如何獲取字符串中的第 n 次出現？

Question

我想獲得2nd出現的ABC的起始 position，如下所示：

var string = "XYZ 123 ABC 456 ABC 789 ABC";
getPosition(string, 'ABC', 2) // --> 16

你會怎么做？

Answer 1

 const string = "XYZ 123 ABC 456 ABC 789 ABC"; function getPosition(string, subString, index) { return string.split(subString, index).join(subString).length; } console.log( getPosition(string, 'ABC', 2) // --> 16 )

Answer 2

您還可以在不創建任何數組的情況下使用字符串 indexOf。

第二個參數是開始尋找下一個匹配項的索引。

function nthIndex(str, pat, n){
    var L= str.length, i= -1;
    while(n-- && i++<L){
        i= str.indexOf(pat, i);
        if (i < 0) break;
    }
    return i;
}

var s= "XYZ 123 ABC 456 ABC 789 ABC";

nthIndex(s,'ABC',3)

/*  returned value: (Number)
24
*/

Answer 3

根據 kennebec 的回答，我創建了一個原型函數，如果沒有找到第 n 次出現而不是 0，它將返回 -1。

String.prototype.nthIndexOf = function(pattern, n) {
    var i = -1;

    while (n-- && i++ < this.length) {
        i = this.indexOf(pattern, i);
        if (i < 0) break;
    }

    return i;
}

Answer 4

因為遞歸總是答案。

function getPosition(input, search, nth, curr, cnt) {
    curr = curr || 0;
    cnt = cnt || 0;
    var index = input.indexOf(search);
    if (curr === nth) {
        if (~index) {
            return cnt;
        }
        else {
            return -1;
        }
    }
    else {
        if (~index) {
            return getPosition(input.slice(index + search.length),
              search,
              nth,
              ++curr,
              cnt + index + search.length);
        }
        else {
            return -1;
        }
    }
}

Answer 5

這是我的解決方案，它只是遍歷字符串，直到找到n匹配項：

String.prototype.nthIndexOf = function(searchElement, n, fromElement) {
    n = n || 0;
    fromElement = fromElement || 0;
    while (n > 0) {
        fromElement = this.indexOf(searchElement, fromElement);
        if (fromElement < 0) {
            return -1;
        }
        --n;
        ++fromElement;
    }
    return fromElement - 1;
};

var string = "XYZ 123 ABC 456 ABC 789 ABC";
console.log(string.nthIndexOf('ABC', 2));

>> 16

Answer 6

此方法創建一個函數，該函數調用存儲在數組中的第 n 次出現的索引

function nthIndexOf(search, n) { 
    var myArray = []; 
    for(var i = 0; i < myString.length; i++) { //loop thru string to check for occurrences
        if(myStr.slice(i, i + search.length) === search) { //if match found...
            myArray.push(i); //store index of each occurrence           
        }
    } 
    return myArray[n - 1]; //first occurrence stored in index 0 
}

Answer 7

更短的方法，我認為更容易，而不會創建不必要的字符串。

const findNthOccurence = (string, nth, char) => {
  let index = 0
  for (let i = 0; i < nth; i += 1) {
    if (index !== -1) index = string.indexOf(char, index + 1)
  }
  return index
}

Answer 8

使用indexOf和Recursion ：

首先檢查傳遞的第 n 個位置是否大於子串出現的總數。 如果通過，則遞歸遍歷每個索引，直到找到第 n 個索引。

var getNthPosition = function(str, sub, n) {
    if (n > str.split(sub).length - 1) return -1;
    var recursePosition = function(n) {
        if (n === 0) return str.indexOf(sub);
        return str.indexOf(sub, recursePosition(n - 1) + 1);
    };
    return recursePosition(n);
};

Answer 9

使用[String.indexOf][1]

var stringToMatch = "XYZ 123 ABC 456 ABC 789 ABC";

function yetAnotherGetNthOccurance(string, seek, occurance) {
    var index = 0, i = 1;

    while (index !== -1) {
        index = string.indexOf(seek, index + 1);
        if (occurance === i) {
           break;
        }
        i++;
    }
    if (index !== -1) {
        console.log('Occurance found in ' + index + ' position');
    }
    else if (index === -1 && i !== occurance) {
        console.log('Occurance not found in ' + occurance + ' position');
    }
    else {
        console.log('Occurance not found');
    }
}

yetAnotherGetNthOccurance(stringToMatch, 'ABC', 2);

