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

问题

I have this code.

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

where q="+str

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

回答1:

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.

回答2:

Append + "&anotherVar=" + anotherString.

来源：https://stackoverflow.com/questions/3065981/using-xmlhttp-open-how-do-i-add-more-than-one-parameter-to-url