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SOCK · 02-26-2008, 02:07 PM

You've confused / swapped the 'open' and 'send' methods. You should be using

Code:

xmlHttp.open("GET", "a.php", true);
xmlHttp.onreadystatechange= responseAsx;
xmlHttp.send(void);

You're also not using the responseAsx callback function correctly; notice how I've altered the code to use the name of the callback, not act as if it were a function call.

Wikipedia.org :: XMLHttpRequest

02-26-2008, 02:07 PM	#2 (permalink)
SOCK The Acquainted Join Date: Nov 2007 Posts: 154 Thanks: 31	You've confused / swapped the 'open' and 'send' methods. You should be using Code: xmlHttp.open("GET", "a.php", true); xmlHttp.onreadystatechange= responseAsx; xmlHttp.send(void); You're also not using the responseAsx callback function correctly; notice how I've altered the code to use the name of the callback, not act as if it were a function call. Wikipedia.org :: XMLHttpRequest __________________ I reject your reality, and substitute my own.