有一些简单的方法来填充字符串在Java?

似乎是一些应该在一些stringutil类API,但我找不到任何东西,这样做。


当前回答

这是一个高效的实用工具类,用于Java中的左填充,右填充,中心填充和零填充字符串。

package com.example;

/**
 * Utility class for left pad, right pad, center pad and zero fill.
 */
public final class StringPadding {

    public static String left(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();
            char[] output = new char[length];

            int delta = length - chars.length;

            for (int i = 0; i < length; i++) {
                if (i < delta) {
                    output[i] = fill;
                } else {
                    output[i] = chars[i - delta];
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String right(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();
            char[] output = new char[length];

            for (int i = 0; i < length; i++) {
                if (i < chars.length) {
                    output[i] = chars[i];
                } else {
                    output[i] = fill;
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String center(String string, int length, char fill) {

        if (string.length() < length) {

            char[] chars = string.toCharArray();

            int delta = length - chars.length;
            int a = (delta % 2 == 0) ? delta / 2 : delta / 2 + 1;
            int b = a + chars.length;

            char[] output = new char[length];
            for (int i = 0; i < length; i++) {
                if (i < a) {
                    output[i] = fill;
                } else if (i < b) {
                    output[i] = chars[i - a];
                } else {
                    output[i] = fill;
                }
            }

            return new String(output);
        }

        return string;
    }

    public static String zerofill(String string, int length) {
        return left(string, length, '0');
    }

    private StringPadding() {
    }

    /**
     * For tests!
     */
    public static void main(String[] args) {

        String string = "123";
        char blank = ' ';

        System.out.println("left pad:    [" + StringPadding.left(string, 10, blank) + "]");
        System.out.println("right pad:   [" + StringPadding.right(string, 10, blank) + "]");
        System.out.println("center pad:  [" + StringPadding.center(string, 10, blank) + "]");
        System.out.println("zero fill:   [" + StringPadding.zerofill(string, 10) + "]");
    }
}

输出如下:

left pad:    [       123]
right pad:   [123       ]
center pad:  [    123   ]
zero fill:   [0000000123]

其他回答

我花了一点时间才想明白。 真正的关键是阅读Formatter文档。

// Get your data from wherever.
final byte[] data = getData();
// Get the digest engine.
final MessageDigest md5= MessageDigest.getInstance("MD5");
// Send your data through it.
md5.update(data);
// Parse the data as a positive BigInteger.
final BigInteger digest = new BigInteger(1,md5.digest());
// Pad the digest with blanks, 32 wide.
String hex = String.format(
    // See: http://download.oracle.com/javase/1.5.0/docs/api/java/util/Formatter.html
    // Format: %[argument_index$][flags][width]conversion
    // Conversion: 'x', 'X'  integral    The result is formatted as a hexadecimal integer
    "%1$32x",
    digest
);
// Replace the blank padding with 0s.
hex = hex.replace(" ","0");
System.out.println(hex);
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...

这里有一个并行版本的你有很长的字符串:-)

int width = 100;
String s = "129018";

CharSequence padded = IntStream.range(0,width)
            .parallel()
            .map(i->i-(width-s.length()))
            .map(i->i<0 ? '0' :s.charAt(i))
            .collect(StringBuilder::new, (sb,c)-> sb.append((char)c), (sb1,sb2)->sb1.append(sb2));

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

我知道这个线程有点老了,最初的问题是为了一个简单的解决方案,但如果它应该是真的很快,你应该使用字符数组。

public static String pad(String str, int size, char padChar)
{
    if (str.length() < size)
    {
        char[] temp = new char[size];
        int i = 0;

        while (i < str.length())
        {
            temp[i] = str.charAt(i);
            i++;
        }

        while (i < size)
        {
            temp[i] = padChar;
            i++;
        }

        str = new String(temp);
    }

    return str;
}

格式化程序解决方案不是最佳的。仅仅构建格式字符串就会创建2个新字符串。

Apache的解决方案可以通过用目标大小初始化sb来改进,从而替换下面的内容

StringBuffer padded = new StringBuffer(str); 

StringBuffer padded = new StringBuffer(pad); 
padded.append(value);

会阻止某人内部缓冲的增长。