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. - Tul --- This email has been checked for viruses by Avast antivirus software. http://www.avast.com