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
  *********

一个简单的解决方案是:

package nl;
public class Padder {
    public static void main(String[] args) {
        String s = "123" ;
        System.out.println("#"+("     " + s).substring(s.length())+"#");
    }
}

不管怎样，我一直在寻找一些可以填充的东西，然后我决定自己编写代码。它非常简洁，你可以很容易地从中推导出padLeft和padRight

    /**
     * Pads around a string, both left and right using pad as the template, aligning to the right or left as indicated.
     * @param a the string to pad on both left and right
     * @param pad the template to pad with, it can be of any size
     * @param width the fixed width to output
     * @param alignRight if true, when the input string is of odd length, adds an extra pad char to the left, so values are right aligned
     *                   otherwise add an extra pad char to the right. When the input is of even length no extra chars will be inserted
     * @return the input param a padded around.
     */
    public static String padAround(String a, String pad, int width, boolean alignRight) {
        if (pad.length() == 0)
            throw new IllegalArgumentException("Pad cannot be an empty string!");
        int delta = width - a.length();
        if (delta < 1)
            return a;
        int half = delta / 2;
        int remainder = delta % 2;
        String padding = pad.repeat(((half+remainder)/pad.length()+1)); // repeating the padding to occupy all possible space
        StringBuilder sb = new StringBuilder(width);
//        sb.append( padding.substring(0,half + (alignRight ? 0 : remainder)));
        sb.append(padding, 0, half + (alignRight ? 0 : remainder));
        sb.append(a);
//        sb.append( padding.substring(0,half + (alignRight ? remainder : 0)));
        sb.append(padding, 0, half + (alignRight ? remainder : 0));

        return sb.toString();
    }

虽然它应该是相当快的，它可能会受益于使用一些韵母在这里和那里。

如此:

"".format("%1$-" + 9 + "s", "XXX").replaceAll(" ", "0")

它会用空白填充你的字符串XXX，最多9个字符。在此之后，所有空格将被替换为0。你可以把空格和0改为任何你想要的…

在番石榴中，这很简单:

Strings.padStart("string", 10, ' ');
Strings.padEnd("string", 10, ' ');

让我给一些情况下的答案，你需要给左/右填充(或前缀/后缀字符串或空格)在你连接到另一个字符串之前，你不想测试长度或任何if条件。

与所选答案相同，我更喜欢Apache Commons的StringUtils，但使用这种方式:

StringUtils.defaultString(StringUtils.leftPad(myString, 1))

解释:

myString:我输入的字符串，可以为空 stringutil的。leftPad(myString, 1):如果string为空，此语句也将返回null 然后使用defaultString给出空字符串，以防止连接null