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
var myString = "This is a string.";
var foundAtPosition = 0;
var Count = 0;
while (foundAtPosition != -1)
{
foundAtPosition = myString.indexOf("is",foundAtPosition);
if (foundAtPosition != -1)
{
Count++;
foundAtPosition++;
}
}
document.write("There are " + Count + " occurrences of the word IS");
请参阅:-count字符串中出现的子字符串,以了解分步说明。
其他回答
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 );
}
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.
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;
};
正则表达式(global的缩写)中的g表示搜索整个字符串,而不仅仅是查找第一个出现的字符串。此匹配是两次:
var temp=“这是一个字符串。”;var count=(temp.match(/is/g)| |[]).length;console.log(计数);
如果没有匹配项,则返回0:
var temp=“Hello World!”;var count=(temp.match(/is/g)| |[]).length;console.log(计数);
你可以试试这个
let count = s.length - s.replace(/is/g, "").length;