130

我想通过以下方式获得2nd发生的起始位置ABC：

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

你会怎么做？

4

13 回答 13

184 

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
)

于 2013-01-23T13:01:09.750 回答
93

您也可以使用字符串 indexOf 而不创建任何数组。

第二个参数是开始寻找下一个匹配的索引。

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
*/
于 2013-01-23T14:31:35.693 回答
20

根据 kennebec 的回答，我创建了一个原型函数，如果未找到第 n 个出现而不是 0，它将返回 -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;
}
于 2013-12-15T19:42:33.097 回答
4

因为递归总是答案。

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;
        }
    }
}
于 2013-01-23T13:15:23.357 回答
3

这是我的解决方案，它只是遍历字符串直到n找到匹配项：

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
于 2013-01-23T14:00:21.617 回答
2

此方法创建一个函数，该函数调用存储在数组中的第 n 个出现的索引

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 
}
于 2016-05-31T15:00:03.717 回答
1

更短的方法，我认为更容易，而不会创建不必要的字符串。

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
}
于 2020-04-21T20:35:11.583 回答
0

I needed a function that could search from the end of the string too so I wrote this:

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
}

Usage:

getPos('x x x', 'x', 1, false) // 2
getPos('x x x', 'x', 0, true) // 4
于 2021-11-08T09:58:01.950 回答
0

一个简单的解决方案只需添加字符串、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
      }

    }
   
  }
}
于 2021-02-03T21:08:18.373 回答
0

使用String.indexOf：

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
于 2018-01-04T11:04:09.583 回答
0

使用indexOf和递归：

首先检查传递的第 n 个位置是否大于子字符串出现的总数。如果通过，递归遍历每个索引，直到找到第 n 个。

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);
};
于 2017-08-30T02:05:27.130 回答
0 
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));
于 2019-02-21T20:50:43.080 回答
-2

我正在为 StackOverflow 上的另一个问题使用以下代码，并认为它可能适合这里。函数 printList2 允许使用正则表达式并按顺序列出所有出现的事件。（printList 是对早期解决方案的尝试，但在许多情况下都失败了。）

<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>

于 2016-06-30T04:00:53.983 回答