ml","r";)
or die("Cannot open!");
print fread($x , 4096);
> -Original Message-
> From: Jay Paulson [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, July 31, 2001 1:40 PM
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP] fopen not
g these or what they mean.
Thanks,
jay
- Original Message -
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Tuesday, July 31, 2001 2:01 AM
Subject: RE: [PHP] fopen not opening url
> Are you still having problems?
>
> Things to
print ($line) ."";
System("echo ''");
flush;
}
return ($line);
}
Let me know how you do.
Lawrence.
-Original Message-
From: Jay Paulson [mailto:[EMAIL PROTECTED]]
Sent: July 31, 2001 5:17 AM
To: [EMAIL PROTECTED]
Original Message -
From: "Jack Dempsey" <[EMAIL PROTECTED]>
To: "'Jay Paulson'" <[EMAIL PROTECTED]>
Sent: Monday, July 30, 2001 1:52 PM
Subject: RE: [PHP] fopen not opening url
> Hi jay,
>
> I've done the exact thing you'
Original Message -
From: "Jack Dempsey" <[EMAIL PROTECTED]>
To: "'Jay Paulson'" <[EMAIL PROTECTED]>
Sent: Monday, July 30, 2001 1:52 PM
Subject: RE: [PHP] fopen not opening url
> Hi jay,
>
> I've done the exact thing you'
hello-
I'm trying to use the fopen() command to open the url below and just read it
into another $var. However, I'm having some problems the warning i get
is below along with the url in the warning. Anyone know what's going on
here??
Thanks,
jay
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