Re: cant view image from database

On Thursday, Feb 12, 2015 at 6:22 pm, Bastien Koert <phpster@gmail.com>, wrote: You're echoing out an image tag, you just need to echo out the image header("Content-type: image/jpeg"); $row = mysqli_fetch_array($result); $img = $row["image"]; echo $img; And at the very least you need to be using mysqli_real_escape_string, and not addslashes! http://php.net/mysqli.real-escape-string -Stuart --  Stuart Dallas 3ft9 Ltd http://3ft9.com/ On Thu Feb 12 2015 at 12:31:23 PM hadi <almarzuki2011@hotmail.com> 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 > it contain > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >