regarding displaying an image 2 ways

[cs7536]

Hello I am working on php as it is a goal of mine to guru at it one day. I have written 2 scripts that seem to work as I wanted.

script 1: uploads an image to a directory on the system and the name of the image in a table in my dtabase.

script 2: uploads an image and image info into my mysql database.

The problem I am having is as followed:

On both scripts I can not get the image to display on the browser. I have attached both scripts in associated order.


example 1:

$db = mysql_connect("localhost", "cyber","cs7536"
mysql_select_db("erotic",$db);

// retrieve image info from database
$result = mysql_query("SELECT * FROM images2 ".
"WHERE img_id='$img_file"

// check that an entry is found
if (mysql_num_rows($result)!=1) {
return false;
}

// get this row
$img_info = mysql_fetch_array($result);
}

// create and return the img-tag
$img = '<img src=&quot;$uploaddir/$img_id(&quot;$img_id&quot border=&quot;0&quot;>';

// display some information
echo &quot;Image uploaded.<br><img src=\&quot;img?f\&quot;><br>&quot;.
&quot;URL: $server$uploaddir$myimage_name&quot;;


return $img;

?>


example 2:

my first way of trying to display image from database to a browser.

<?php
if ($id_files) {
include &quot;open_db.inc&quot;;
$sql = &quot;SELECT bin_data, filetype, filename, filesize FROM tbl_Files WHERE id_files=$id_files&quot;;

$result = @mysql_query($sql, $db);
$data = @mysql_result($result, 0, &quot;bin_data&quot;
$name = @mysql_result($result, 0, &quot;filename&quot;
$size = @mysql_result($result, 0, &quot;filesize&quot;
$type = @mysql_result($result, 0, &quot;filetype&quot;

header(&quot;Content-type: $type&quot;
header(&quot;Content-length: $size&quot;
header(&quot;Content-Disposition: attachment; filename=$name&quot;
header(&quot;Content-Description: PHP Generated Data&quot;
echo $data;
}
?>

my second way of trying to display image from database to a browser.

<?php

//open the connection
$conn = mysql_connect(&quot;localhost&quot;, &quot;cyber&quot;, &quot;cs7536&quot;

//selecting the database
mysql_select_db(&quot;binary_files&quot;,$conn);

$result = (&quot;SELECT * FROM tbl_Files WHERE id_files=$id_files&quot;
$data = mysql_query($result);
$row = mysql_fetch_array($data);
$name = $row['id_files'];
echo &quot;<p>$name</p>&quot;;



echo &quot;
header (\&quot;Content-type: image/jpeg\&quot;
<div id=\&quot;under\&quot; style=\&quot;left: 225px; top: 132px; position: absolute\&quot;> <img src=\&quot;../FRAME.jpg\&quot; width=\&quot;300\&quot; height=\&quot;375\&quot;> </div>
<div id=\&quot;top\&quot; style=\&quot;left: 261px; top: 166px; position: absolute\&quot;> <img src=\&quot;../$name\&quot; width=\&quot;228\&quot; height=\&quot;308\&quot;> </div>
<BR><BR>
&quot;

?>

I am sure you can see I am on the right track but unsure can someone please look at my code and point out what I am doing wrong. please help with solution.

THanks
Max

 

[vbkris]

what is the image source displayed in the first method (right click on the page and select 'view source' and check whether the path is as u want it).

Known is handfull, Unknown is worldfull