我是 Java 新手，想根据输入的最大长度将用户输入的消息拆分为多行。我将如何重复它？这是我目前所拥有的：
int rem = m - maxlength;
System.out.println(message.substring(0, message.length() - rem));
System.out.println(message.substring(message.length() - rem));
【问题讨论】：
正则表达式版本（最简单）：
String text = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book.";
int lineMaxLength = 10;
System.out.println(java.util.Arrays.toString(
text.split("(?<=\\G.{"+lineMaxLength+"})")
));
哪些打印：
[Lorem Ipsu, m is simpl, y dummy te, xt of the , printing a, nd typeset, ting indus, try. Lorem, Ipsum has, been the , industry's, standard , dummy text, ever sinc, e the 1500, s, when an, unknown p, rinter too, k a galley, of type a, nd scrambl, ed it to m, ake a type, specimen , book.]
没有正则表达式：
import java.util.List;
import java.util.ArrayList;
public class MyClass {
public static void main(String args[]) {
String text = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book.";
int lineMaxLength = 10;
List<String> lines = new ArrayList<>();
int length = text.length();
StringBuilder s = new StringBuilder();
for(int i = 0; i < length; i++){
if (i % lineMaxLength == 0){
if(i != 0){
lines.add(s.toString());
}
s = new StringBuilder();
}
s.append(text.charAt(i));
}
int linesLength = lines.size();
for(int i= 0; i < linesLength; i++){
System.out.println(lines.get(i));
}
}
}
哪些打印：
Lorem Ipsu
m is simpl
y dummy te
xt of the
printing a
nd typeset
ting indus
try. Lorem
Ipsum has
been the
industry's
standard
dummy text
ever sinc
e the 1500
s, when an
unknown p
rinter too
k a galley
of type a
nd scrambl
ed it to m
ake a type
specimen
【讨论】：
这类作品。我选择了一个非常短的 10 个字符的“行”长度，您可能应该增加它。但它展示了如何在没有太多代码的情况下做到这一点。将正则表达式中的“10”更改为不同的数字以增加行长。
（我稍微改了一下代码，更明显的是这个方法不会在中间分割单词，而是总是在空格处分割。）
public static void main( String[] args ) {
String s = "This is a test. This is a test. This is a test. This is a test. This is a test. This is a test. ";
String regex = ".{1,10}\\s";
Matcher m = Pattern.compile( regex ).matcher( s );
ArrayList<String> lines = new ArrayList<>();
while( m.find() ) {
lines.add( m.group() );
}
System.out.println( String.join( "\n", lines ) );
}
运行：
This is a
test. This
is a test.
This is a
test. This
is a test.
This is a
test. This
is a test.
BUILD SUCCESSFUL (total time: 0 seconds)
【讨论】：
像这样使用 WordIterator。 BreakIterator 类对语言环境敏感。您可以使用不同的语言。
public static void main(String[] args) {
String msg = "This is a long message. Message is too long. Long is the message";
int widthLimit = 15;
printWithLimitedWidth(msg, widthLimit);
}
static void printWithLimitedWidth(String s, int limit) {
BreakIterator br = BreakIterator.getWordInstance(); //you can get locale specific instance to handle different languages.
br.setText(s);
int currLenght = 0;
int start = br.first();
int end = br.next();
while (end != BreakIterator.DONE) {
String word = s.substring(start,end);
currLenght += word.length();
if (currLenght <= limit) {
System.out.print(word);
} else {
currLenght = 0;
System.out.print("\n"+word);
}
start = end;
end = br.next();
}
}
输出：
This is a long
message. Message is
too long. Long is
the message
这是一个示例代码。根据您的需要处理空格和其他特殊字符。 更多信息请参考https://docs.oracle.com/javase/tutorial/i18n/text/about.html
【讨论】：
int maxlength = 10;
String message = "1234567890ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String transformedMessage = message.replaceAll("(?<=\\G.{" + maxlength + "})", "\n");
System.out.println(transformedMessage);
/*Output
1234567890
ABCDEFGHIJ
KLMNOPQRST
UVWXYZ
*/
【讨论】：