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
