Using xmlhttp.open() how do I add more than one parameter to url?

I have this code.

xml开发者_开发技巧http.open("GET","getuser.php?q="+str,true);

where q="+str

I want to pass a second var how do I do this?

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xmlhttp.open("GET","getuser.php?q=" + q + "&r=" + r, true);

Note that this will not properly escape your parameters if they contains special characters. You might want to use something like encodeURIComponent(q) instead.

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Append + "&anotherVar=" + anotherString.