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
一种简单的方法是将字符串拆分为所需单词，即我们要计算出现次数的单词，然后从部分数中减去1：
function checkOccurences(string, word) {
return string.split(word).length - 1;
}
const text="Let us see. see above, see below, see forward, see backward, see left, see right until we will be right";
const count=countOccurences(text,"see "); // 2
其他回答
试试看
<?php
$str = "33,33,56,89,56,56";
echo substr_count($str, '56');
?>
<script type="text/javascript">
var temp = "33,33,56,89,56,56";
var count = temp.match(/56/g);
alert(count.length);
</script>
只需编码打高尔夫球丽贝卡·切尔诺夫的解决方案：-）
alert(("This is a string.".match(/is/g) || []).length);
//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;
}
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"));
一种简单的方法是将字符串拆分为所需单词，即我们要计算出现次数的单词，然后从部分数中减去1：
function checkOccurences(string, word) {
return string.split(word).length - 1;
}
const text="Let us see. see above, see below, see forward, see backward, see left, see right until we will be right";
const count=countOccurences(text,"see "); // 2
