;
fclose($fp);
echo $content;
- Original Message -
From: "Jack Dempsey" <[EMAIL PROTECTED]>
To: "'Jay Paulson'" <[EMAIL PROTECTED]>
Sent: Monday, July 30, 2001 1:52 PM
Subject: RE: [PHP] fopen not opening url
> Hi jay,
>
> I've d
Original Message -
From: "Jack Dempsey" <[EMAIL PROTECTED]>
To: "'Jay Paulson'" <[EMAIL PROTECTED]>
Sent: Monday, July 30, 2001 1:52 PM
Subject: RE: [PHP] fopen not opening url
> Hi jay,
>
> I've done the exact thing you'
...
if ($datei = @fopen($file[$i], "r+"))
...
On Thu, 26 Jul 2001, Vanessa wrote:
> Hello List,
>
> this is probably a very stupid question, but I dont know how to solve this
> little problem:
> I have a script with which text files (exported access db data sheets) can
> be uploaded to the mys
That works!
Thanks a lot :)
Cheers, Nessi
At 17:11 26/07/01 , you wrote:
>try @fopen("categories.txt","r+");
>
>
>- Original Message -
>From: Vanessa <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Thursday, July 26, 2001 12:05 PM
>Subject: [PHP] fopen - warnings
>
>
>Hello List,
>
Assuming that your code works fine apart from the warning message you can
use @ so suppress any warning messages
e.g. @fopen(
-Stewart
-Original Message-
From: Vanessa [mailto:[EMAIL PROTECTED]]
Sent: 26 July 2001 17:06
To: [EMAIL PROTECTED]
Subject: [PHP] fopen - warnings
Hello Li
try @fopen("categories.txt","r+");
- Original Message -
From: Vanessa <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, July 26, 2001 12:05 PM
Subject: [PHP] fopen - warnings
Hello List,
this is probably a very stupid question, but I dont know how to solve this
little proble
On Thu, May 10, 2001 at 10:42:10PM -0300, Christian Dechery wrote:
> as I've seen PHP running as Apache module cannot create or update files...
> it gives me 'permission denied'...
> so I've tried these solutions:
>
> * fopen("ftp://user:[EMAIL PROTECTED]/directory/newfile.ext","w";);
> this actu
even when you are trying to access a COM port?
""Kelvin"" <[EMAIL PROTECTED]> wrote in message
9d6dc8$eug$[EMAIL PROTECTED]">news:9d6dc8$eug$[EMAIL PROTECTED]...
> Hi sean,
>
>try to put a file name not a directory or path.
> $file = fopen("filename.withextension","r+"
>
> Kelvin.
>
>
> <
Hi sean,
try to put a file name not a directory or path.
$file = fopen("filename.withextension","r+"
Kelvin.
<[EMAIL PROTECTED]> wrote in message
9d6apq$eka$[EMAIL PROTECTED]">news:9d6apq$eka$[EMAIL PROTECTED]...
> hi
>
> I am running the following on my redhat 7.0 / php4 / apache box and
On Friday 06 April 2001 03:12, you wrote:
> Technically yes.
>
> In an HTTP session, the server returns a header "Content-length" which
> lists the more or less exact size of the file being sent.
Caution: The server is not required to send this header - PHP for example
doesn't do it automaticall
"Steve Werby" <[EMAIL PROTECTED]> wrote:
> acutally i found my problem, what was happening was that script1 that was
> on the server that file resides had that file open at the same time that
> scirpt2 was trying to open via http, when i change that files name, the
> permissions changed, and scrip
Technically yes.
In an HTTP session, the server returns a header "Content-length" which lists
the more or less exact size of the file being sent.
But there isn't really an easy way of reading HTTP headers in php, yet. I
really wish there were.
You can use fsockopen to make your own http session
acutally i found my problem, what was happening was that script1 that was
on the server that file resides had that file open at the same time that
scirpt2 was trying to open via http, when i change that files name, the
permissions changed, and script1 was unable to open the file, and so
script2 wa
No, you receive the handle to the file. You cannot rename the handle.
http://www.latest.txt";
$fd = fopen ($filename, "r");
$contents = fread ($fd, filesize ($filename));
echo $contents;
fclose ($fd); ?>
Should do the trick.. or something.. Modify it or whatever.
- Richard
"Jerry" <[EMA
"Jerry" <[EMAIL PROTECTED]> wrote:
> im trying to fopen a url with something like to following :
>
> fopen("something.com/something.txt", r);
>
> this returns a "0"
If you're trying to open an external file make sure you prepend the URL with
'http://'. If that's a local file and it's not finding
> Now i get a error window saying: "Document contained no data"
That usually means PHP crashed.
Check your web-server error logs.
Also, try adding some debug output before each statement to pinpoint the
crash line.
> --code-
>
> $noaa = "http://weather.noaa.gov/cgi-bin
Basically, you shouldn't be checking for '1' as an answer.
Either you got 0, and it didn't work, or you didn't, and the URL is valid.
--
Visit the Zend Store at http://www.zend.com/store/
Wanna help me out? Like Music? Buy a CD: http://l-i-e.com/artists.htm
Volunteer a little time: http://chat
, 2001 4:58 PM
> To: PHP
> Subject: Re: [PHP] fopen to validate a URL
>
>
> Well I changed the code to this expecting a true or false response (1 or 0)
>
> $url=$row[url];
> echo "Old URL = ".$url."";
>
> $fp = fopen("$url","r
Well I changed the code to this expecting a true or false response (1 or 0)
$url=$row[url];
echo "Old URL = ".$url."";
$fp = fopen("$url","r");
echo "fp = ".$fp.""; //Debug line to see what $fp is returning
switch ($fp){
case 1:
echo "Valid URL";
break;
case 0:
echo "URL IS NOT VALID";
$url="
> header( "Content-type: image/gif" );
>
> $img = fopen($DOCUMENT_ROOT . "/store/include/TestingImage.gif","rb");
> print($img);
> ?>
> prints this:
> Resource id #1
> I'm trying to print the image as binary or as an image. Have I missed a
> step? Hope i'm not being too confusing.
$img is a "f
> I'm trying to fopen a URL that I have no problems getting to
> if I just past it into the address field of my browser. However,
> when I use it in my fopen() function call, I'm getting an "Error
> 0" (zero) message.
> I've looked all over the documentation and I could not find what
> 'Error 0'
On Wed, 21 Feb 2001 15:23, Ben Weinberger wrote:
> Hi~
> We're working on a page located at
> http://www.manageasy.com/request_offer.php3. The page comes up with a
> bunch of errors, and we don't know why-- the page worked on our old
> server, and we didn't change anything from there... we think i
@ sign is your friend.
if ($file = @fopen('...', 'r'))
{
} else
{
}
--
Chris Lee
Mediawaveonline.com
[EMAIL PROTECTED]
""Ade Smith"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi
>
> I am trying to open a file which exists remotely(code works), b
Yeah, in the following fragment...
$fp = fopen("http://www.parentprofiles.com/clicker.php?profile_id=" .
$this-profile_id
. "&code=1","r");
You set up $fp to be a pointer to the object, in this case, a file. Now you
need to pull data out of it. Um, unfortunately, I don't remember how to do
t
> $fp = fopen("http://www.parentprofiles.com/clicker.php?profile_id=" .
$this->profile_id
> . "&code=1","r");
> echo $fp;
> ?>
>
> to do it. however when I do all I get is a response that says:
> Resource id #1
> instead of giving me the page. Any ideas?
Basically, that Resource id # 1 is like
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