Javascript 检查字符串在字符串中的次数

我得到了这个字符串：

var longText="This is a superuser test, super user is is super important!";

我想知道字符串“su”在长文中出现了多少次，以及每个“su”的位置

我试着：

var nr4 = longText.replace("su", "").length;

主文本和nr4之间的长度差除以“su”lenght being 2会导致多次重复，但我打赌有更好的方法。可以做到：

var indexesOf = function(baseString, strToMatch){

    var baseStr = new String(baseString);

    var wordLen = strToMatch.length;
    var listSu = [];
    // Number of strToMatch occurences
    var nb = baseStr.split(strToMatch).length - 1;

    for (var i = 0, len = nb; i < len; i++){
        var ioF = baseStr.indexOf(strToMatch);
        baseStr = baseStr.slice(ioF + wordLen, baseStr.length);
        if (i > 0){
            ioF = ioF + listSu[i-1] + wordLen;
        }
        listSu.push(ioF);
    }
    return listSu;
}

indexesOf("This is a superuser test, super user is is super important!","su");
return [10, 26, 43]

var indexesOf=function（baseString，strotomatch）{
var baseStr=新字符串（baseString）；
var-wordLen=strToMatch.length；
var listSu=[]；
//发生strToMatch的次数
var nb=基线分割（strToMatch）.length-1；
对于（变量i=0，len=nb；i0）{
ioF=ioF+listSu[i-1]+wordLen；
}
listSu.push（ioF）；
}
返回列表SU；
}
indexesOf（“这是一个超级用户测试，超级用户非常重要！”，“su”）；
返回[10,26,43]

例如

var parts=longText.split("su");
alert(parts.length-1); // length will be two if there is one "su"

更多详细信息请使用

---

使用。根据MDN代码修改的示例

len

包含

su

出现的次数

var myRe = /su/g;
var str = "This is a superuser test, super user is is super important!";
var myArray, len = 0;
while ((myArray = myRe.exec(str)) !== null) {
  len++;
  var msg = "Found " + myArray[0] + ".  ";
  msg += "Next match starts at " + myRe.lastIndex;
  console.log(msg, len);
}

// "Found su.  Next match starts at 12" 1
// "Found su.  Next match starts at 28" 2
// "Found su.  Next match starts at 45" 3

---

var howmany=longText.split（“su”）.length

或

var res=longText.match（/su/g）正常，如何检查每个“su”？“su”的位置。拆分（“su”）。不建议使用长度=2的破坏方法。
var myRe = /su/g;
var str = "This is a superuser test, super user is is super important!";
var myArray, len = 0;
while ((myArray = myRe.exec(str)) !== null) {
  len++;
  var msg = "Found " + myArray[0] + ".  ";
  msg += "Next match starts at " + myRe.lastIndex;
  console.log(msg, len);
}

// "Found su.  Next match starts at 12" 1
// "Found su.  Next match starts at 28" 2
// "Found su.  Next match starts at 45" 3

var longText="This is a superuser test, super user is is super important!";
var count = 0;
while(longText.indexOf("su") != -1) { // NB the indexOf() method is case sensitive!
    longText = longText.replace("su",""); //replace first occurence of 'su' with a void string
    count++;
}