将char放入一个java字符串中，每N个字符
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Putting char into a java string for each N characters
提问by Giancarlo
I have a java string, which has a variable length.
我有一个长度可变的java字符串。
I need to put the piece "<br>"into the string, say each 10 characters.
我需要把这件作品"<br>"放入字符串中，比如每 10 个字符。
For example this is my string:
例如，这是我的字符串：
`this is my string which I need to modify...I love stackoverlow:)`
How can I obtain this string?:
我怎样才能获得这个字符串？：
`this is my<br> string wh<br>ich I nee<br>d to modif<br>y...I love<br> stackover<br>flow:)`
Thanks
谢谢
采纳答案by Jon Skeet
Something like:
就像是：
public static String insertPeriodically(
String text, String insert, int period)
{
StringBuilder builder = new StringBuilder(
text.length() + insert.length() * (text.length()/period)+1);
int index = 0;
String prefix = "";
while (index < text.length())
{
// Don't put the insert in the very first iteration.
// This is easier than appending it *after* each substring
builder.append(prefix);
prefix = insert;
builder.append(text.substring(index,
Math.min(index + period, text.length())));
index += period;
}
return builder.toString();
}
回答by cletus
Try:
尝试：
String s = // long string
s.replaceAll("(.{10})", "<br>");
EDIT:The above works... most of the time. I've been playing around with it and came across a problem: since it constructs a default Pattern internally it halts on newlines. to get around this you have to write it differently.
编辑：以上工作......大部分时间。我一直在玩它并遇到一个问题：因为它在内部构造了一个默认模式，所以它在换行符处停止。要解决这个问题，您必须以不同的方式编写它。
public static String insert(String text, String insert, int period) {
Pattern p = Pattern.compile("(.{" + period + "})", Pattern.DOTALL);
Matcher m = p.matcher(text);
return m.replaceAll("" + insert);
}
and the astute reader will pick up on another problem: you have to escape regex special characters (like "$1") in the replacement text or you'll get unpredictable results.
精明的读者会发现另一个问题：您必须在替换文本中转义正则表达式特殊字符（如“$1”），否则您将得到不可预测的结果。
I also got curious and benchmarked this version against Jon's above. This one is slower by an order of magnitude (1000 replacements on a 60k file took 4.5 seconds with this, 400ms with his). Of the 4.5 seconds, only about 0.7 seconds was actually constructing the Pattern. Most of it was on the matching/replacement so it doesn't even ledn itself to that kind of optimization.
我也很好奇并根据上面的 Jon 对这个版本进行了基准测试。这个要慢一个数量级（60k 文件上的 1000 次替换需要 4.5 秒，他需要 400 毫秒）。在这 4.5 秒中，只有大约 0.7 秒实际构建了模式。大部分是在匹配/替换上，所以它甚至没有导致这种优化。
I normally prefer the less wordy solutions to things. After all, more code = more potential bugs. But in this case I must concede that Jon's version--which is really the naive implementation (I mean that in a good way)--is significantly better.
我通常更喜欢不那么冗长的解决方案。毕竟，更多的代码 = 更多潜在的错误。但在这种情况下，我必须承认 Jon 的版本——这确实是天真的实现（我的意思是在一个很好的方式）——明显更好。
回答by DJClayworth
StringBuilder buf = new StringBuilder();
for (int i = 0; i < myString.length(); i += 10) {
buf.append(myString.substring(i, i + 10);
buf.append("\n");
}
You can get more efficient than that, but I'll leave that as an exercise for the reader.
您可以获得比这更高的效率，但我会将其作为练习留给读者。
回答by Outlaw Programmer
How about 1 method to split the string every N characters:
每隔 N 个字符拆分字符串的 1 种方法如何：
public static String[] split(String source, int n)
{
List<String> results = new ArrayList<String>();
int start = 0;
int end = 10;
while(end < source.length)
{
results.add(source.substring(start, end);
start = end;
end += 10;
}
return results.toArray(new String[results.size()]);
}
Then another method to insert something after each piece:
然后是另一种在每件作品后插入内容的方法：
public static String insertAfterN(String source, int n, String toInsert)
{
StringBuilder result = new StringBuilder();
for(String piece : split(source, n))
{
result.append(piece);
if(piece.length == n)
result.append(toInsert);
}
return result.toString();
}
回答by Outlaw Programmer
To avoid chopping off the words...
为了避免砍掉单词......
