Head Butting the Monitor

sunit_modi

OK guys it's 5.19am and I need help.

Basically i am trying to retrieve and image stored into a database to display it onto a screen.

$query = "select bin_data,filetype from inventory where item_no='$item_no'";
$result = @("$query");

$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");
Header( "Content-type: $type");
echo $data;

and it works great when i test it by itself, but falls over and dies when i place it my script... i get the usual :

Warning: Cannot add header information - headers already sent by (output started at c:\apache\htdocs\project\admin\inventory_full_validate.php:207) in c:\apache\htdocs\project\admin\inventory_full_validate.php on line 276

blah blah blah

right... so i put the heading at the beginning of the script with a specific heading i.e.

Header( "Content-type: img/pjpeg");

since i know thats the only files i have in the database and the value in $type is img/pjpeg in the database table anyway.

now when i run it, the browser (IE6) ask me if i want to dowload the php file, whats that all about?

what am i doing wrong? is there anyway of getting around it without using the header.

In addition, if omit the header line of code, and run the scipt, the browser crashes.

Many Thanks

Sunit

😕 🙁

Weedpacket

Can we see the top part of the script (from the start of the file up to the header() line)? It sounds like your script is producing output you don't know about (like blank lines).

Ticktaku

Have you printed any message at all before that code if you have, that's the problem. If you haven't I don't know...

Rayn

make sure you are not using include / require that calls from a script that already has header() within it.

I would suggest you connect to the database from within the script, like I have done when I worked with files.

sunit_modi

OK guys, ive done the following instead and made the display element in a new file.

$item_no=21547980;
$DB_SERVER="localhost"; // Database Server machine
$DB_LOGIN="admin"; // Database login
$HTTP_HOST="//localhost"; // HTTP Host
$DOCROOT="project"; // Path, where application is installed
$ADMIN_FULL_PATH="localhost/project/admin"; // The path to the admin folder

if (!($link = mysql_pconnect($DB_SERVER, $DB_LOGIN, $DB_PASSWORD))){
	   DisplayErrMsg(sprintf("internal error %d:%s\n", mysql_errno(), mysql_error()));
   exit() ;
}

if (!(mysql_select_db($DB,$link))) {
	echo("<br>");
	echo("<br>");
	echo ("<font color=red> Cannot Connect To Database Table");
	exit();
}
$query = "select bin_data,filetype from inventory where item_no='$item_no'";
$search = @MYSQL_QUERY("$query");
$data = @MYSQL_RESULT($search,0,"bin_data");
$type = @MYSQL_RESULT($search,0,"filetype");
$search="select * from inventory where item_no='$item_no';";
$query = mysql_query("$search");
$myrow = mysql_fetch_row(mysql_query($search,$link));
Header( "Content-type: $type");
echo ("$data");
echo ("<tr><td>Image Upload:</td><td>");
echo ("<table align=center border=1>");
echo ("<tr><td>Item no:</td><td>$myrow[0] </td></tr>");
echo ("<tr><td>Title:</td><td>$myrow[6] </td></tr>");
echo ("<tr><td>Artist:</td><td>$myrow[7] </td></tr>");
echo ("<tr><td>Category:</td><td>$myrow[1] </td></tr>");
echo ("<tr><td>Format: </td> <td>$myrow[8] </td></tr>");
echo ("<tr><td>No Of CD's:</td><td>$myrow[9] </td></tr>");
echo ("<tr><td>Price:</td> <td>$myrow[12] </td></tr>");
echo ("<tr><td>Sale Price:</td><td>$myrow[13] </td></tr>");
echo ("<tr><td>Stock Level:</td><td>$myrow[14] </td></tr>");
echo ("<tr><td>Reorder Level:</td><td>$myrow[15]  </td></tr>");
echo ("</td></tr>");
echo ("<tr><td>Track Info:</td><td>$myrow[10] </td></tr>");
echo ("<tr><td>Description:</td><td> $myrow[11]</td></tr></table>");
echo ("<table align=center><tr><td><form method=post action=inventory_create2.php?item_no=$item_no><input type=submit value='Go Back'></form></table>");

But when i call the script, the image file is displayed only... nothing else happens i.e. the table is not displayed, any thoughts why it only displays the image?

I have commented the lines:

Header( "Content-type: $type");
echo ("$data");

to see what happens and the table gets displayed
Have i missed something

In addition, is there away of displaying the image so that i can see it as a thimb nail? i tried

echo ("<img src='$data' height=300 width=300 />");

but only the machine code came on the screen.

Any help will be great.

Many Thanks

Sunit

😕 🙁

Rayn

Heh...well I can say the adding HTML below it will cause a error since html is not a image compiled into code.. I would suggest you take all html out and just have the image. But I know what you are shooting for for do this instead.

make a new script with all the database connection, header, and call to the image data from the database. Within the query have a "WHERE id=$id"

now in another script put all the code for the imahe information to be displayed and then to call the imageadd this...

echo("<img src=\"image_script.php?id=$myrow[0]\">");

and it should show the image is the image itself was uploaded correctly.

sunit_modi

NICE ONE MATE.

I was a bit sceptical at first but followed your advice and it work.

Somone buy this guy a beer.

Many Thanks - My life is much better all of a sudden.

Sunit

Rayn

Originally posted by sunit_modi
NICE ONE MATE.

I was a bit sceptical at first but followed your advice and it work.

Somone buy this guy a beer.

Many Thanks - My life is much better all of a sudden.

Sunit

NICE ONE!!! 😃

Glad it worked and thanks.