在 Java 中逐字地将字符串包裹在多个字符之后

Wrap the string after a number of characters word-wise in Java

提问by Suvonkar

I have this code:

我有这个代码：

    String s = "A very long string containing " +
                   "many many words and characters. " +
                   "Newlines will be entered at spaces.";

    StringBuilder sb = new StringBuilder(s);

    int i = 0;
    while ((i = sb.indexOf(" ", i + 20)) != -1) {
        sb.replace(i, i + 1, "\n");
    }

    System.out.println(sb.toString());

The output of the code is:

代码的输出是：

A very long string containing
many many words and
characters. Newlines
will be entered at spaces.

The above code is wrapping the string after the next space of every 30 characters, but I need to wrap the string after the previous space of every 30 characters, like for the first line it will be:

上面的代码是在每 30 个字符的下一个空格之后包装字符串，但我需要在每 30 个字符的前一个空格之后包装字符串，就像第一行一样：

A very long string

A very long string

And the 2nd line will be

第二行将是

containing many

Please give some proper solution.

请给出一些适当的解决方案。

采纳答案by Emil

You can use Apache-common'sWordUtils.wrap().

您可以使用Apache-common 的WordUtils.wrap()。

回答by vitaut

Use lastIndexOfinstead of indexOf, e.g.

使用lastIndexOf代替indexOf，例如

StringBuilder sb = new StringBuilder(s);

int i = 0;
while (i + 20 < sb.length() && (i = sb.lastIndexOf(" ", i + 20)) != -1) {
    sb.replace(i, i + 1, "\n");
}

System.out.println(sb.toString());

This will produce the following output:

这将产生以下输出：

A very long string
containing many
many words and
characters.
Newlines will be
entered at spaces.

回答by Yalovali

You can try the following:

您可以尝试以下操作：

public static String wrapString(String s, String deliminator, int length) {
    String result = "";
    int lastdelimPos = 0;
    for (String token : s.split(" ", -1)) {
        if (result.length() - lastdelimPos + token.length() > length) {
            result = result + deliminator + token;
            lastdelimPos = result.length() + 1;
        }
        else {
            result += (result.isEmpty() ? "" : " ") + token;
        }
    }
    return result;
}

call as wrapString("asd xyz afz","\n",5)

调用 wrapString("asd xyz afz","\n",5)

回答by Sainath Patwary karnate

public static void main(String args[]) {

         String s1="This is my world. This has to be broken.";
         StringBuffer buffer=new StringBuffer();

         int length=s1.length();
         int thrshld=5; //this valueis threshold , which you can use 
         int a=length/thrshld;

         if (a<=1) {
             System.out.println(s1);
         }else{
            String split[]=s1.split(" ");
            for (int j = 0; j < split.length; j++) {
                buffer.append(split[j]+" "); 

                if (buffer.length()>=thrshld) { 

                    int lastindex=buffer.lastIndexOf(" ");

                    if (lastindex<buffer.length()) { 

                        buffer.subSequence(lastindex, buffer.length()-1);
                        System.out.println(buffer.toString()); 
                        buffer=null;
                        buffer=new StringBuffer();
                    }
                }
            }
         }
     }

this can be one way to achieve

这可能是实现的一种方式

回答by TimWalter

"\n" makes a wordwrap.

"\n" 进行自动换行。

String s = "A very long string containing \n" +  
"many many words and characters. \n" +
"Newlines will be entered at spaces.";

this will solve your problem

这将解决您的问题

回答by Aaron Bar

I know it's an old question, but . . . Based on another answer I found here, but can't remember the posters name. Kuddos to him/her for pointing me in the right direction.

我知道这是一个老问题，但是。. . 根据我在此处找到的另一个答案，但不记得海报名称。感谢他/她为我指明了正确的方向。

    public String truncate(final String content, final int lastIndex) {
        String result = "";
        String retResult = "";
        //Check for empty so we don't throw null pointer exception
        if (!TextUtils.isEmpty(content)) {
            result = content.substring(0, lastIndex);
            if (content.charAt(lastIndex) != ' ') {
                //Try the split, but catch OutOfBounds in case string is an
                //uninterrupted string with no spaces
                try {
                    result = result.substring(0, result.lastIndexOf(" "));
                } catch (StringIndexOutOfBoundsException e) {
                    //if no spaces, force a break
                    result = content.substring(0, lastIndex);
                }
                //See if we need to repeat the process again
                if (content.length() - result.length() > lastIndex) {
                    retResult = truncate(content.substring(result.length(), content.length()), lastIndex);
                } else {
                    return result.concat("\n").concat(content.substring(result.length(), content.length()));
                }
            }
            //Return the result concatenating a newline character on the end
            return result.concat("\n").concat(retResult);;
            //May need to use this depending on your app
            //return result.concat("\r\n").concat(retResult);;
        } else {
            return content;
        }
    }