Re: Multiple send() calls in XmlHttpRequest

> You need to invoke open() to set the object in the proper state.

Ok, good trick (didn't know it, thanks :-) ), but I was asking about
with just one call to .open(), being able to call several times to
.send(). That's why I did the reference to KeepAlive header.


-- 
"Si quieres viajar alrededor del mundo y ser invitado a hablar en un
monton de sitios diferentes, simplemente escribe un sistema operativo
Unix."
– Linus Tordvals, creador del sistema operativo Linux

Received on Wednesday, 4 June 2014 13:30:30 UTC