如何计算特定字符串在另一个字符串中出现的次数。例如,这就是我试图在Javascript中做的事情:
var temp = "This is a string.";
alert(temp.count("is")); //should output '2'
如何计算特定字符串在另一个字符串中出现的次数。例如,这就是我试图在Javascript中做的事情:
var temp = "This is a string.";
alert(temp.count("is")); //should output '2'
当前回答
非正则表达式版本:
var string='这是一个字符串',searchFor='is',计数=0,pos=string.indexOf(searchFor);而(位置>-1){++计数;pos=string.indexOf(searchFor,++pos);}console.log(计数);//2.
其他回答
您可以使用match来定义这样的函数:
String.prototype.count = function(search) {
var m = this.match(new RegExp(search.toString().replace(/(?=[.\\+*?[^\]$(){}\|])/g, "\\"), "g"));
return m ? m.length:0;
}
对于将来找到此线程的任何人,请注意,如果您对其进行概括,则接受的答案不会总是返回正确的值,因为它会阻塞正则表达式运算符,如$和。。这里有一个更好的版本,可以处理任何针头:
function occurrences (haystack, needle) {
var _needle = needle
.replace(/\[/g, '\\[')
.replace(/\]/g, '\\]')
return (
haystack.match(new RegExp('[' + _needle + ']', 'g')) || []
).length
}
看到这篇帖子。
let str = 'As sly as a fox, as strong as an ox';
let target = 'as'; // let's look for it
let pos = 0;
while (true) {
let foundPos = str.indexOf(target, pos);
if (foundPos == -1) break;
alert( `Found at ${foundPos}` );
pos = foundPos + 1; // continue the search from the next position
}
相同的算法可以被布置得更短:
let str = "As sly as a fox, as strong as an ox";
let target = "as";
let pos = -1;
while ((pos = str.indexOf(target, pos + 1)) != -1) {
alert( pos );
}
var countInstances=函数(主体,目标){var全局计数器=0;var concatstring=“”;for(var i=0,j=target.length;i<body.length;i++){concatstring=body.substring(i-1,j);if(concatstring===目标){全局计数器+=1;concatstring='';}}返回全局计数器;};console.log(countInstance('abcabc','abc'));//==>2.console.log(countInstance('ababa','aba'));//==>2.console.log(countInstance('aaabbb','ab'));//==>1.
//Try this code
const countSubStr = (str, search) => {
let arrStr = str.split('');
let i = 0, count = 0;
while(i < arrStr.length){
let subStr = i + search.length + 1 <= arrStr.length ?
arrStr.slice(i, i+search.length).join('') :
arrStr.slice(i).join('');
if(subStr === search){
count++;
arrStr.splice(i, search.length);
}else{
i++;
}
}
return count;
}