我更喜欢这样的代码:

public final class StrMgr {

    public static String rightPad(String input, int length, String fill){                   
        String pad = input.trim() + String.format("%"+length+"s", "").replace(" ", fill);
        return pad.substring(0, length);              
    }       

    public static String leftPad(String input, int length, String fill){            
        String pad = String.format("%"+length+"s", "").replace(" ", fill) + input.trim();
        return pad.substring(pad.length() - length, pad.length());
    }
}

然后:

System.out.println(StrMgr.leftPad("hello", 20, "x")); 
System.out.println(StrMgr.rightPad("hello", 20, "x"));

String str = "129018";
StringBuilder sb = new StringBuilder();

for (int toPrepend=10-str.length(); toPrepend>0; toPrepend--) {
    sb.append('0');
}

sb.append(str);
String result = sb.toString();

老问题了，但我也有两个方法。

对于固定(预定义的)长度:

    public static String fill(String text) {
        if (text.length() >= 10)
            return text;
        else
            return "0000000000".substring(text.length()) + text;
    }

对于可变长度:

    public static String fill(String text, int size) {
        StringBuilder builder = new StringBuilder(text);
        while (builder.length() < size) {
            builder.append('0');
        }
        return builder.toString();
    }

    int number = -1;
    int holdingDigits = 7;
    System.out.println(String.format("%0"+ holdingDigits +"d", number));

刚刚在采访中问过这个问题........

我的答案如下，但这个(上面提到的)要好得多——>

管柱。format(“% 05d”,全国矿工工会);

我的回答是:

static String leadingZeros(int num, int digitSize) {
    //test for capacity being too small.

    if (digitSize < String.valueOf(num).length()) {
        return "Error : you number  " + num + " is higher than the decimal system specified capacity of " + digitSize + " zeros.";

        //test for capacity will exactly hold the number.
    } else if (digitSize == String.valueOf(num).length()) {
        return String.valueOf(num);

        //else do something here to calculate if the digitSize will over flow the StringBuilder buffer java.lang.OutOfMemoryError 

        //else calculate and return string
    } else {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < digitSize; i++) {
            sb.append("0");
        }
        sb.append(String.valueOf(num));
        return sb.substring(sb.length() - digitSize, sb.length());
    }
}

我更喜欢这样的代码:

public final class StrMgr {

    public static String rightPad(String input, int length, String fill){                   
        String pad = input.trim() + String.format("%"+length+"s", "").replace(" ", fill);
        return pad.substring(0, length);              
    }       

    public static String leftPad(String input, int length, String fill){            
        String pad = String.format("%"+length+"s", "").replace(" ", fill) + input.trim();
        return pad.substring(pad.length() - length, pad.length());
    }
}

然后:

System.out.println(StrMgr.leftPad("hello", 20, "x")); 
System.out.println(StrMgr.rightPad("hello", 20, "x"));

如果你的字符串只包含数字，你可以让它成为一个整数，然后做填充:

String.format("%010d", Integer.parseInt(mystring));

如果没有，我想知道如何才能做到。