拆分全名

Question

我有字符串，例如“John Daws Black”与空格分开，我需要将它们分成两部分，这样就会有像“John Daws”这样的名字部分和像“Black”这样的姓氏部分。然而，正面名称部分可以是任何长度，如“John Erich Daws Black”。我的代码可以得到最后一部分：

public String getSurname(String fullName){
    String part = "";
     for (String retval: fullName.split(" "))
         part = retval;
         return part;
}

但我不知道如何获得前部。

Answer 1

找到 last 空间，然后使用$str=file_get_contents('sample.css'); //replace all new lines and spaces $str = str_replace("\n", "", $str); $str = str_replace(" ", "", $str); //write the entire string file_put_contents('sample.css', $str); 手动拆分。

substring()

Answer 2

试试这个。

public String getName(String fullName){
    return fullName.split(" (?!.* )")[0];
}

public String getSurname(String fullName){
    return fullName.split(" (?!.* )")[1];
}

Answer 3

分割后获取数组的最后一个元素：

function _see(sq, fen, depth, maxDepth, color, chess) {
    "use strict";
    if (chess.fen() !== fen) {
        console.error("s fen/chess sync error");
        chess.load(fen);
    }

    if (chess.in_checkmate() || chess.game_over()) {
        return MATE;
    } else if (chess.in_check()) {
        return 0; // ????
    }

    var value = 0, moves, index, move_score, tfen, foo, bar;

    if (depth < maxDepth) {
        moves = chess.moves({
            square: sq,
            verbose: true
        });
        if (moves.length > 0) {
            counter.seeNodes = counter.seeNodes + 1;
            moves = _.chain(moves)
                //only captures
                .reject(function (e) {
                    return !e.hasOwnProperty('captured');
                })
                //material MVV
                .sortBy(function (s) {
                    return evalPiece(s.piece);
                })
                //captures LVA
                .sortBy(function (s) {
                    return -evalPiece(s.captured);
                })
                .value();
            //counter.sDepth = Math.max(depth, counter.sDepth);
            //counter.maxSDepth = Math.max(maxDepth, counter.maxSDepth);        console.error(JSON.stringify(moves));

            for (index = 0; index < moves.length; index += 1) {
                foo = chess.move(moves[index]);
                if (foo === null) {
                    console.error("see move generated error, aborting loop");
                    break;
                }
                tfen = chess.fen();
                value = Math.max(0, evalPiece(foo.captured) - _see(sq, tfen, depth + 1, maxDepth, -color, chess));
                bar = chess.undo();
                if (bar === null) {
                    console.error("see: bar=null");
                }
            }

        }

    }
    return value;
}

输出：

    String fullName= "John Daws Black";
    String surName=fullName.split(" ")[fullName.split(" ").length-1];
    System.out.println(surName);

修改 对于前部，使用子字符串：

Black

输出：

    String fullName= "John Daws Black";
    String surName=fullName.split(" ")[fullName.split(" ").length-1];
    String firstName = fullName.substring(0, fullName.length() - surName.length());
    System.out.println(firstName );

Answer 4

//Split all data by any size whitespace 
final Pattern whiteSpacePattern = Pattern.compile("\\s+");
final List<String> splitData = whiteSpacePattern.splitAsStream(inputData)
.collect(Collectors.toList());

//Create output where first part is everything but the last element
if(splitData.size() > 1){
    final int lastElementIndex = splitData.size() - 1;
    //connect all names excluding the last one
    final String firstPart = IntStream.range(0,lastElementIndex).
    .mapToObj(splitData::get)
    .collect(Collectors.joining(" "));

    final String result = String.join(" ",firstPart,
    splitData.get(lastElementIndex));
}

Answer 5

以下方法适合我。它考虑到如果存在中间名，那么它将包含在名字中：

private static String getFirstName(String fullName) {
        int index = fullName.lastIndexOf(" ");
        if (index > -1) {
            return fullName.substring(0, index);
        }
        return fullName;
    }

    private static String getLastName(String fullName) {
        int index = fullName.lastIndexOf(" ");
        if (index > -1) {
            return fullName.substring(index + 1 , fullName.length());
        }
        return "";
    }

Answer 6

只需为上面的 Andreas 答案进行重写，但要使用正确的方法来获取名字和姓氏

    public static String get_first_name(String full_name)
    {
      int last_space_index = full_name.lastIndexOf(' ');
      if (last_space_index == -1) // single name
          return full_name;
      else
          return full_name.substring(0, last_space_index);
    }

    public static String get_last_name(String full_name)
    {
      int last_space_index = full_name.lastIndexOf(' ');
      if (last_space_index == -1) // single name
          return null;
      else
          return full_name.substring(last_space_index + 1);
    }

Answer 7

这是我之前使用过的一种解决方案，它也考虑了后缀。.（末尾3个字符）

/**
 * Get First and Last Name : Convenience Method to Extract the First and Last Name.
 *
 * @param fullName
 *
 * @return Map with the first and last names.
 */
public static Map<String, String> getFirstAndLastName(String fullName) {

    Map<String, String> firstAndLastName = new HashMap<>();

    if (StringUtils.isEmpty(fullName)) {
        throw new RuntimeException("Name Cannot Be Empty.");
    }

    String[] nameParts = fullName.trim().split(" ");

    /*
     * Remove Name Suffixes.
     */
    if (nameParts.length > 2 && nameParts[nameParts.length - 1].length() <= 3) {
        nameParts = Arrays.copyOf(nameParts, nameParts.length - 1);
    }

    if (nameParts.length == 2) {
        firstAndLastName.put("firstName", nameParts[0]);
        firstAndLastName.put("lastName", nameParts[1]);
    }

    if (nameParts.length > 2) {
        firstAndLastName.put("firstName", nameParts[0]);
        firstAndLastName.put("lastName", nameParts[nameParts.length - 1]);
    }

    return firstAndLastName;

}

Answer 8

尝试一下：

String[] name = fullname.split(" ")
if(name.size > 1){
    surName = name[name.length - 1]
    for (element=0; element<(name.length - 1); element++){
        if (element == (name.length - 1)) 
            firstName = firstName+name[element]
        else 
            firstName = firstName+name[element]+" "
    }
}
else 
    firstName = name[0]