Javascript 在javascript中查找字符串中第n次出现的字符
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Finding the nth occurrence of a character in a string in javascript
提问by sathish kumar
I am working on a javascript code to find the nth occurrence of a character in a string. Using the indexOf()function we are able to get the first occurrence of the character. Now the challenge is to get the nth occurrence of the character. I was able to get the second third occurrence and so on using the code given below:
我正在编写一个 javascript 代码来查找字符串中第 n 次出现的字符。使用该indexOf()函数，我们可以获得该字符的第一次出现。现在的挑战是获得该字符的第 n 次出现。我能够使用下面给出的代码获得第二次第三次出现，依此类推：
function myFunction() {
var str = "abcdefabcddesadfasddsfsd.";
var n = str.indexOf("d");
document.write("First occurence " +n );
var n1 = str.indexOf("d",parseInt(n+1));
document.write("Second occurence " +n1 );
var n2 = str.indexOf("d",parseInt(n1+1));
document.write("Third occurence " +n2 );
var n3 = str.indexOf("d",parseInt(n2+1));
document.write("Fourth occurence " +n3);
// and so on ...
}
The result is given below
结果如下
First occurence 3
Second occurence 9
Third occurence 10
Fourth occurence 14
Fifth occurence 18
Sixth occurence 19
I would like to generalize the script so that I am able to find the nth occurrence of the character as the above code requires us to repeat the script n times. Let me know if there is a better method or alternative to do the same. It would be nice if we just give the occurrence (at run time) to get the index of that character.
我想概括脚本，以便我能够找到第 n 次出现的字符，因为上面的代码要求我们重复脚本 n 次。让我知道是否有更好的方法或替代方法来做同样的事情。如果我们只给出出现次数（在运行时）来获取该字符的索引，那就太好了。
The following are some of my questions:
以下是我的一些问题：
- How do we do it in JavaScript?
- Does any framework provide any functionality to do the same implementation in an easier way or what are the alternate methods to implement the same in other frameworks /languages?
- 我们如何在 JavaScript 中做到这一点？
- 是否有任何框架提供任何功能以更简单的方式执行相同的实现，或者在其他框架/语言中实现相同的替代方法是什么？
采纳答案by Marten
function nth_occurrence (string, char, nth) {
var first_index = string.indexOf(char);
var length_up_to_first_index = first_index + 1;
if (nth == 1) {
return first_index;
} else {
var string_after_first_occurrence = string.slice(length_up_to_first_index);
var next_occurrence = nth_occurrence(string_after_first_occurrence, char, nth - 1);
if (next_occurrence === -1) {
return -1;
} else {
return length_up_to_first_index + next_occurrence;
}
}
}
// Returns 16. The index of the third 'c' character.
nth_occurrence('aaaaacabkhjecdddchjke', 'c', 3);
// Returns -1. There is no third 'c' character.
nth_occurrence('aaaaacabkhjecdddhjke', 'c', 3);
回答by Nelson
You can do it easily by implementing a function using charAt(), like this:
您可以通过使用 实现函数来轻松完成charAt()，如下所示：
function nth_ocurrence(str, needle, nth) {
for (i=0;i<str.length;i++) {
if (str.charAt(i) == needle) {
if (!--nth) {
return i;
}
}
}
return false;
}
alert( nth_ocurrence('aaaaacabkhjecdddchjke', 'c', 3) );//alerts 16
Thanks to CQQL for let me know what OP really wanted. I updated a bit my original function to achieve the new behaviour.
感谢 CQQL 让我知道 OP 真正想要什么。我更新了一点我原来的功能来实现新的行为。
回答by kennebec
indexOf takes a second argument, the character index in the string to begin the search.
indexOf 接受第二个参数，即开始搜索的字符串中的字符索引。
function nthChar(string, character, n){
var count= 0, i=0;
while(count<n && (i=string.indexOf(character,i)+1)){
count++;
}
if(count== n) return i-1;
return NaN;
}
var s= 'abcbbasdbgasdnnaabaasdert';
nthChar(s,'a',7);
回答by foomip
So a nice way to do this is to extend the string class like so:
所以一个很好的方法是像这样扩展字符串类：
(function() {
String.prototype.nthOccurrenceIndex = function(charToMatch, occurrenceIndex) {
var char, index, matches, _i, _len;
matches = 0;
index = 0;
for (_i = 0, _len = this.length; _i < _len; _i++) {
char = this[_i];
if (char === charToMatch) {
matches += 1;
if (matches === occurrenceIndex) {
return index;
}
}
index += 1;
}
return -1;
};
}).call(this);
The much more concise CoffeeScript version:
更简洁的 CoffeeScript 版本：
String.prototype.nthOccurrenceIndex = (charToMatch, occurrenceIndex)->
matches = 0
index = 0
for char in @
if char is charToMatch
matches += 1
return index if matches is occurrenceIndex
index += 1
-1
So now you can do stuff like:
所以现在你可以做这样的事情：
"abcabc".nthOccurrenceIndex('a', 1)
# -> 0"abcabc".nthOccurrenceIndex('a', 2)
# -> 3"abcabc".nthOccurrenceIndex('a', 3)
# -> -1
"abcabc".nthOccurrenceIndex('a', 1)
# -> 0"abcabc".nthOccurrenceIndex('a', 2)
# -> 3"abcabc".nthOccurrenceIndex('a', 3)
# -> -1
回答by Sharon Choe
function nthIndexOf(search, n) {
var myArray = [];
for(var i = 0; i < myStr.length; i++) {
if(myStr.slice(i, i + search.length) === search) {
myArray.push(i);
}
}
return myArray[n - 1];
}
回答by HymanRed
A maybe clearer function. Recursive and copy the mechanism of indexOf:
一个可能更清晰的功能。递归复制机制indexOf：
- Doesn't cause an error if an incorrectnumber for nth (ie <= 0). It will return
-1like you can give a negative number (or greater than the length of the string) asfromIndextoindexOf. - Can take a
fromIndexargument (the same than forindexOf: An integer representing the index at which to start the search; the default value is 0.)
- 如果第 n 个数字不正确（即 <= 0），则不会导致错误。它将返回
-1就像你可以给（除弦的长度或更大）负数作为fromIndex对indexOf。 - 可以接受一个
fromIndex参数（与 for 相同indexOf：表示开始搜索的索引的整数；默认值为 0。）
function indexOfNth (string, char, nth, fromIndex=0) {
let indexChar = string.indexOf(char, fromIndex);
if (indexChar === -1){
return -1;
} else if (nth === 1) {
return indexChar;
} else {
return indexOfNth(string, char, nth-1, indexChar+1);
}
}
let test = 'string for research purpose';
console.log('first s:', indexOfNth(test, 's', 1));
console.log('second s:', indexOfNth(test, 's', 2));
console.log('15th s:', indexOfNth(test, 's', 15));
console.log('first z:', indexOfNth(test, 'z', 1));
console.log('-1th s:', indexOfNth(test, 's', -1));
console.log('first s starting from index=1:', indexOfNth(test, 's', 1, 1));