Replacing ALL spaces using the zip.replace function

[sperry]

Is there a way within this formula to replace ALL spaces, instead of just the first one it finds?

 

var zip = Field("Name");

return zip.replace(" ","<span><z newsize=9> </span>");

[step]

var zip = Field("Name");
return zip.replace(/\s/g,"<span><z newsize=9> </span>");