JavaScript Non-regex Replace

Question

Do any of the existing JavaScript frameworks have a non-regex replace() function, or has this already been posted on the web somewhere as a one-off function?

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Answer 1

You can do it with or without ignoring case sensitivity.
Sadly, JavaScript's indexOf doesn't take locale vs. invariant as argument, so you'll have to replace toLowerCase with toLocaleLowerCase if you want to preserve culture-specifity.

function replaceAll(str, find, newToken, ignoreCase)
{
    var i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
}

or as prototype:

if (!String.prototype.replaceAll ) {  
String.prototype.replaceAll = function (find, replace) {
    var str = this, i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
};
}