Trying to call .jpg from database

crash3909

Hi,

I used a tutorial to upload a picture to the dbase which worked fine. The problem is calling it and getting it to print the picture and not the binary mish-mosh. Below is what I have so far. The other problem is the header. It keeps telling me Cannot add header information - headers already sent by... Any help would be greatly appreciated. Thanks...

if($id) {

@MYSQL_CONNECT("localhost","name","password");

@mysql_select_db("dbase_name");

$query = "select bin_data,filetype from table_name where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");

#$getdata.php3?id=1;

header("Content-type: image/pjpeg"); 
echo $data;

}

<form method="post" action="<?php echo $PHP_SELF; ?>" enctype="multipart/form-data">
ID Number:<br>
<input type="text" name="id" size="3">
<INPUT TYPE="hidden" name="MAX_FILE_SIZE" value="1000000">
<br>
<p><input type="submit" name="submit" value="submit">

</form>

nnichols

The script must not output anything other than the header and the image data.

You need to get rid of the form you have tacked on the end of the script.

Hope this helps 😉

crash3909

Thanks for your reply. I am now confused. How does the user call the id # if there is no form to do it with? Thanks again for your help.

nnichols

Typically it would be passed to the script in the querystring -

script.php?id=13

crash3909

I understand what you said, but with the script above, would the following be correct to call the .jpg with id #1:

If it correct, it is not working for me.

Thanks again.

nnichols

Yes that is how you would do it. You might have to access the 'id' variable through the HTTP_GET_VARS array.

The header should be -

header("Content-type: image/jpeg");

Hope this helps 😉