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
太老了,但我今天需要做这样的事情,只想事后检查一下。对我来说工作很快。
String.prototype.count = function(substr,start,overlap) {
overlap = overlap || false;
start = start || 0;
var count = 0,
offset = overlap ? 1 : substr.length;
while((start = this.indexOf(substr, start) + offset) !== (offset - 1))
++count;
return count;
};
其他回答
var temp=“这是一个字符串。”;console.log((temp.match(新RegExp(“is”,“g”))||[]).length);
function substrCount( str, x ) {
let count = -1, pos = 0;
do {
pos = str.indexOf( x, pos ) + 1;
count++;
} while( pos > 0 );
return count;
}
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.
第二次迭代次数较少(仅当子字符串的第一个字母匹配时),但循环仍使用2:
function findSubstringOccurrences(str, word) {
let occurrences = 0;
for(let i=0; i<str.length; i++){
if(word[0] === str[i]){ // to make it faster and iterate less
for(let j=0; j<word.length; j++){
if(str[i+j] !== word[j]) break;
if(j === word.length - 1) occurrences++;
}
}
}
return occurrences;
}
console.log(findSubstringOccurrences("jdlfkfomgkdjfomglo", "omg"));
添加了此优化:
如何计算字符串中的字符串出现次数?
这可能是这里最快的实现,但如果您将“++pos”替换为“pos+=searchFor.length”,则会更快汉森里克
function occurrences(str_, subStr) {
let occurence_count = 0
let pos = -subStr.length
while ((pos = str_.indexOf(subStr, pos + subStr.length)) > -1) {
occurence_count++
}
return occurence_count
}