errno=2

bogdani

Hello everybody,

Can you help me with this? From where is generated this error (having in mind that the code is okay and it worked nicely on my localhost server and on another remote server).

Warning: filesize() [function.filesize]: Stat failed for (errno=2 - No such file or directory) in /usr/home/admin/sites/my/_admin/do_upload.php on line 4

Warning: fread(): supplied argument is not a valid stream resource in /usr/home/admin/sites/my/_admin/do_upload.php on line 4

Thanks!

bogdani

Line 4:
$data2 = addslashes(fread(fopen($form_data2, "r"), filesize($form_data2)));

It is a form which uploads an image to dataase...

swr

The error message contains a fairly clear explanation (among other things): "no such file or directory". It can't get the size of the file because there is no file.

Try echo'ing $form_data2 and see what it contains.

If it works on one server but not another, the difference is probably that you are depending on the register_globals setting being enabled. It probably was on for the servers where it worked, but is off where it is not working. Details here:

http://www.php.net/register_globals

bogdani

the problem was from register globals which was off,... now I upload the image using $_FILES and it's okay. thanks for the clue 😉

but when I'm trying to display it, it just give me "Array" instead of content of the image,.. do you know what code should I use ?

swr

Use $_FILES['form_post2']['tmp_name'].

More info here:
http://www.php.net/manual/en/features.file-upload.php

bogdani

Originally posted by swr
Use $_FILES['form_post2']['tmp_name'].

More info here:
http://www.php.net/manual/en/features.file-upload.php

thanks for your answer!!
and how can I actually read the image from database ?

I'm using the following code and it's not working:

if($id) {
    mysql_select_db($database_ft, $ft);
    $query = "select id,photo_small1 from products where id=$id";
    $result = MYSQL_QUERY($query);
    $data = MYSQL_RESULT($result,0,"photo_small1");
    Header( "Content-type: image/JPEG");
    echo $data;
};

bogdani

now appears only the name of the temporary file created,.. when I echo data,.. any help ?

swr

Looks like you inserted the file name into the database instead of the file itself. When you receive the file, use this to get the data (untested; assuming your old way was otherwise correct):

$form_data2 = $_FILES['form_data2']['tmp_name'];
$data2 = addslashes(fread(fopen($form_data2, "r"), filesize($form_data2)));

You might want to use mysql_escape_string() instead of addslashes(). Mysql_escape_string is specifically for escaping data for a MySQL query; addslashes just happens to do the same thing most (all?) of the time.

Make sure the field in the database is of a type that supports binary data. If it isn't you might run in to trouble.

bogdani

If I'm using $data2 = addslashes(fread(fopen($form_data2, "r"), filesize($form_data2))); again will not work because the server has register globals set to off,... so I'm back on my old problem: working on my site not working on the other,..

Thanks

swr

Have you tried what I posted?

Having "$form_data2 = $_FILES['form_data2']['tmp_name'];" gives you a valid $form_data2 even if register_globals is off.

bogdani

Originally posted by swr
Have you tried what I posted?

Having "$form_data2 = $_FILES['form_data2']['tmp_name'];" gives you a valid $form_data2 even if register_globals is off.

I tried using:

$form_data2 = $_FILES['form_data2']['tmp_name']; 
$data2 = addslashes(fread(fopen($form_data2, "r"), filesize($form_data2)))

Works perfectly, but on the other server (which has register globals off) make the upload but is not displaying the images,.. it displays broken images.

But something else strange: if I try to connect from that server to my database (where the uploading / viewing is working fine) it just can't display the images for some reasons (all are broken images).

Do you have an IM so we can talk maybe? Thanks!

swr

If you're depending on register_globals for one thing, you're almost certainly depending on it for other stuff. You'll need to fix the rest of your code too.

http://www.php.net/register_globals

bogdani

It seems that how I retrieve data from DB is the problem,..

Using this:

if($id) {
    mysql_select_db($database_ft, $ft);
    $query = "select id,photo_small1 from products where id=$id";
    $result = MYSQL_QUERY($query);
    $data = MYSQL_RESULT($result,0,"photo_small1");

Header( "Content-type: image/JPEG");
echo $data;
}

I'm wondering if there is another possibility? Or this should work?

Thanks!

bogdani

I have an idea: the problem is not given by the $id variable which is taken from url but nowhere is defined (because is assuming that we are using register globals on and variables are auto-registred) ?

bogdani

Yes, that was the problem 🙂

Thanks to everybody for helping me !

If somebody will be in the same situation I recommend using this code:

mysql_select_db($database_ft, $ft);
$image_id = $_GET['id'];

$query = "select id,photo_small1 from products where id=$image_id";
$carimage = mysql_query($query) or die(mysql_error());
$row_carimage = mysql_fetch_assoc($carimage);

Header("Content-type: image/pjpeg");
echo $row_carimage['photo_small1'];