檢查字符串多少次是一個字符串

Question

裏面我得到這個字符串：

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

我想知道字符串「蘇」有多少次是在LONGTEXT，每個「蘇」的位置。

我與努力：

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

而且lenght的正文和NR4的「蘇」 lenght beeing 2導致的重複次數之差除以但我敢打賭，有一個更好的辦法做到這一點。

Answer 1

例如

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

更多細節使用exec

FIDDLE

var re =/su/g, pos=[]; 
while ((result = re.exec(longText)) !== null) { 
    pos.push(result.index); 
} 
if (pos.length>0) alert(pos.length+" found at "+pos.join(",")); 

---

Answer 2

可以這樣做：

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] 

Answer 3

使用exec。從MDN代碼修改的示例。 len包含出現次數su。 。

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 

DEMO

Answer 4

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++; 
}