How Do I Display Images Stored As Blob Data Type In Mysql With Php?

How do i display an image stored in mySQL database as BLOB ? What it tried so far: 1. Created a new php function/file to get picture (getpicture.php). 2. In the html, I have the

Solution 1:

I answered my own question, it's working now..

Below is the getpicture.php:

<?php$db = new MySQLi('localhost', '', '', 'mydatabase');

if ($db->connect_errno) {
    echo'Connection to database failed: '. $db->connect_error;
    exit();
}

if (isset($_GET['id'])) {

$id = $db->real_escape_string($_GET['id']);

$query = "SELECT `Picture` FROM member WHERE `Id` = '$id'";

$result = $db->query($query);

while($row = mysqli_fetch_array($result)) {
    $imageData = $row['Picture'];
    header("Content-type:image/jpeg");
    echo$imageData;
}

} ?>

The php script which retrieve the getpicture.php above looks like this:

echo'<img src="getpicture.php?id=' . htmlspecialchars($_GET["id"]) . '"border ="0" height="250" width="250" />';

Thaank you all for the help

Solution 2:

This is wrong:

$query = mysqli_query("SELECT * FROM 'people' WHERE 'People_Id' = '$id'");

you use wrong quotes for table name (must be backtick instead of single quote (see tylda ~ key). See docs: http://dev.mysql.com/doc/refman/5.1/en/reserved-words.html

Also this is wrong too

header("content-type: image/jpeg");
echo$imageData;
echo$id;

get rid of last echo $i; and replace it with exit(); otherwise you corrupt the image data stream you just sent.

Solution 3:

Plenty is wrong.

1. your SQL is wrong. Remove the single quotes from the table and column and replace with back ticks.
2. Although it may still work, you should have a couple of new lines after your header
3. You shouldn't echo out your $id after you echo your image data.
4. You're checking for the wrong value when you check isset
5. Also, you should be using OOP for mysqli
6. Since your image data is only a single row, you don't need to wrap it in a while loop

Here is an updated code example

<?php$db = new MySQLi('localhost', 'user', 'password', 'myDatabase');

if ($db->connect_errno) {
    echo'Connection to database failed: '. $db->connect_error;
    exit();
}

if (isset($_GET['id'])) {

    $id = $db->real_escape_string($_GET['id']);
    $result = $db->query("SELECT * FROM `people` WHERE `People_Id` = '$id'");

    $row = $result->fetch_assoc();
    $imageData = $row['image'];

    header("Content-type: image/jpeg\n\n");
    echo$imageData;
}

else {
    echo"Error!";
}