Java中的多行字符串连接

Question

我正在寻求帮助。在Java中连接多行字符串并在之后打印它的最简单方法是什么？

例如：我有两个字符串：

String turtle1 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r";
String turtle2 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r";

我想在 Java Eclipse 控制台中得到这个结果：

         _            _
     .-./*)       .-./*)
   _/___\/      _/___\/
     U U          U U

我已经尝试了一些算法来将我的字符串分成不同的部分，然后重新连接它。但它没有成功。 我知道有 StringBuffer 类和 StringBuilder 类，但经过一番研究，我没有找到符合我需要的东西。

提前感谢您的帮助。

Answer 1

看我下面的例子，应该是自我解释的。

public class Turtle {

    private static final String returnpattern = "\r\n";

    public static void main(String[] args) {

        // the data to run through
        String turtle1 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r\n";
        String turtle2 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r\n";

        // split the data into individual parts
        String[] one = turtle1.split(returnpattern);
        String[] two = turtle2.split(returnpattern);

        // find out the longest String in data set one
        int longestString = 0;
        for (String s : one) {
            if (longestString < s.length()) {
                longestString = s.length();
            }
        }

        // loop through parts and build new string
        StringBuilder b = new StringBuilder();
        for (int i = 0; i < one.length; i++) {
            String stringTwo = String.format("%1$" + longestString + "s", two[i]); // left pad the dataset two to match
                                                                                   // length
            b.append(one[i]).append(stringTwo).append(returnpattern);
        }

        // output
        System.out.println(b);
    }
}

Answer 2

只是为了好玩，这里是另一个使用流的解决方案，准备让两个以上的海龟并排显示：

public static void main(String[] args) {
    String turtle1 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r";
    String turtle2 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r";

    // split lines into fragments
    List<List<String>> fragments = Stream.of(turtle1, turtle2)
            .map(x -> Stream.of(x.split("\\r\\n?|\\n")).collect(Collectors.toList()))
            .collect(Collectors.toList());

    // make all lists same length by adding empty lines as needed
    int lines = fragments.stream().mapToInt(List::size).max().orElse(0);
    fragments.forEach(x -> x.addAll(Collections.nCopies(lines - x.size(), "")));

    // pad all fragments to maximum width (per list)
    List<List<String>> padded = fragments.stream().map(x -> {
        int width = x.stream().mapToInt(String::length).max().orElse(0);
        return x.stream().map(y -> String.format("%-" + width + "s", y)).collect(Collectors.toList());
    }).collect(Collectors.toList());

    // join corresponding fragments to result lines, and join result lines
    String result = IntStream.range(0, lines)
            .mapToObj(i -> padded.stream().map(x -> x.get(i)).collect(Collectors.joining()))
            .collect(Collectors.joining(System.lineSeparator()));

    System.out.println(result);
}

Answer 3

不是很漂亮，但很有效：

String turtle1 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r\n";
String turtle2 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r\n";
String[] turtle1Lines = turtle1.split("\r\n");
String[] turtle2Lines = turtle2.split("\r\n");
StringBuilder sb = new StringBuilder();
int turtle1Width = 0;
for (int i = 0; i < 4; i++) {
    if (turtle1Lines[i].length() > turtle1Width) {
        turtle1Width = turtle1Lines[i].length();
    }
}
for (int i = 0; i < 4; i++) {
    sb.append(turtle1Lines[i]);
    for (int j = turtle1Width - turtle1Lines[i].length(); j > 0; j--) {
        sb.append(' ');
    }
    sb.append(turtle2Lines[i]);
    sb.append("\r\n");
}
String turtles = sb.toString();

Answer 4

我也在；）

public class Test {
    static String turtle1 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r".replace("\r", "");
    static String turtle2 = "         _\r\n     .-./*)\r\n   _/___\\/\r\n     U U\r".replace("\r", "");

    public static int countRows(String string){
        return string.length() - string.replace("\n", "").length() + 1;
    }

    public static int getMaxLength(String string){
        int maxLength = 0;
        int currentLength = 0;
        char[] data = string.toCharArray();
        for(Character c : data){
            if(c != '\n'){
                if(++currentLength > maxLength) {
                    maxLength = currentLength;
                }
            }else{
                currentLength = 0;
            }
        }
        return maxLength;
    }

    public static String[] toStringArray(String string){
        int length = getMaxLength(string);
        int rows = countRows(string);
        String[] result = new String[rows];
        int last = 0;
        for(int i = 0; i < rows; i++){
            int temp = string.indexOf("\n", last);
            String str;
            if(temp != -1) {
                str = string.substring(last, temp);
            }else{
                str = string.substring(last);
            }
            while(str.length() < length){
                str += " ";
            }
            result[i] = str;
            last = temp + 1;
        }
        return result;
    }

    public static String concatMultilineStrings(String first, String second){
        StringBuilder sb = new StringBuilder();
        String[] arrayFirst = toStringArray(first);
        String[] arraySecond = toStringArray(second);
        if(arrayFirst.length != arraySecond.length){
            System.exit(69);
        }
        for(int i = 0; i < arrayFirst.length; i++){
            sb.append(arrayFirst[i]);
            sb.append(arraySecond[i]);
            sb.append("\n");
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        System.out.println(concatMultilineStrings(turtle1, turtle2));
    }
}