RE: [PHP] cant view image from database

> -----Original Message----- > From: Maciek Sokolewicz [mailto:tularis@gmail.com] On Behalf Of Maciek > Sokolewicz > Sent: Thursday, February 12, 2015 10:18 PM > To: hadi; 'Stuart Dallas'; 'Bastien Koert' > Cc: php-general@lists.php.net > Subject: Re: [PHP] cant view image from database > > On 12-2-2015 20:00, hadi wrote: > >>> Hi, > >> > >>> > >> > >>> I can't view my image from mysql database. > >> > >>> I used curl to download the image to mysql database, and im trying > >>> to view > >> > >>> the image, im getting error from firefox debugger. > >> > >>> > >> > >>> Here is my code for downloading the image > >> > >>> > >> > >>> <?php > >> > >>> > >> > >>> $servername = "localhost"; > >> > >>> $username = "recorduser"; > >> > >>> $password = "password123"; > >> > >>> $database = "record"; > >> > >>> > >> > >>> // Create connection > >> > >>> $conn = mysqli_connect($servername, $username, $password, > >> $database); > >> > >>> // Check connection > >> > >>> if (!$conn) { > >> > >>> die("Connection failed: " . mysqli_connect_error()); > >> > >>> } > >> > >>> > >> > >>> > >> > >>> > >> > >>> > >> > >>> > >> > >>> $ch = curl_init > >> > >>> > >>> ("http://www.albaldnews.com/upimages/news/thumb_albald11- > >> > >>> 04-2014-993734.jpg" > >> > >>> ); > >> > >>> curl_setopt($ch, CURLOPT_HEADER, 0); > >> > >>> curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); > >> > >>> curl_setopt($ch, CURLOPT_BINARYTRANSFER,1); > >> > >>> $rawdata=curl_exec ($ch); > >> > >>> curl_close ($ch); > >> > >>> > >> > >>> $rawdata = addslashes($rawdata); > >> > >>> > >> > >>> > >> > >>> $sql = "INSERT INTO picture (image)VALUES ('$rawdata')"; > >> > >>> $result = mysqli_query($conn, $sql); > >> > >>> > >> > >>> ?> > >> > >>> > >> > >>> And the code for viewing the image as flow; > >> > >>> > >> > >>> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" > >> > >>> > >>> "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> > >> > >>> <html > >>> xmlns="http://www.w3.org/1999/xhtml"> > >> > >>> <head> > >> > >>> <title>dispaly image</title> > >> > >>> <meta http-equiv="Content-Type" content="text/html; > >>> charset=utf-8" /> > >> > >>> </head> > >> > >>> > >> > >>> <body> > >> > >>> > >> > >>> > >> > >>> <?php > >> > >>> #header("Content-type: image/jpeg"); > >> > >>> > >> > >>> $servername = "localhost"; > >> > >>> $username = "recorduser"; > >> > >>> $password = "password123"; > >> > >>> $database = "record"; > >> > >>> > >> > >>> // Create connection > >> > >>> $conn = mysqli_connect($servername, $username, $password, > >> $database); > >> > >>> // Check connection > >> > >>> if (!$conn) { > >> > >>> die("Connection failed: " . mysqli_connect_error()); > >> > >>> } > >> > >>> > >> > >>> $sql = "SELECT * FROM picture"; > >> > >>> $result = mysqli_query($conn, $sql); > >> > >>> > >> > >>> header("Content-type: image/jpeg"); > >> > >>> $row = mysqli_fetch_array($result); > >> > >>> $img = $row["image"]; > >> > >>> > >> > >>> echo '<img src="'.$img.'" />'; > >> > >>> > >> > >>> > >> > >>> ?> > >> > >>> > >> > >>> Firefox error code: > >> > >>> > >> > >>> <img > >>> src="http://192.168.206.129/rssfeed/showimage.php" > >>> alt="The > >>> image > >> > >>> > >>> "http://192.168.206.129/rssfeed/showimage.php" cannot be > >>> displayed > >> because > > well of course you are... > > Let's establish a few baselines: > 1 - an HTML page is basically a bunch of text data, which may contain > *links* to other data via URIs (such as an URL like > http://www.example.com/image.jpg or /images/image.jpg) > 2 - an image is basically a bunch of BINARY data (so NOT text(!!)) on its own. > 3 - combining BINARY data INSIDE an HTML file does NOT work (except under > very specific circumstances). > > So... what did you do wrong? > Well, simply put: > 1 - you altered your image data by use of addslashes() to binary data. > Which means you've just corrupted the image-data, making it impossible to > show without undoing the damage. Using Stuart's suggestion you would not > have to do this. > 2 - You are putting BINARY data inside an HTML tag. The tag however > requires an URL and NOT raw data. > > How to fix this? > 1 - get rid of the addslashes and use a prepared statement to insert the > image data. > 2 - instead of echoing the data, link it to a new PHP page (for instance > showImage.php) > 3 - inside this showImage.php file retrieve the image data just as you have > done above, but do NOT echo or print ANYTHING except this image data > from the database and a header with the content-type. Hi Tul, I did exactly what you suggest me. Created link.html pointing to showimage1.php. but still not working im unable to view the picture. also replaced addslashes with mysqli_real_escape_string Link.html <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Comet php backend</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> </head> <body> <a href="showimage1.php">view picture</a> </body> </html> showimage1.php <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>dispaly image</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> </head> <body> <?php error_reporting(E_ALL); ini_set('display_errors', 1); $servername = "localhost"; $username = "recorduser"; $password = "password123"; $database = "record"; // Create connection $conn = mysqli_connect($servername, $username, $password, $database); // Check connection if (!$conn) { die("Connection failed: " . mysqli_connect_error()); } $sql = "SELECT * FROM picture"; $result = mysqli_query($conn, $sql); header("Content-type: image/jpeg"); $row = mysqli_fetch_array($result); $img = $row["image"]; echo $img; ?> </body> </html>