我使用Java从用户获得一个字符串输入。我试着让输入的第一个字母大写。

我试了一下:

String name;

BufferedReader br = new InputStreamReader(System.in);

String s1 = name.charAt(0).toUppercase());

System.out.println(s1 + name.substring(1));

这导致了以下编译错误:

类型不匹配:不能从InputStreamReader转换为BufferedReader 不能在基本类型char上调用toUppercase()


当前回答

试试这个

这个方法做的是,考虑单词“hello world”这个方法把它变成“hello world”每个单词的开头大写。

 private String capitalizer(String word){

        String[] words = word.split(" ");
        StringBuilder sb = new StringBuilder();
        if (words[0].length() > 0) {
            sb.append(Character.toUpperCase(words[0].charAt(0)) + words[0].subSequence(1, words[0].length()).toString().toLowerCase());
            for (int i = 1; i < words.length; i++) {
                sb.append(" ");
                sb.append(Character.toUpperCase(words[i].charAt(0)) + words[i].subSequence(1, words[i].length()).toString().toLowerCase());
            }
        }
        return  sb.toString();

    }

其他回答

再举一个例子,如何让用户输入的第一个字母大写:

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String string = br.readLine();
// handle supplementary characters
IntStream.concat(
        IntStream.of(string.codePointAt(0))
                .map(Character::toUpperCase), string.codePoints().skip(1)
)
.forEach(cp -> System.out.print(Character.toChars(cp)));

为了避免异常(IndexOutOfBoundsException或NullPointerException当使用子字符串(0,1)为空或空字符串时),您可以使用regex ("^.") (自Java 9起):

    try(BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) {
        String name = reader.readLine();

        name = Pattern.compile("^.")    // regex for the first character of a string
                .matcher(name)
                .replaceFirst(matchResult -> matchResult.group().toUpperCase());

        System.out.println(name);
    } catch(IOException ignore) {}

你可以试试这个

/**
 * capitilizeFirst(null)  -> ""
 * capitilizeFirst("")    -> ""
 * capitilizeFirst("   ") -> ""
 * capitilizeFirst(" df") -> "Df"
 * capitilizeFirst("AS")  -> "As"
 *
 * @param str input string
 * @return String with the first letter capitalized
 */
public String capitilizeFirst(String str)
{
    // assumptions that input parameter is not null is legal, as we use this function in map chain
    Function<String, String> capFirst = (String s) -> {
        String result = ""; // <-- accumulator

        try { result += s.substring(0, 1).toUpperCase(); }
        catch (Throwable e) {}
        try { result += s.substring(1).toLowerCase(); }
        catch (Throwable e) {}

        return result;
    };

    return Optional.ofNullable(str)
            .map(String::trim)
            .map(capFirst)
            .orElse("");
}

其中一个答案的正确率是95%,但它在我的unitTest中失败了@Ameen Maheen的解决方案几乎完美。除了在输入被转换为String数组之前,您必须修剪输入。所以最完美的一个:

private String convertStringToName(String name) {
        name = name.trim();
        String[] words = name.split(" ");
        StringBuilder sb = new StringBuilder();
        if (words[0].length() > 0) {
            sb.append(Character.toUpperCase(words[0].charAt(0)) + words[0].subSequence(1, words[0].length()).toString().toLowerCase());
            for (int i = 1; i < words.length; i++) {
                sb.append(" ");
                sb.append(Character.toUpperCase(words[i].charAt(0)) + words[i].subSequence(1, words[i].length()).toString().toLowerCase());
            }
        }
        return sb.toString();
    }

因为您第一次从原始字符串中获得Char。你不能在char上使用String属性,所以先使用to upper,然后使用charAt

String s1 = name.toUppercase().charAt(0);