我希望得到2nd出现ABC的起始位置，如下所示：
var string = "XYZ 123 ABC 456 ABC 789 ABC";
getPosition(string, 'ABC', 2) // --> 16
你会怎么做？
答案 0 :(得分：132)
function getPosition(string, subString, index) {
return string.split(subString, index).join(subString).length;
}
答案 1 :(得分：51)
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
*/
答案 2 :(得分：15)
根据kennebec的回答，我创建了一个原型函数，如果找不到第n个，则返回-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;
}
答案 3 :(得分：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;
}
}
}
答案 4 :(得分：2)
这是我的解决方案，它只是迭代字符串，直到找到n匹配：
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
答案 5 :(得分：2)
此方法创建一个函数，该函数调用存储在数组
中的第n次出现的索引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
}
答案 6 :(得分：2)
更紧凑的方式，我认为更容易，而无需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
}
答案 7 :(得分：0)
使用indexOf和递归：
首先检查传递的第n个位置是否大于子串出现的总数。如果通过，则递归遍历每个索引，直到找到第n个索引。
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);
};
答案 8 :(得分：0)
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
答案 9 :(得分：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））;
答案 10 :(得分：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
}
}
}
}
答案 11 :(得分：-2)
我正在使用以下代码来解决StackOverflow上的另一个问题，并认为它可能适用于此处。函数printList2允许使用正则表达式并按顺序列出所有事件。 （printList是对早期解决方案的尝试，但在许多情况下失败了。）
<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>