On 12 Feb 2015 at 21:58, hadi <almarzuki2011@xxxxxxxxxxx> wrote: > Hi Tul, > > I did exactly what you suggest me. No you didn't. > 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> Remove all the above html from showimage1.php - all of it. showimage1.php should start with the <?php line that follows here. You must NOT output anything before the header() below. The html above is "output". > <?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; And you don't need the html after here either. -- Cheers -- Tim
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