        String string = scnr.nextLine();
        result.append(string + "\n");
    }
    return result.toString();

Answer 1

根据搜索，这是Olathe's String inspection method转换为使用StringBuilder和lucasvc's Scanner contents retriever的组合：

import java.util.Scanner;

public class InspectScanner {
  public static String inspect(final String str) {
    StringBuilder result = new StringBuilder("\"");
    for (int i = 0; i < str.length(); i++) {
      final char ch = str.charAt(i);
      if      (ch >= 128) result.append(String.format("\\u%04x", (int) ch));
      else if (ch == '"') result.append("\\\"");
      else if (ch >=  32) result.append(ch);
      else switch (ch) {
        case '\b':        result.append("\\b"); break;
        case '\f':        result.append("\\f"); break;
        case '\n':        result.append("\\n"); break;
        case '\r':        result.append("\\r"); break;
        case '\t':        result.append("\\t"); break;
        default:        final int iNext = i + 1;
                        if (iNext == str.length())
                          result.append(String.format("\\%o", (int) ch));
                        else {
                          final char nextCh = str.charAt(iNext);
                          result.append(String.format((nextCh >= '0' && nextCh <= '9') ? "\\%03o" : "\\%o", (int) ch));
                        }
      }
    }
    result.append("\"");
    return result.toString();
  }

  public static String getRemainingScannerContents(final Scanner scanner) {
    scanner.useDelimiter("\\A");
    return scanner.hasNext() ? scanner.next() : "";
  }

  public static final void main(final String[] args) {
    Scanner scanner = new Scanner("test\ntest\n12345");
    System.out.println(inspect(getRemainingScannerContents(scanner)));
  }
}

This produces on stdout：

"test\ntest\n12345"

使scnr.nextLine（）忽略“ \ n”

1 个答案: