lol, in my book plumbers are experts and make a nice living!

You have been generous with your time and I appreciate it. This function is very close to working and I believe the "parse" is causing an issue because it is an actual JavaScript method so perhaps the JS engine is Safari is upset that it is being declared as a variable in this function.

Plus, the actual debugging error in Safari says "Can't find variable: parse"

So I tried changing the name (to "parced" and "place" but it still says it can't find that variable.

I then declared it as a variable, var parsed; (undefined), but same result.