简单的正则表达式问题：用'？'代替单词

Question

Alright, here's my current test function:

function make_void( str )
{
    var str_arr = str.split( /[\W]+/ );
    var voidstr;
    var newstr = "";

    for ( var i = 0; i < str_arr.length; i++ )
    {
        voidstr = str_arr[i];
        // if ( Math.random() <= 0.9 )
        // {
            voidstr = voidstr.replace( /\w/gi, "?" );
        // }

        newstr += voidstr + " ";
    }

    document.writeln( newstr );
}

The problem? Punctuations is lost.

What's a good way to revise that such that they aren't?

Answer 1

Split on whitespace ( \\s ) not on non-word ( \\W ) and you will retain punctuation.

function make_void( str )
{
        var str_arr = str.split( /\s+/ ); //  !!!THIS LINE CHANGED!!!
        var voidstr;
        var newstr = "";

        for ( var i = 0; i < str_arr.length; i++ )
        {
                voidstr = str_arr[i];
                // if ( Math.random() <= 0.9 )
                // {
                        voidstr = voidstr.replace( /\w/gi, "?" );
                // }

                newstr += voidstr + " ";
        }

        document.writeln( newstr );
}

update: example snippet using Array.join() method:

for ( var i = 0; i < str_arr.length; i++ )
{
    // if ( Math.random() <= 0.9 )
    // {
        str_arr[i] = str_arr[i].replace( /\w/gi, "?" );
    // }
}

var newstr = str_arr.join(' ');

Answer 2

Some sample text of what you're trying to match against might help. (What do you actually want to keep ?)

For now, the following regex might help:

[\w\d,.?:;"'-()]

This matches words, digits, and a number of punctuation characters (though by no means all).