CODEWITHSUNDEEP
phpmysql
Shihan
Shihan

Reputation: 21

Retrieving images from MySQL database

I am new to php and am trying to retrieve images from my MySQL database.

Upon running the below script, I get "Please check the ID!" message. Am I missing something?

Database structure

CREATE TABLE tbl_images(
    id tinyint( 3 ) unsigned NOT NULL AUTO_INCREMENT ,
    image blob NOT NULL ,
    PRIMARY KEY ( id ) 
)

PHP

<?php
if(isset($_GET['id']) && is_numeric($_GET['id'])) {
    # Connect to DB
    $link = mysqli_connect("$host", "$username", "$password") or die("Could not connect: " . mysqli_error());
    mysqli_select_db("$database") or die(mysqli_error());

    # SQL Statement
    $sql = "SELECT `image` FROM `tbl_images` WHERE id=" . mysqli_real_escape_string($_GET['id']) . ";";
    $result = mysqli_query("$sql") or die("Invalid query: " . mysqli_error());

    # Set header
    header("Content-type: image/png");
    echo mysqli_result($result, 0);
} else
    echo 'Please check the ID!';
?>

Upvotes: 1

Views: 319

Answers (3)

Chinmay235
Chinmay235

Reputation: 3414

Here is complete code -

CREATE TABLE IF NOT EXISTS `tbl_images` (
  `id` tinyint(3) unsigned NOT NULL AUTO_INCREMENT,
  `image` longblob NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

PHP

<?php
function mysqli_result($res, $row, $field=0) {
    $res->data_seek($row);
    $datarow = $res->fetch_array();
    return $datarow[$field];
}

if(isset($_GET['id']) && is_numeric($_GET['id'])) {
    # Connect to DB
    $link = mysqli_connect("$host", "$username", "$password") or die("Could not connect: " . mysqli_error());
    mysqli_select_db("$database") or die(mysqli_error());

    # SQL Statement
    $sql = "SELECT `image` FROM `tbl_images` WHERE id=" . mysqli_real_escape_string($_GET['id']) . ";";
    $result = mysqli_query("$sql") or die("Invalid query: " . mysqli_error());

    # Set header
    $res1= mysqli_result($result, 0);
    echo '<img src="data:image/png;base64,' . base64_encode($res1) . '" />';
} else
    echo 'Please check the ID!';
?>

Upvotes: 0

Chinmay235
Chinmay235

Reputation: 3414

Make your url = http://www.domain.com?id=[YOUR_IMAGE_ID]

Pass the ?id=your_id querystring you will get the id with $_GET['id']

Remove the header("Content-type: image/png");

add this two line below your $result variable

$res1= mysqli_result($result, 0);
echo '<img src="data:image/png;base64,' . base64_encode($res1) . '" />';

I am also unable to find mysqli_result function in your above code. If you have not declare

Please use this- 

function mysqli_result($res, $row, $field=0) {
    $res->data_seek($row);
    $datarow = $res->fetch_array();
    return $datarow[$field];
}

Upvotes: 0

Dolph
Dolph

Reputation: 50650

So you are either not providing an ID, or you are providing an ID and it's not "numeric." Check the value of $_GET['id'].

Upvotes: 3

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