如何在 Java 中填充字符串？
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How can I pad a String in Java?
提问by pvgoddijn
Is there some easy way to pad Strings in Java?
有没有一些简单的方法可以在 Java 中填充字符串？
Seems like something that should be in some StringUtil-like API, but I can't find anything that does this.
似乎应该在一些类似 StringUtil 的 API 中，但我找不到任何可以做到这一点的东西。
采纳答案by GaryF
Apache StringUtilshas several methods: leftPad, rightPad, centerand repeat.
阿帕奇StringUtils有几种方法：leftPad，rightPad，center和repeat。
But please note that — as others have mentioned and demonstrated in this answer— String.format()and the Formatterclasses in the JDK are better options. Use them over the commons code.
但请注意-如其他人所说的，并证明了这个答案-String.format()和FormatterJDK中的类是更好的选择。在公共代码上使用它们。
回答by Miserable Variable
Besides Apache Commons, also see String.formatwhich should be able to take care of simple padding (e.g. with spaces).
除了 Apache Commons，还要看看String.format哪些应该能够处理简单的填充（例如空格）。
回答by Arne Burmeister
Have a look at org.apache.commons.lang.StringUtils#rightPad(String str, int size, char padChar).
看看org.apache.commons.lang.StringUtils#rightPad(String str, int size, char padChar)。
But the algorithm is very simple (pad right up to size chars):
但是算法非常简单（填充到大小字符）：
public String pad(String str, int size, char padChar)
{
StringBuffer padded = new StringBuffer(str);
while (padded.length() < size)
{
padded.append(padChar);
}
return padded.toString();
}
回答by joel.neely
You can reduce the per-call overhead by retaining the padding data, rather than rebuilding it every time:
您可以通过保留填充数据来减少每次调用的开销，而不是每次都重建它：
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);
}
}
As an alternative, you can make the result length a parameter to the pad(...)method. In that case do the adjustment of the hidden padding in that method instead of in the constructor.
作为替代方法，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。
(Hint: For extra credit, make it thread-safe! ;-)
（提示：为了额外的功劳，让它线程安全！;-)
回答by Tom Hawtin - tackline
java.util.Formatterwill do left and right padding. No need for odd third party dependencies (would you want to add them for something so trivial).
java.util.Formatter将做左右填充。不需要奇怪的第三方依赖项（您是否想为如此微不足道的东西添加它们）。
[I've left out the details and made this post 'community wiki' as it is not something I have a need for.]
[我忽略了细节并将这篇文章称为“社区维基”，因为它不是我需要的东西。]
回答by RealHowTo
Since Java 1.5, String.format()can be used to left/right pad a given string.
从 Java 1.5 开始，String.format()可用于左/右填充给定的字符串。
public static String padRight(String s, int n) {
return String.format("%-" + n + "s", s);
}
public static String padLeft(String s, int n) {
return String.format("%" + n + "s", s);
}
...
public static void main(String args[]) throws Exception {
System.out.println(padRight("Howto", 20) + "*");
System.out.println(padLeft("Howto", 20) + "*");
}
And the output is:
输出是：
Howto *
Howto*
回答by ef_oren
you can use the built in StringBuilder append() and insert() methods, for padding of variable string lengths:
您可以使用内置的 StringBuilder append() 和 insert() 方法来填充可变字符串长度：
AbstractStringBuilder append(CharSequence s, int start, int end) ;
For Example:
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());
回答by ck.
i know this thread is kind of old and the original question was for an easy solution but if it's supposed to be really fast, you should use a char array.
我知道这个线程有点老，最初的问题是为了一个简单的解决方案，但如果它应该非常快，你应该使用一个字符数组。
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;
}
the formatter solution is not optimal. just building the format string creates 2 new strings.
格式化程序解决方案不是最佳的。只需构建格式字符串即可创建 2 个新字符串。
apache's solution can be improved by initializing the sb with the target size so replacing below
可以通过使用目标大小初始化 sb 来改进 apache 的解决方案，因此在下面替换
StringBuffer padded = new StringBuffer(str);
with
和
StringBuffer padded = new StringBuffer(pad);
padded.append(value);
would prevent the sb's internal buffer from growing.
会阻止某人的内部缓冲区增长。
回答by Nthalk
This took me a little while to figure out. The real key is to read that Formatter documentation.
这花了我一点时间才弄明白。真正的关键是阅读格式化程序文档。
// 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$][flags][width]conversion
// Conversion: 'x', 'X' integral The result is formatted as a hexadecimal integer
"%1x",
digest
);
// Replace the blank padding with 0s.
hex = hex.replace(" ","0");
System.out.println(hex);
回答by Sebastian S.
This works:
这有效：
"".format("%1$-" + 9 + "s", "XXX").replaceAll(" ", "0")
It will fill your String XXX up to 9 Chars with a whitespace. After that all Whitespaces will be replaced with a 0. You can change the whitespace and the 0 to whatever you want...
它将用空格填充最多 9 个字符的 String XXX。之后，所有空格都将替换为 0。您可以将空格和 0 更改为您想要的任何内容...