Try:
尝试：
int wrapLength = 10;
String wrapString = new String();
String remainString = "The quick brown fox jumps over the lazy dog The quick brown fox jumps over the lazy dog";
while(remainString.length()>wrapLength){
int lastIndex = remainString.lastIndexOf(" ", wrapLength);
wrapString = wrapString.concat(remainString.substring(0, lastIndex));
wrapString = wrapString.concat("\n");
remainString = remainString.substring(lastIndex+1, remainString.length());
}
System.out.println(wrapString);
回答by Dave Jarvis
If you don't mind a dependency on a third-party libraryand don't mind regexes:
import com.google.common.base.Joiner;
/**
* Splits a string into N pieces separated by a delimiter.
*
* @param text The text to split.
* @param delim The string to inject every N pieces.
* @param period The number of pieces (N) to split the text.
* @return The string split into pieces delimited by delim.
*/
public static String split( final String text, final String delim, final int period ) {
final String[] split = text.split("(?<=\G.{"+period+"})");
return Joiner.on(delim).join(split);
}
Then:
然后：
split( "This is my string", "<br/>", 5 );
This doesn't split words at spaces but the question, as stated, doesn't ask for word-wrap.
这不会在空格处拆分单词，但如上所述，问题不要求自动换行。
回答by Qarj
I've unit tested this solution for boundary conditions:
我已经针对边界条件对该解决方案进行了单元测试：
public String padded(String original, int interval, String separator) {
String formatted = "";
for(int i = 0; i < original.length(); i++) {
if (i % interval == 0 && i > 0) {
formatted += separator;
}
formatted += original.substring(i, i+1);
}
return formatted;
}
Call with:
致电：
padded("this is my string which I need to modify...I love stackoverflow:)", 10, "<br>");
回答by Mira_Cole
The following method takes three parameters. The first is the text that you want to modify. The second parameter is the text you want to insert every n characters. The third is the interval in which you want to insert the text at.
下面的方法需要三个参数。第一个是您要修改的文本。第二个参数是你想要每 n 个字符插入的文本。第三个是您想要插入文本的间隔。
private String insertEveryNCharacters(String originalText, String textToInsert, int breakInterval) {
String withBreaks = "";
int textLength = originalText.length(); //initialize this here or in the start of the for in order to evaluate this once, not every loop
for (int i = breakInterval , current = 0; i <= textLength || current < textLength; current = i, i += breakInterval ) {
if(current != 0) { //do not insert the text on the first loop
withBreaks += textToInsert;
}
if(i <= textLength) { //double check that text is at least long enough to go to index i without out of bounds exception
withBreaks += originalText.substring(current, i);
} else { //text left is not longer than the break interval, so go simply from current to end.
withBreaks += originalText.substring(current); //current to end (if text is not perfectly divisible by interval, it will still get included)
}
}
return withBreaks;
}
You would call to this method like this:
你会像这样调用这个方法：
String splitText = insertEveryNCharacters("this is my string which I need to modify...I love stackoverlow:)", "<br>", 10);
The result is:
结果是：
this is my<br> string wh<br>ich I need<br> to modify<br>...I love <br>stackoverl<br>ow:)
^This is different than your example result because you had a set with 9 characters instead of 10 due to human error ;)
^这与您的示例结果不同，因为由于人为错误，您有一个包含 9 个字符而不是 10 个字符的集合；)
回答by Kashif Ibrahim
You can use the regular expression '..' to match each two characters and replace it with "$0 " to add the space:
您可以使用正则表达式 '..' 来匹配每两个字符并将其替换为“$0”以添加空格：
s = s.replaceAll("..", "$0 "); You may also want to trim the result to remove the extra space at the end.
s = s.replaceAll("..", "$0"); 您可能还想修剪结果以删除末尾的额外空间。
Alternatively you can add a negative lookahead assertion to avoid adding the space at the end of the string:
或者，您可以添加否定前瞻断言以避免在字符串末尾添加空格：
s = s.replaceAll("..(?!$)", "$0 ");
s = s.replaceAll("..(?!$)", "$0");
For example:
例如：
String s = "23423412342134";
s = s.replaceAll("....", "$0<br>");
System.out.println(s);
String s = "23423412342134";
s = s.replaceAll("....", "$0<br>");
System.out.println(s);
Output: 2342<br>3412<br>3421<br>34
输出： 2342<br>3412<br>3421<br>34
回答by Vincent Mimoun-Prat
I got here from String Manipulation insert a character every 4th characterand was looking for a solution on Android with Kotlin.
我从String Manipulation来到这里，每 4 个字符插入一个字符，并正在寻找使用 Kotlin 在 Android 上的解决方案。
Just adding a way to do that with Kotlin (you got to love the simplicity of it)
只需使用 Kotlin 添加一种方法即可（您必须喜欢它的简单性）
val original = "123498761234"
val dashed = original.chunked(4).joinToString("-") // 1234-9876-1234
