> 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
On Wed, Jun 4, 2014 at 3:05 PM, pira...@gmail.com wrote:
>> The send() flag is only set as long as the fetch is ongoing. Once the
>> fetch has finished, it is no longer set. So this is already possible.
>
> I did some tests some months ago and I was not be able to achieve it,
> nor found any info
> The send() flag is only set as long as the fetch is ongoing. Once the
> fetch has finished, it is no longer set. So this is already possible.
I did some tests some months ago and I was not be able to achieve it,
nor found any info or examples on Internet about if it's or it's not
possible.
So,
On Wed, Jun 4, 2014 at 2:45 PM, pira...@gmail.com wrote:
> According to the spec, the send() method of XmlHttpRequest objects
> enable an internal 'send()' flag showing that the request has been
> done, and not allowing to do several requests on the same XHR object,
> although the current API with
According to the spec, the send() method of XmlHttpRequest objects
enable an internal 'send()' flag showing that the request has been
done, and not allowing to do several requests on the same XHR object,
although the current API with an open() method leads to think that's
possible. Due to this, I p
