Ive tried to use fget and fgetsv in different formats to parse my file
but keep getting:
Warning: fgetcsv(): 8 is not a valid stream resource in
My code for now looks like this:
$handle = fopen($thefile, 'r');
while(($data = fgetcsv($handle, 4096, ',')) !== FALSE)
It fails at the while part
How about a file without any spaces in the name?
-Original Message-
From: Joris Willekens [mailto:[EMAIL PROTECTED]
Sent: 04 June 2008 15:30
To: php-windows@lists.php.net
Subject: Re: [PHP-WIN] mime_content_type remote file
Yes, but with the same result
""Brereton, Stephen"" <[EMAIL PR
Yes, but with the same result
""Brereton, Stephen"" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
A 404 error only ever means the remote file cannot be found.
Being as you're swapping out the encoded spaces, I'd be inclined to
start with that as the possible cause - did you try t
A 404 error only ever means the remote file cannot be found.
Being as you're swapping out the encoded spaces, I'd be inclined to
start with that as the possible cause - did you try the script with the
encoding still in place?
-Original Message-
From: Joris Willekens [mailto:[EMAIL PROTECT
Hello
I am using following to determine the mime_type of a remote file
$filename = str_replace("%20", " ",$_GET["image"]);
echo "Filetype: ". mime_content_type($filename) .
"\n\n\n\n\n\n\n";
When I view the link in browser, it's show ok. When i run the script, I get
following message:
Warning:
Hello
I am using following to determine the mime_type of a remote file
$filename = str_replace("%20", " ",$_GET["image"]);
echo "Filetype: ". mime_content_type($filename) .
"\n\n\n\n\n\n\n";
When I view the link in browser, it's show ok. When i run the script, I get
following message:
Warning:
