字符串首字母大写的更短/更快的版本代码是:

String name  = "stackoverflow"; 
name = name.substring(0,1).toUpperCase() + name.substring(1).toLowerCase();

name的值为Stackoverflow

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();
    }

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("");
}

给出的答案是只大写一个单词的第一个字母。使用以下代码将整个字符串大写。

public static void main(String[] args) {
    String str = "this is a random string";
    StringBuilder capitalizedString = new StringBuilder();
    String[] splited = str.trim().split("\\s+");

    for (String string : splited) {         
        String s1 = string.substring(0, 1).toUpperCase();
        String nameCapitalized = s1 + string.substring(1);

        capitalizedString.append(nameCapitalized);
        capitalizedString.append(" ");
    }
    System.out.println(capitalizedString.toString().trim());
}

输出: 这是一个随机字符串

这只是为了告诉你，你没有错。

BufferedReader br = new InputstreamReader(System.in);
// Assuming name is not blank
String name = br.readLine(); 

//No more error telling that you cant convert char to string
String s1 = (""+name.charAt(0)).toUppercase());
// Or, as Carlos prefers. See the comments to this post.
String s1 = Character.toString(name.charAt(0)).toUppercase());

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

注意:这根本不是最好的方法。这只是为了告诉OP，它也可以使用charAt()来完成。；）

使用common .lang. stringutils，最好的答案是:

public static String capitalize(String str) {  
    int strLen;  
    return str != null && (strLen = str.length()) != 0 ? (new StringBuffer(strLen)).append(Character.toTitleCase(str.charAt(0))).append(str.substring(1)).toString() : str;  
}

我发现它很聪明，因为它用StringBuffer包装字符串。您可以随心所欲地操作StringBuffer，尽管使用相同的实例。