// Output: Occurance found in 16 position

yetAnotherGetNthOccurance(stringToMatch, 'ABC', 20);

// Output: Occurance not found in 20 position

yetAnotherGetNthOccurance(stringToMatch, 'ZAB', 1)

// Output: Occurance not found

Answer 10

function getStringReminder(str, substr, occ) {
   let index = str.indexOf(substr);
   let preindex = '';
   let i = 1;
   while (index !== -1) {
      preIndex = index;
      if (occ == i) {
        break;
      }
      index = str.indexOf(substr, index + 1)
      i++;
   }
   return preIndex;
}
console.log(getStringReminder('bcdefgbcdbcd', 'bcd', 3));

Answer 11

一個簡單的解決方案只是添加字符串、字符和 idx：

function getCharIdx(str,char,n){
  let r = 0
  for (let i = 0; i<str.length; i++){
    if (str[i] === char){
      r++
      if (r === n){
        return i
      }

    }
   
  }
}

Answer 12

我需要一個可以從字符串末尾搜索的函數，所以我寫了這個：

function getPos(str, char, index, backwards) {
  var split = str.split(char);
  var result = 0;
  var done = false;

  split.forEach(function (item, i) {
    if (done) {return}
    result += item.length
    if (!backwards && i === index) {
      done = true
      return
    } else if (backwards && i === split.length - index - 2) {
      done = true
      return
    }  
    result += char.length
  })
  return result
}

用法：

getPos('x x x', 'x', 1, false) // 2
getPos('x x x', 'x', 0, true) // 4

Answer 13

var getPosition = function(string, subStr, index) {
    if(!string.includes(subStr)) return null;

    let arrs = string.split(subStr);
    if(arrs.length < index) return null;

    let result = 0;
    for (let i = 0; i < index; i++) {
        result += arrs[i].length;
    }
    result += (index - 1) * subStr.length;
    return result;
}

Answer 14

我正在為 StackOverflow 上的另一個問題使用以下代碼，並認為它可能適合這里。 函數 printList2 允許使用正則表達式並按順序列出所有出現的情況。 （printList 是對早期解決方案的嘗試，但在許多情況下都失敗了。）

 <html> <head> <title>Checking regex</title> <script> var string1 = "123xxx5yyy1234ABCxxxabc"; var search1 = /\\d+/; var search2 = /\\d/; var search3 = /abc/; function printList(search) { document.writeln("<p>Searching using regex: " + search + " (printList)</p>"); var list = string1.match(search); if (list == null) { document.writeln("<p>No matches</p>"); return; } // document.writeln("<p>" + list.toString() + "</p>"); // document.writeln("<p>" + typeof(list1) + "</p>"); // document.writeln("<p>" + Array.isArray(list1) + "</p>"); // document.writeln("<p>" + list1 + "</p>"); var count = list.length; document.writeln("<ul>"); for (i = 0; i < count; i++) { document.writeln("<li>" + " " + list[i] + " length=" + list[i].length + " first position=" + string1.indexOf(list[i]) + "</li>"); } document.writeln("</ul>"); } function printList2(search) { document.writeln("<p>Searching using regex: " + search + " (printList2)</p>"); var index = 0; var partial = string1; document.writeln("<ol>"); for (j = 0; j < 100; j++) { var found = partial.match(search); if (found == null) { // document.writeln("<p>not found</p>"); break; } var size = found[0].length; var loc = partial.search(search); var actloc = loc + index; document.writeln("<li>" + found[0] + " length=" + size + " first position=" + actloc); // document.writeln(" " + partial + " " + loc); partial = partial.substring(loc + size); index = index + loc + size; document.writeln("</li>"); } document.writeln("</ol>"); } </script> </head> <body> <p>Original string is <script>document.writeln(string1);</script></p> <script> printList(/\\d+/g); printList2(/\\d+/); printList(/\\d/g); printList2(/\\d/); printList(/abc/g); printList2(/abc/); printList(/ABC/gi); printList2(/ABC/i); </script> </body> </html>