Hugh,
That script works fine.
Thank you.
What I am trying to achieve is to pull a .jpeg from a remote URL and then
resize and/or crop it after it is placed in a variable.
Your script is able to open and read the original file for output.
In the following script, I have the variable $contents hol
At 04:22 11.03.2003, Anthony Ritter said:
[snip]
>I'm trying to test the following script to display the contents of the
>following URL but it will not output.
>
>Any advice will be greatly appreciated.
>Thank you.
>Tony Ritter
>.
Anthony,
Oh, I forgot to close the file, and it's an important step!
fclose($fp);
Hugh
- Original Message -
From: "Anthony Ritter" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, March 10, 2003 7:22 PM
Subject: [PHP] newbie: contents will not output
> I'm trying to test the
Anthony,
Try:
$file_to_open="http://www.weather.com";;
$fp=fopen($file_to_open,"r");
$contents=fread($fp,1); //reads to eof or ~10K whichever comes first
print $contents;
Hope this helps.
Hugh
- Original Message -
From: "Anthony Ritter" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED
> I'm trying to test the following script to display the contents of the
> following URL but it will not output.
>
> Any advice will be greatly appreciated.
> Thank you.
> Tony Ritter
> ...
>
>
>
>
>
> $file_handler = fopen("http://www.w
I've never opened a URL with fopen.
Have you looked for examples in the manual? http://php.net/fopen ???
however, I can see one problem...
filesize($file) -- where is $file
Justin French
on 11/03/03 2:22 PM, Anthony Ritter ([EMAIL PROTECTED])
wrote:
> I'm trying to test the following s
6 matches
Mail list logo
