문자열에서 n 번째 발생을 얻는 방법?

문자열에서 n 번째 발생을 얻는 방법?

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다음과 같이 2nd발생 시작 위치를 얻고 ABC싶습니다.

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

어떻게 하시겠습니까?

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function getPosition(string, subString, index) {
   return string.split(subString, index).join(subString).length;
}

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배열을 만들지 않고 indexOf 문자열을 사용할 수도 있습니다.

두 번째 매개 변수는 다음 일치 항목을 찾기 시작하는 색인입니다.

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
*/

---

kennebec의 답변으로 작업하면서 n 번째 발생이 0이 아닌 발견되지 않으면 -1을 반환하는 프로토 타입 함수를 만들었습니다.

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;
}

---

재귀가 항상 답이기 때문입니다.

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;
        }
    }
}

---

다음은 n일치 항목이 발견 될 때까지 문자열을 반복하는 솔루션입니다 .

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

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이 메서드는 배열에 저장된 n 번째 항목의 인덱스를 호출하는 함수를 만듭니다.

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 
}

---

사용 indexOf및 재귀 :

First check if the nth position passed is greater than the total number of substring occurrences. If passed, recursively go through each index until the nth one is found.

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);
};

---

Using [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

---

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));

---

I was playing around with the following code for another question on StackOverflow and thought that it might be appropriate for here. The function printList2 allows the use of a regex and lists all the occurrences in order. (printList was an attempt at an earlier solution, but it failed in a number of cases.)

<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>

참고URL : https://stackoverflow.com/questions/14480345/how-to-get-the-nth-occurrence-in-a-string