CODEWITHSUNDEEP
phpandroidmysql
Ken CHAN
Ken CHAN

Reputation: 13

PHP - getting image from database

I want to save user's image to a database.

p9_member (member_no ,member_name , .... , member_photo)

Now, I am using a BLOB to save the image for "member_photo" column
What I want to do now is that:
1. I have an application that need to connect the database and get the image from the database.
2. I have also a website doing the same things.
I am implementing the application now. As I need to send the image with some information of the member like name or id. So I create an object to save all the things. This is my PHP code:

<?PHP
   Class Friends{
var $no;
var $name;
var $photo;
var $status;
var $admin;
public function __construct($no,$name,$photo,$admin,$status){
    $this->name = $name;
    $this->photo = $photo;
    $this->no = $no;
    $this->status=$status;
    $this->admin=$admin;
}
};
include ("../../server_config.php");
$friends = array();
$SQL = "SELECT p9_member.member_no,p9_member.photo AS     PHOTO,p9_member.name,p9_member.photo FROM p9_member, p9_friends WHERE p9_member.member_no = p9_friends.member_no2 AND p9_friends.member_no1 =".$_POST['memberno'];
$result = mysqli_query($con,$SQL);
if (mysqli_num_rows($result)>0) {
        while($row = mysqli_fetch_array($result)){
            $friends[] = new Friends($row['member_no'],$row['name'],$row['photo'],null,null);
        }
    }else
        $friends[] = 0;
    echo json_encode($friends);

?>

The SQL CODE:

SELECT p9_member.member_no,p9_member.photo AS PHOTO
FROM p9_member, p9_friends 
WHERE p9_member.member_no = p9_friends.member_no2 
AND p9_friends.member_no1 =".$_POST['memberno']

I cannot get the data back from the application after json decode. What is the best solution for me?

Upvotes: 0

Views: 74

Answers (1)

Vetrivel
Vetrivel

Reputation: 1149

TRY the below code it will work.. 

$image_qry=mysql_query("SELECT * FROM tabel_name WHERE your condition");
$image=mysql_fetch_assoc($image_qry);
$get_image=$image['image_column']; //BLOB DATA TYPE
header("Content-type: image/jpeg"); // need to add the header content type in your code
echo $get_image;

Upvotes: 2

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