填充到10个字符:

String.format("%10s", "foo").replace(' ', '*');
String.format("%-10s", "bar").replace(' ', '*');
String.format("%10s", "longer than 10 chars").replace(' ', '*');

输出:

  *******foo
  bar*******
  longer*than*10*chars

密码字符显示“*”:

String password = "secret123";
String padded = String.format("%"+password.length()+"s", "").replace(' ', '*');

输出与密码字符串长度相同:

  secret123
  *********

你可以使用内置的StringBuilder append()和insert()方法， 对于可变字符串长度的填充:

AbstractStringBuilder append(CharSequence s, int start, int end) ;

例如:

private static final String  MAX_STRING = "                    "; //20 spaces

    Set<StringBuilder> set= new HashSet<StringBuilder>();
    set.add(new StringBuilder("12345678"));
    set.add(new StringBuilder("123456789"));
    set.add(new StringBuilder("1234567811"));
    set.add(new StringBuilder("12345678123"));
    set.add(new StringBuilder("1234567812234"));
    set.add(new StringBuilder("1234567812222"));
    set.add(new StringBuilder("12345678122334"));

    for(StringBuilder padMe: set)
        padMe.append(MAX_STRING, padMe.length(), MAX_STRING.length());

简单的东西:

该值应该是字符串。如果不是，就转换成字符串。比如"" + 123或Integer.toString(123)

// let's assume value holds the String we want to pad
String value = "123";

子字符串从值length char索引开始，直到填充的结束长度:

String padded="00000000".substring(value.length()) + value;

// now padded is "00000123"

更精确的

垫:

String padded = value + ("ABCDEFGH".substring(value.length())); 

// now padded is "123DEFGH"

垫左:

String padString = "ABCDEFGH";
String padded = (padString.substring(0, padString.length() - value.length())) + value;

// now padded is "ABCDE123"

你可以通过保留填充数据来减少每次调用的开销，而不是每次都重新构建:

public class RightPadder {

    private int length;
    private String padding;

    public RightPadder(int length, String pad) {
        this.length = length;
        StringBuilder sb = new StringBuilder(pad);
        while (sb.length() < length) {
            sb.append(sb);
        }
        padding = sb.toString();
   }

    public String pad(String s) {
        return (s.length() < length ? s + padding : s).substring(0, length);
    }

}

作为一种替代方法，您可以将结果长度作为pad(…)方法的参数。在这种情况下，在该方法中而不是在构造函数中调整隐藏填充。

(提示:为了获得额外的学分，让它是线程安全的!: -)

public static String LPad(String str, Integer length, char car) {
  return (str + String.format("%" + length + "s", "").replace(" ", String.valueOf(car))).substring(0, length);
}

public static String RPad(String str, Integer length, char car) {
  return (String.format("%" + length + "s", "").replace(" ", String.valueOf(car)) + str).substring(str.length(), length + str.length());
}

LPad("Hi", 10, 'R') //gives "RRRRRRRRHi"
RPad("Hi", 10, 'R') //gives "HiRRRRRRRR"
RPad("Hi", 10, ' ') //gives "Hi        "
RPad("Hi", 1, ' ')  //gives "H"
//etc...

为此编写自己的函数可能比其他答案更简单。

private static String padRight(String str, String padChar, int n) {

    String paddedString = str;

    while (paddedString.length() < n) {

        paddedString = padChar + str;

    }

    return paddedString;

}

private static String padLeft(String str, String padChar, int n) {

    String paddedString = str;

    while (paddedString.length() < n) {

        paddedString += padChar;

    }

    return paddedString;

}