> -----Original Message----- > From: Ashley Sheridan [mailto:ash@xxxxxxxxxxxxxxxxxxxx] > Sent: Friday, February 13, 2015 1:17 AM > To: hadi > Cc: 'Maciek Sokolewicz'; 'Stuart Dallas'; 'Bastien Koert'; php- > general@xxxxxxxxxxxxx > Subject: Re: cant view image from database > > On Fri, 2015-02-13 at 00:58 +0300, hadi wrote: > > > > > -----Original Message----- > > > From: Maciek Sokolewicz [mailto:tularis@xxxxxxxxx] On Behalf Of > > > Maciek Sokolewicz > > > Sent: Thursday, February 12, 2015 10:18 PM > > > To: hadi; 'Stuart Dallas'; 'Bastien Koert' > > > Cc: php-general@xxxxxxxxxxxxx > > > Subject: Re: 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> > > > > > > > > > > > > > > Why are you outputting loads of HTML with the image in your > showimage1.php script? That's what's making the image invalid. Hi Ash, I removed html loads, but still not working. > It really looks like you're trying to run before you can walk given some of the > questions you've asked over the past couple of weeks. As a start you should > at least learn HTML as that would help you a lot here, in terms of > understanding what an image is and how you use it within a web page. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
