在第 10 个字符之前的空白处拆分数组中的字符串

Question

我想将一个字符串拆分为 10 个字符。但我不想拆分为一个单词。

因此字符串“还有九个字符 - 然后还有更多”将被拆分为 ["Nine", "characters", "to go -", "then some", "more"]。
超过 10 个字符的单词可以拆分。

我使用正则表达式最接近的是.{1,10}(?<=\s)。
这会将“九个字符-然后更多”拆分为["Nine ", "haracters ", "to go - ", "then some "]。
但行为很奇怪。在示例字符串中，它完全跳过了字符“c”。在其他测试字符串中，它只会在单独的数组项中添加“-”，而只有破折号适合它之前的数组项。所以它在空格之后分裂。

我还尝试在空格 (.split(' ')) 上使用 .split()，并使用 .reduce() 或 for 循环将数组项连接到其他最多 10 个字符的数组项中。

for ( i = 0; i < splitArray.length; i++ ) {
  if ( i === 0 ) {
    // add first word in new array. Doesn't take into account yet that word can be longer than 10 characters
    newArray.push( splitArray[i] );
  } else {
    if ( newArray[ newArray.length - 1 ].length + splitArray[i].length + 1 < 10 ) {
      // if next word fits with the word in the new array (taking space into account), add it to it
      newArray[ newArray.length - 1 ] = newArray[ newArray.length - 1 ] + " " + splitArray[i];
    } else if ( newArray[ newArray.length - 1 ].length + splitArray[i].length + 1 >= 10 ) {
      // next word doesn't fit
      // split word and add only part to it and add the rest in separate item in newArray
      const index = 9 - newArray[ newArray.length - 1 ].length
      const prev = splitArray[i].slice( 0, index );
      const next = splitArray[i].slice( index, splitArray[i].length );
      newArray[ newArray.length - 1 ] = newArray[ newArray.length - 1 ] + " " + prev;
      newArray.push( next );
    } else {
      // push new item in newArray
      newArray.push( splitArray[i] );
    }
  }
}

结果：["Nine chara", "cters to g", "o - then s", "ome more"].
没有else if：["Nine", "characters", "to go -", "then some", "more"]
没有else if 其他字符串：["Paul van", "den Dool", "-", "Alphabet", "- word"]
这很接近，但“字母”不会与连字符连接，因为它们不适合。我尝试用else if 声明来捕捉它，但这又破坏了我不想破坏的单词，并且与上面的正则表达式的结果相同。

我在这个问题上已经筋疲力尽了，我需要蜂巢思维来解决这个问题。因此，非常感谢您对此提供任何帮助。

上下文
我正在尝试在具有最小字体大小的有限大小框中显示画布上的文本。我的解决方案是在必要时将用户可以输入的字符串分成多行。为此，我需要将字符串拆分为一个数组，对其进行循环并相应地定位文本。

Answer 1

使用

console.log(
  "Nine characters to go - then some more"
     .match(/.{1,10}(?=\s|$)/g)
     .map(z => z.trim())
);

使用.match(/.{1,10}(?=\s|$)/g)，项目长度为 1 到 10 个字符，(?=\s|$) 将确保匹配空格或字符串结尾。

Answer 2

const string = "Nine characters to go - then some more"
let arr = string.split(" ");
for(let i = 1; i < arr.length; i++) {
  if(arr[i].length >= 10 || arr[i].length + arr[i-1].length >= 10) {
     continue;
  }
  if(arr[i].length < 10 && arr[i].length + arr[i-1].length <= 10) {
    arr[i] = arr[i - 1] + " " + arr[i];
    arr[i-1] = false;
  }

}
arr = arr.filter(string => string)

console.log(arr);

Answer 3

如果需要拆分，使用.split():

const str = 'Nine characters to go - then some more',
      
      result = str.split(/(.{1,10})\s/).filter(Boolean)
      
console.log(result)

Answer 4

我可以使用一个简单的for 循环来解决这个问题：

const str = "Nine characters to go - then some more";

// make an array with each words
const arr = str.trim().split(' ');

// This is the length, how much we want to take the length of the words
const length = 10;
const res = [];

/**
 * Put the first word into the result array
 * because if the word greater or less than
 * the `length` we have to take it.
 */
res.push(arr[0]);

// Result array's current index
let index = 0;

for (let i = 1, l = arr.length; i < l; i++) {
  /**
   * If the length of the concatenation of the
   * last word of the result array
   * and the next word is less than or equal to the length
   * then concat them and put them as the last value
   * of the resulting array.
   */
  if ((res[index] + arr[i]).length <= length) {
    res[index] += ' ' + arr[i];
  } else {
    /**
     * Otherwise push the current word
     * into the resulting array
     * and increase the last index of the
     * resulting array.
     */
    res.push(arr[i]);
    index++;
  }
}

console.log(res);

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