对于输入字符串 'this is a sentence',当位置为6或7时,它必须返回 'is'。当位置为0、1、2、3或4时,结果必须为'this'。最简单的方法是什么?
function getWordAt (str, pos) {
// check ranges
if ((pos < 0) || (pos > str.length)) {
return '';
}
// Perform type conversions.
str = String(str);
pos = Number(pos) >>> 0;
// Search for the word's beginning and end.
var left = str.slice(0, pos + 1).search(/\S+$/), // use /\S+\s*$/ to return the preceding word
right = str.slice(pos).search(/\s/);
// The last word in the string is a special case.
if (right < 0) {
return str.slice(left);
}
// Return the word, using the located bounds to extract it from the string.
return str.slice(left, right + pos);
}
/\S+$/
/\s/
""
;因为空格本身不是单词的一部分。如果你希望函数返回前面的单词,可以将/\S+$/
改为/\S+\s*$/
。
"这是一个句子。"
0: This
1: This
2: This
3: This
4:
5: is
6: is
7:
8: a
9:
10: sentence.
// ...
18: sentence.
0: This
1: This
2: This
3: This
4: This
5: is
6: is
7: is
8: a
9: a
10: sentence.
// ...
18: sentence.
$
标记。我相信/\S+\s*$/
应该能按预期工作。 - Tanner Stern我在撰写时最受欢迎的答案中遇到了一些奇怪的行为,如果位置位于不是最后一个单词的最后一个字符,则无法获取该单词。
这是我的版本:
[position,position]
。实际上这是有意义的,因为该位置没有单词:它是一个长度为0的单词。function getWordBoundsAtPosition(str, position) {
const isSpace = (c) => /\s/.exec(c);
let start = position - 1;
let end = position;
while (start >= 0 && !isSpace(str[start])) {
start -= 1;
}
start = Math.max(0, start + 1);
while (end < str.length && !isSpace(str[end])) {
end += 1;
}
end = Math.max(start, end);
return [start, end];
}
要插入到子字符串中,只需拆分返回的边界。
const myString = 'This is a sentence.';
const position = 7;
const [start, end] = getWordBoundsAtPosition(myString, position);
const wordAtPosition = myString.substring(start, end); // => 'is'
function getWordBoundsAtPosition(str, position) {
const isSpace = (c) => /\s/.exec(c);
let start = position - 1;
let end = position;
while (start >= 0 && !isSpace(str[start])) {
start -= 1;
}
start = Math.max(0, start + 1);
while (end < str.length && !isSpace(str[end])) {
end += 1;
}
end = Math.max(start, end);
return [start, end];
}
function analyzeStringWithCursor(str, bounds, cursorIdx) {
const [start, end] = bounds;
document.getElementById("cursor-position").textContent = String(cursorIdx);
document.getElementById("fn-call").textContent = `getWordBoundsAtPosition("${str}", ${cursorIdx})`;
document.getElementById("fn-result").textContent = `[${start}, ${end}]`;
document.getElementById("substring-call").textContent = `"${str}".substring(${start}, ${end})`;
document.getElementById("substring-result").textContent = `"${str.substring(start, end)}"`;
document.getElementById("viz").textContent = ` ${"0123456789".repeat(Math.floor((str.length - 1) / 10) + 1).substring(0, str.length)}
"${str}"
${" ".repeat(start) + "↗" + " ".repeat(end - start) + "↖"}`;
}
analyzeStringWithCursor(
"",
[0, 0],
0
);
const update = (e) => {
analyzeStringWithCursor(
e.target.value,
getWordBoundsAtPosition(e.target.value, e.target.selectionStart),
e.target.selectionStart
);
};
document.getElementById("input").addEventListener('keyup', update);
document.getElementById("input").addEventListener('click', update);
<p>Type some words below. Your cursor position will be used as the position argument to the function.</p>
<input id="input" placeholder="Start typing some words..." />
<pre id="viz" style="font-size: 2rem; margin: 0.25rem"></pre>
<ul>
<li>Cursor Position: <code id="cursor-position"></code></li>
<li><code id="fn-call"></code>: <code id="fn-result"></code></li>
<li><code id="substring-call"></code>: <code id="substring-result"></code></li>
</ul>
var str = "this is a sentence";
function GetWordByPos(str, pos) {
var left = str.substr(0, pos);
var right = str.substr(pos);
left = left.replace(/^.+ /g, "");
right = right.replace(/ .+$/g, "");
return left + right;
}
alert(GetWordByPos(str, 6));
注:尚未经过彻底测试,也没有错误处理。
function getWordAt(str, pos) {
// Sanitise input
str = str + "";
pos = parseInt(pos, 10);
// Snap to a word on the left
if (str[pos] == " ") {
pos = pos - 1;
}
// Handle exceptional cases
if (pos < 0 || pos >= str.length-1 || str[pos] == " ") {
return "";
}
// Build word
var acc = "";
for ( ; pos > 0 && str[pos-1] != " "; pos--) {}
for ( ; pos < str.length && str[pos] != " "; pos++) {
acc += str[pos];
}
return acc;
}
alert(getWordAt("this is a sentence", 6));
像这样的东西。一定要彻底测试循环逻辑;我没有。
parseInt(pos,10)
或Math.floor(pos)
- 永远不要在未指定基数的情况下使用parseInt。 - PleaseStandfunction getWordAt(s, pos) {
// make pos point to a character of the word
while (s[pos] == " ") pos--;
// find the space before that word
// (add 1 to be at the begining of that word)
// (note that it works even if there is no space before that word)
pos = s.lastIndexOf(" ", pos) + 1;
// find the end of the word
var end = s.indexOf(" ", pos);
if (end == -1) end = s.length; // set to length if it was the last word
// return the result
return s.substring(pos, end);
}
getWordAt("this is a sentence", 4);
最终我采用了自己的解决方案。
注1:这些正则表达式并未经过彻底测试。
注2:该解决方案速度很快(即使对于大字符串),而且即使在单词中间也可以工作(即使位置(pos
参数)在单词中间)。
function getWordByPosition(str, pos) {
let leftSideString = str.substr(0, pos);
let rightSideString = str.substr(pos);
let leftMatch = leftSideString.match(/[^.,\s]*$/);
let rightMatch = rightSideString.match(/^[^.,\s]*/);
let resultStr = '';
if (leftMatch) {
resultStr += leftMatch[0];
}
if (rightMatch) {
resultStr += rightMatch[0];
}
return {
index: leftMatch.index,
word: resultStr
};
}