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。
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
Java联机程序,没有花哨的库。
// 6 characters padding example
String pad = "******";
// testcases for 0, 4, 8 characters
String input = "" | "abcd" | "abcdefgh"
左Pad,不要限制
result = pad.substring(Math.min(input.length(),pad.length())) + input;
results: "******" | "**abcd" | "abcdefgh"
右移,不要限制
result = input + pad.substring(Math.min(input.length(),pad.length()));
results: "******" | "abcd**" | "abcdefgh"
左衬垫,限制衬垫长度
result = (pad + input).substring(input.length(), input.length() + pad.length());
results: "******" | "**abcd" | "cdefgh"
右垫,限制垫的长度
result = (input + pad).substring(0, pad.length());
results: "******" | "abcd**" | "abcdef"
其他回答
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...
使用该函数。
private String leftPadding(String word, int length, char ch) {
return (length > word.length()) ? leftPadding(ch + word, length, ch) : word;
}
如何使用?
leftPadding(month, 2, '0');
输出: 01 02 03 04 ..11日12
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));
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());
@ck和@Marlon Tarak的答案是唯一使用char[]的答案,对于每秒有几个填充方法调用的应用程序来说,这是最好的方法。然而,它们没有利用任何数组操作优化,而且对我来说有点覆盖;这完全不需要循环。
public static String pad(String source, char fill, int length, boolean right){
if(source.length() > length) return source;
char[] out = new char[length];
if(right){
System.arraycopy(source.toCharArray(), 0, out, 0, source.length());
Arrays.fill(out, source.length(), length, fill);
}else{
int sourceOffset = length - source.length();
System.arraycopy(source.toCharArray(), 0, out, sourceOffset, source.length());
Arrays.fill(out, 0, sourceOffset, fill);
}
return new String(out);
}
简单测试方法:
public static void main(String... args){
System.out.println("012345678901234567890123456789");
System.out.println(pad("cats", ' ', 30, true));
System.out.println(pad("cats", ' ', 30, false));
System.out.println(pad("cats", ' ', 20, false));
System.out.println(pad("cats", '$', 30, true));
System.out.println(pad("too long for your own good, buddy", '#', 30, true));
}
输出:
012345678901234567890123456789
cats
cats
cats
cats$$$$$$$$$$$$$$$$$$$$$$$$$$
too long for your own good, buddy