javascript中的ReplaceAll函数

我得到一个字符串

“test+test1+asd.txt”

，我想把它转换成

“testtest1 asd.txt”

我试图使用函数

str=str.replace（“/+//g”，”）

但这是行不通的

问候,，
hemant
+1如果您打算使用正则表达式，但对于单个字符替换，您可以轻松使用：

yourString = yourString.split("+").join(" ");

下面是一个简单的javascript函数，用于替换所有：

function replaceAll (originalstring, exp1, exp2) {
//Replaces every occurrence of exp1 in originalstring with exp2 and returns the new string.

    if (exp1 == "") {
        return;  //Or else there will be an infinite loop because of i = i - 1 (see later).
        }

    var len1 = exp1.length;
    var len2 = exp2.length;
    var res = "";  //This will become the new string

    for (i = 0; i < originalstring.length; i++) {
        if (originalstring.substr(i, len1) == exp1) {  //exp1 found
            res = res + exp2;  //Append to res exp2 instead of exp1
            i = i + (len1 - 1);  //Skip the characters in originalstring that have been just replaced
        }
        else {//exp1 not found at this location; copy the original character into the new string
            res = res + originalstring.charAt(i);
        }
    }
    return res;
}

函数replaceAll（originalstring，exp1，exp2）{
//将originalstring中出现的每个exp1替换为exp2并返回新字符串。
如果（exp1==“”）{
return；//否则，由于i=i-1，将有一个无限循环（见下文）。
}
变量len1=exp1.length；
var len2=exp2.length；
var res=”“；//这将成为新字符串
对于（i=0；i
在正则表达式中，
+
有特殊的含义。所以，你必须用反斜杠来逃避它。这就是s.Mark的正则表达式起作用的原因<代码>\+

并且在Javascript中没有将正则表达式放在字符串中，这是替换失败的第二个原因。+1但是IMHO需要解释，比如Edmassa:）如果正则表达式只有一个固定字符，为什么要使用正则表达式？

function replaceAll (originalstring, exp1, exp2) {
//Replaces every occurrence of exp1 in originalstring with exp2 and returns the new string.

    if (exp1 == "") {
        return;  //Or else there will be an infinite loop because of i = i - 1 (see later).
        }

    var len1 = exp1.length;
    var len2 = exp2.length;
    var res = "";  //This will become the new string

    for (i = 0; i < originalstring.length; i++) {
        if (originalstring.substr(i, len1) == exp1) {  //exp1 found
            res = res + exp2;  //Append to res exp2 instead of exp1
            i = i + (len1 - 1);  //Skip the characters in originalstring that have been just replaced
        }
        else {//exp1 not found at this location; copy the original character into the new string
            res = res + originalstring.charAt(i);
        }
    }
    return res;
}