Hi there,
I'm new to these groups so forgive me if I'm asking at the wrong place (tell me where
then :)
Ok,
I'm not PHP guru but I need to create a simple script that would do this:
Include a content from another URL into the current output. I have done something like
this:
-- code start
ht
* Thus wrote Nik ([EMAIL PROTECTED]):
> Include a content from another URL into the current output. I have done
> something like
> this:
>
> -- code start
> include 'http://myotherurl.com:8080';
> ?>
> -- code end
You want to use readfile() instead.
http://php.net/readfile
Curt
--
"
Great!
Worked as expected. Thanks.
Now it is time to ask how do I handle form (HTTP POST request) data into the
URL.
Can the same fopen() work as well?
Thanks,
Nik
"Chris Boget" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> > Include a content from another URL into the current out
> Include a content from another URL into the current output. I have done
> something like this:
Look into the fopen() function. I also believe that the file() function can
also take a URL, but I'm not sure.
Chris
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Hi there,
I'm new to these groups so forgive me if I'm asking at the wrong place (tell
me where then :)
Ok,
I'm not PHP guru but I need to create a simple script that would do this:
Include a content from another URL into the current output. I have done
something like
this:
-- code start
htt
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