如何截断一定长度的字符串但在截断后包含完整的单词

Question

我想从字符串中截断最多 60 个字符的子字符串，但也想在子字符串中获取完整的单词。这是我正在尝试的。

String originalText =" Bangladesh's first day of Test cricket on Indian soil has not been a good one. They end the day having conceded 71 runs in the last 10 overs, which meant they are already staring at a total of 356. M Vijay was solid and languid as he made his ninth Test century and third of the season. ";
String afterOptimized=originalText.substring(0, 60);
System.out.println("This is text . "+afterOptimized);

这是输出

This is text .  Bangladesh's first day of Test cricket on Indian soil has n

但是我的要求是不要把中间的单词删掉。我怎么知道60个字符之后有没有完整的单词。

Answer 1

您可以为此使用正则表达式，最多占用 60 个字符并在单词边界处结束：

Pattern pattern = Pattern.compile("(.{1,60})(\\b|$)(.*)");
Matcher m = pattern.match(originalText);
If (m.matches())
    afterOptimized = m.group(1);

或者，在一个循环中：

Pattern pattern = Pattern.compile("\\s*(.{1,60})(\\b|$)");
Matcher m = pattern.matcher(originalText);
int last = 0;
while (m.find()) {
    System.out.println(m.group(1));
    last = m.end();
}
if (last != originalText.length())
    System.out.println(originalText.substring(last));

如果您只想在空格而不是单词边界处换行（可能在逗号、点等之前换行），您可能需要将 \b 替换为 \s。

Answer 2

如果原始字符串在位置 60（第 61 个字符）有一个字符，表示您要剪切一个单词，或者一个单词正在开始，则从位置 59（第 60 个字符）开始搜索并停止，当您找个空间。然后我们可以在该位置对字符串进行子串化。如果字符串不超过 60 个字符，我们将原样返回。

public void truncateTest() {
    System.out.println(truncateTo("Bangladesh's first day of Test cricket on Indian soil has not been a good one. They end the day having conceded 71 runs in the last 10 overs, which meant they are already staring at a total of 356. M Vijay was solid and languid as he made his ninth Test century and third of the season. ", 60));
    System.out.println(truncateTo("Bangladesh's first day.", 60));
    System.out.println(truncateTo("They end the day having conceded 71 runs in the last 10 overs, which meant they are already staring at a total of 356. M Vijay was solid and languid as he made his ninth Test century and third of the season.", 60));
}

public String truncateTo(String originalText, int len) {
    if (originalText.length() > len) {
        if (originalText.charAt(len) != ' ') {
            for (int x=len-1;x>=0;x--) {
                if (Character.isWhitespace(originalText.charAt(x))) {
                    return originalText.substring(0, x);
                }
            }
        }
        // default if none of the conditions are met
        return originalText.substring(0, len);
    }
    return originalText;
}

结果...

Bangladesh's first day of Test cricket on Indian soil has
Bangladesh's first day.
They end the day having conceded 71 runs in the last 10

我认为我的 +1 / -1 索引逻辑是正确的 :)

总结印度的击球，Pujara 是耐心的缩影，Vijay 的投篮令人蔑视，队长 Kohli 以完全不屑的方式结束了比赛，结果证明印度队完全控制了比赛。

Answer 3

String originalText=" 孟加拉国在印度土地上测试板球的第一天并不是一个好日子。他们在过去 10 轮比赛中丢了 71 球，这意味着他们已经盯着总共 356 球。M Vijay 在他的第九个测试世纪和本赛季的第三个赛季中表现稳定而慵懒。”;

//trim the string to 60 characters

String  trimmedString = originalText.substring(0, 60);

//re-trim if we are in the middle of a word and to get full word instead of brolken one

String result=trimmedString.substring(0, Math.min(trimmedString.length(), trimmedString.lastIndexOf(" ")));

System.out.println(result);

Answer 4

 int cutoff = originalText.substring(0,60).lastIndexOf(" ");
 String afterOptimized = originalText.substring(0, cutoff);

代码打印如下：“Bangladesh's first day of Test cricket on Indian soil has”

Answer 5

假设您的文本在两个单词之间有空格，只需剪切文本并检查 char 的结尾 + 在 end char 之前 + 在 end char 之后以确定我们需要剪切的内容：

if (char[i] != ' ') {
    if(i+1 == length || (i+1 < length && char[i+1] == ' '))
         return mString; // [I'm loser] bla ==> [I'm loser]
    if(i-1 > -1 && char[i-1] == ' ')
         return subHeadString(mString, 2); // return mString which has length = length - 2, ex: [I'm loser b]la ==> [I'm loser]
    return findBackStringWithSpace(mString, i); // coming back until has space char and return that sub string 
// [I'm loser bl]a ==> [I'm loser] 
} else {
    return mString;
}