我如何在JavaScript中取代一个字符串的所有事件?

根據一條線:

s = "Test abc test test abc test test test abc test test abc";

这似乎只是在上面的行中删除ABC的第一次出现:

s = s.replace('abc', '');

如何替代所有事件?

当前回答

我已经阅读了这个问题和答案,但我找不到一个合适的解决方案,虽然答案是相当有用的,我决定创建自己的解决方案从切割。

例如,对于搜索结果,我需要用相同的案例取代它,如果我处理内部HTML,我可以轻松地损害HTML标签(例如,在href属性中出现 hr)。

因此,我写的功能,以突出搜索结果在一个表,在那里表数据细胞可能有链接在内部,以及其他HTML标签。

我決定為所有人提供相同的問題的解決方案. 當然,你可以用它不僅為表,但為任何元素。

/* Iterate over table data cells to insert a highlight tag */
function highlightSearchResults(textFilter) {
  textFilter = textFilter.toLowerCase().replace('<', '&lt;').replace('>', '&gt;');
  let tds;
  tb = document.getElementById('sometable'); //root element where to search
  if (tb) {
    tds = tb.getElementsByTagName("td"); //sub-elements where to make replacements
  }
  if (textFilter && tds) {
    for (td of tds) {
      //specify your span class or whatever you need before and after
      td.innerHTML = insertCaseInsensitive(td.innerHTML, textFilter, '<span class="highlight">', '</span>');
    }
  }
}

/* Insert a highlight tag */
function insertCaseInsensitive(srcStr, lowerCaseFilter, before, after) {
  let lowStr = srcStr.toLowerCase();
  let flen = lowerCaseFilter.length;
  let i = -1;
  while ((i = lowStr.indexOf(lowerCaseFilter, i + 1)) != -1) {
    if (insideTag(i, srcStr)) continue;
    srcStr = srcStr.slice(0, i) + before + srcStr.slice(i, i+flen) + after + srcStr.slice(i+flen);
    lowStr = srcStr.toLowerCase();
    i += before.length + after.length;
  }
  return srcStr;
}

/* Check if an ocurrence is inside any tag by index */
function insideTag(si, s) {
  let ahead = false;
  let back = false;
  for (let i = si; i < s.length; i++) {
    if (s[i] == "<") {
      break;
    }
    if (s[i] == ">") {
      ahead = true;
      break;
    }
  }
  for (let i = si; i >= 0; i--) {
    if (s[i] == ">") {
      break;
    }
    if (s[i] == "<") {
      back = true;
      break;
    }
  }
  return (ahead && back);
}

2021-12-08 21:24:04

其他回答

就像上面的分裂/合并解决方案一样,下面的解决方案与逃避字符没有任何问题,与常规表达方法不同。

function replaceAll(s, find, repl, caseOff, byChar) {
    if (arguments.length<2)
        return false;
    var destDel = ! repl;       // If destDel delete all keys from target
    var isString = !! byChar;   // If byChar, replace set of characters
    if (typeof find !== typeof repl && ! destDel)
        return false;
    if (isString && (typeof find !== "string"))
        return false;

    if (! isString && (typeof find === "string")) {
        return s.split(find).join(destDel ? "" : repl);
    }

    if ((! isString) && (! Array.isArray(find) ||
        (! Array.isArray(repl) && ! destDel)))
        return false;

    // If destOne replace all strings/characters by just one element
    var destOne = destDel ? false : (repl.length === 1);

    // Generally source and destination should have the same size
    if (! destOne && ! destDel && find.length !== repl.length)
        return false

    var prox, sUp, findUp, i, done;
    if (caseOff)  { // Case insensitive

    // Working with uppercase keys and target
    sUp = s.toUpperCase();
    if (isString)
       findUp = find.toUpperCase()
    else
       findUp = find.map(function(el) {
                    return el.toUpperCase();
                });
    }
    else { // Case sensitive
        sUp = s;
        findUp = find.slice(); // Clone array/string
    }

    done = new Array(find.length); // Size: number of keys
    done.fill(null);

    var pos = 0;  // Initial position in target s
    var r = "";   // Initial result
    var aux, winner;
    while (pos < s.length) {       // Scanning the target
        prox  = Number.MAX_SAFE_INTEGER;
        winner = -1;  // No winner at the start
        for (i=0; i<findUp.length; i++) // Find next occurence for each string
            if (done[i]!==-1) { // Key still alive

                // Never search for the word/char or is over?
                if (done[i] === null || done[i] < pos) {
                    aux = sUp.indexOf(findUp[i], pos);
                    done[i] = aux;  // Save the next occurrence
                }
                else
                    aux = done[i]   // Restore the position of last search

                if (aux < prox && aux !== -1) { // If next occurrence is minimum
                    winner = i; // Save it
                    prox = aux;
                }
        } // Not done

        if (winner === -1) { // No matches forward
            r += s.slice(pos);
            break;
        } // No winner

        // Found the character or string key in the target

        i = winner;  // Restore the winner
        r += s.slice(pos, prox); // Update piece before the match

        // Append the replacement in target
        if (! destDel)
            r += repl[destOne ? 0 : i];
        pos = prox + (isString ? 1 : findUp[i].length); // Go after match
    }  // Loop

    return r; // Return the resulting string
}

文档如下:

替代All Syntax ====== 替代All(s, find, [repl, caseOff, byChar) 参数 ==========“s” 是替代序列的目标. “find” 可以是序列或序列的序列. “repl” 应该是相同的类型“find” 或空的如果“find” 是序列,它是一个简单的替代所有“find” 事件在“s” 由序列“repl” 如果“find” 是序列,它将取代

function l() {
    return console.log.apply(null, arguments);
}

var k = 0;
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa"]));  // 1
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do"])); // 2
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"])); // 3
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
     "aeiou", "", "", true)); // 4
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "a", "", true)); // 5
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoiea", "", true)); // 6
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoi", "", true)); // 7
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa", "leg"])); // 8
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"])); // 9
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"], true)); // 10
return;

2017-08-16 20:57:47

您可以在没有Regex的情况下做到这一点,但如果替代文本包含搜索文本,则要小心。

吉。

replaceAll("nihIaohi", "hI", "hIcIaO", true)

因此,这里是一个合适的替代All 选项,包括字符串的原型:

function replaceAll(str, find, newToken, ignoreCase)
{
    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
}

或者,如果你想有一个字符串原型功能:

String.prototype.replaceAll = function (find, replace) {
    let str = this;

    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
};

2019-07-11 12:45:37

这里是一个非常简单的解决方案. 您可以将一个新方法分配给一个 String 对象

String.prototype.replaceAll = function(search, replace){
   return this.replace(new RegExp(search, 'g'), replace)
}

var str = "Test abc test test abc test test test abc test test abc";
str = str.replaceAll('abc', '');

console.log(str) // -> Test  test test  test test test  test test

2020-06-22 06:30:27

方法1

尝试执行一个常见的表达式:

“测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试

方法2

与ABC分开并加入空空间。

“测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试ABC测试

2019-01-29 05:05:24

要替换一次,使用: