This is my HTML code for displaying image stored as blob object which is stored in a database.
<tr><td>photo</td><td><img src="image.php?id=<?php echo $employee['id']; ?>" /></td></tr>
This is the image.php file:
<?php
include db.php;
$employee_id = (isset($_GET['id']) && is_numeric($_GET['id'])) ? intval($_GET['id']) : 0;
try
{
$sql = "SELECT photo
FROM employee where id = $employee_id limit 1";
$s = $pdo->prepare($sql);
$s->execute();
}
catch (PDOException $e)
{
$error = 'Employee not found' . $e->getMessage();
include 'error.html.php';
exit();
}
$employee = $s->fetch();
$image = $employee['photo'];
header('Content-Type: image/jpeg');
echo $image;
?>
When I look at the HTML code in Firebug, I can see
<img src="image.php?id=9">
But no image is displayed in the table row. Do you know how to find out what is wrong?
When I open the page ...image.php?id=9 in Firefox
It shows a message
The image ...image.php?id=9 cannot be
displayed because it contains errors.
It is strange that every image I have contains an error. I inserted only images with
size smaller than the blob object.
do you get an error?
try to use "exit;" after the echo and remove the closing php tag.
it's optional and you don't really need it (http://php.net/manual/en/language.basic-syntax.instruction-separation.php).
header('Content-Type: image/jpeg');
echo $image;
exit;
Related
I tried to display image stored in oracle database
I get it as decode data
I tried to this code but not work
first way
$img= studimage::select('studimage')->where('studnum',$id)->first();
header("Content-type: image/jpeg");
echo ($img->studimage) ;
sec-way
echo '<img src="data:image/jpg;base64,'. base64_encode($img->studimage). '" />';
the two ways does not work :(
I'm using this same functionality in one of my project, please check my code below
you can either make a page that will render the image
<img src="image.php?id=123" />
That image.php page would have this:
$sql = "SELECT image FROM images WHERE image_id = " . (int) $_GET['id'];
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
if (!$row) {
header('Status: 404 Not Found');
} else {
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
}
Or, you could base64 encode it into the src (note, not all browsers handle this well):
<img src="data:image/jpeg;base64,<?php echo base64_encode($img); ?>" />
I was trying to figure out how to upload an image into mysql then display it. I got the uploading part set. The problem I am have is displaying it. When I go to display it does one of the following:
Displays no image but only has a little image icon 10x10px image on top left
Displays the whole string of the image and then properly renders the image.
When I remove/comment out the echo $contentIMG statement, nothing displays.
<?php
include 'connections/conn.php';
error_reporting(E_ALL);
// some basic sanity checks
if(isset($_GET['id']) && is_numeric($_GET['id'])) {
$contentIMG = $row_Recordset1['content'];
echo $row_Recordset1['content'];
//echo $contentIMG;
// set the header for the image
//header("Content-type: image/jpeg");
header("Content-type: image/jpeg\n\n");
echo "where is the image #1<br>";
//echo '<img src="data:image/jpeg;base64,'.$contentIMG.'"/>';
echo "where is the image #2<br>";
echo '<img src="data:image/jpeg;base64,'.base64_encode( $contentIMG ).'"/>';
if (isset($contentIMG)){
//echo '<img src="data:image/jpg;base64,'.base64_encode($contentIMG).'" />';
}
//echo mysql_result($result, 0);
// close the db link
mysql_close($result);
}
else {
echo 'Please use a real id number';
}
echo "break <br>";
$img=base64_encode($row_Recordset1['content']);
?>
I looks like this
enter image description here
I may be wrong but I think that when you set your header Content-type to image/jpeg, you should just return the image data (assuming it is stored as a blob in your database)
<?php
header('Content-type: image/jpeg');
echo $contentIMG;
?>
Your code should look like this:
<?php
include 'connections/conn.php';
error_reporting(E_ALL);
// some basic sanity checks
if(isset($_GET['id']) && is_numeric($_GET['id'])) {
$contentIMG = $row_Recordset1['content'];
header('Content-type: image/jpeg');
echo $row_Recordset1['content'];
// close the db link
mysql_close($result);
}
else {
echo 'Please use a real id number';
}
?>
thank's for help. I have problem displaying images retrieving from my database.
I cant see the image when loading image.php in img src or directly from the page. When i display the variable without header('Content-type: image/jpeg'); i can see all the code inside, as i put this line all goes off.
I have a table called TABLE with id, title, img stored as longblob directly uploaded inside phpmyadmin.
Can anyone help me?
index.php
<?php
session_start();
include "admin/include/connection2.php";
$data = new MysqlClass();
$data->connect();
$query_img ="SELECT * FROM table ORDER BY data ASC LIMIT 4";
$post_sql = $data->query($query_img);
if(mysql_num_rows($post_sql) > 0){
while($post_obj = $data->estrai($post_sql)){
$id = $post_obj->id;
$titolo = stripslashes($post_obj->title);
$data_articolo = $post_obj->data;
$immagine = $post_obj->img;
// visualizzazione dei dati
echo "<h2>".$titolo."</h2>";
echo "Autore <b>". $autore . "</b>";
echo "<br />";
echo '<'.'img src="image.php?id='.$post_sql['id'].'">';
echo $id;
echo "<hr>";
}
}else{
echo "no post aviable.";
}
// here is the image.php code
<?php
include "admin/include/connection2.php";
$data = new MysqlClass();
// connect
$data->connetti();
$id = $_GET['id'];
echo $id;
$query = mysql_query("SELECT * FROM articoli_news WHERE id='".$id."'"; //even tried to send id='1' but not working
echo $query;
$row = mysql_fetch_array($query);
echo $row['id']; //correct displaying
$content = base64_decode($query['img']);
header('Content-type: image/jpeg');
echo $content;
?>
Delete all "echo" commands except "echo $content;" because there are also appear in the output, and damage your image.
And use ob_start(); in the begining of the script, and check out your script file not contain any of whitespace characters before or after the php begint and close tags .
I am refactoring some old code, including rewriting basic mysql queries to use PDO.
The following works brilliantly in all browsers and for all image types:
$query = 'SELECT image FROM image WHERE imageid=' . $image_id;
$result = mysql_query($query, $db_conn); querycheck($result);
header("Content-type: image");
echo mysql_result($result, 0);
Unfortunately, however I rewrite it using PDO, it doesn't work. I've been through the entire PDO documentation and the standard web search, but none of the advice/solutions work.
How can one easily fetch and image from MySQL using PDO and display it?
Edit 1:
Matthew Ratzloff gives what should be the obvious answer below, but it does not work.
Here is the actual code that I test using PDO (and I have tried many variants/parameters):
$connectstring_temp = 'mysql:host=' . $A . ';dbname=' .$B;
$dbh_temp = new PDO($connectstring_temp, $login, $password);
#$dbh_temp->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
#$dbh_temp->setAttribute(PDO::MYSQL_ATTR_USE_BUFFERED_QUERY,true);
$sql = "SELECT image FROM image WHERE imageid=" . $image_id;
$query = $dbh_temp->prepare($sql);
$query->execute();
$query->bindColumn(1, $image, PDO::PARAM_LOB);
$query->fetch(PDO::FETCH_BOUND);
header("Content-Type: image");
echo $image;
I've kept the same syntax, although for the final code $image_id needs to be passed as a parameter. The code above does NOT work. PDO works fine for all other queries of all types.
You need to paramaterize the imageid value and bind the parameter to PDO::PARAM_LOB:
$sql = "SELECT image FROM image WHERE imageid=:id";
$query = $db_conn->prepare($sql);
$query->execute(array(':id' => $image_id));
$query->bindColumn(1, $image, PDO::PARAM_LOB);
$query->fetch(PDO::FETCH_BOUND);
header("Content-Type: image");
echo $image;
Of course, you'll also want to specify the complete, correct content type (e.g., image/png).
My personal approach to Image Blobs is using a php file to serve the image, in combination with a GET argument... It's 100% effective and it doesn't involve file creation... For instance, the file to serve the image from the blob would be:
image.php:
<?php
$db = new PDO('mysql:host=localhost;dbname=anything; charset=utf8', 'account','password');
if (isset($_GET['imageid'])) {
$result = $db->prepare("SELECT image FROM image where imageid = ?");
if ($result->execute(array($_GET['imageid']))) {
$row=$result->fetch();
echo $row['pic']; //this prints the image data, transforming the image.php to an image
}
} // you can put an "else" here to do something on error...
?>
This can be called from you main script... For instance:
<?php
$db = new PDO('mysql:host=localhost;dbname=anything; charset=utf8', 'account','password');
//... some code to do your job, where you get your imageid from
$imageid=...
?>
<img src="./image.php?imageid=<?php echo $imageid;?>">
You can use this code to get image from database using PDO:
public function getImage($id){
$sql = "SELECT * FROM images WHERE id = ?";
$sth = $this->dbh->prepare($sql);
$sth->bindParam(1,$id);
$sth->execute();
$num = $sth->rowCount();
if( $num ){
$row = $sth->fetch(PDO::FETCH_ASSOC);
header("Content-type: ".$row['type']);
print $row['image'];
exit;
}else{
return null;
}
}
type - data type(column name) such as image/png or image/gif
image - image data(column name) stored in table as LOB
$this->dbh connection handler
It works for me but now I need to find out how to use it with JS because result of this function is passed to JavaScript code - so called ajax
This code displays all the images in the dabase so you can change it and make it do what you want it to do
getMessage();
}
?>
<?php
#the folder where the images are saved
$target = "image_uploads/";
$image_name = (isset($_POST['image_name']));
$query = ("SELECT user_id ,image_name FROM tish_images");
$image_show= $con-> prepare($query);
$image_show->execute();
while($record =$image_show->fetch(PDO::FETCH_ASSOC)) {
#this is the Tannery operator to replace a pic when an id do not have one
$photo = ($record['image_name']== null)? "me.png":$record['image_name'];
#display image
echo '<img src="'.$target.$photo.'">';
echo $record['user_id'];
}
?>
Im creating a blog and the little bit of code below is where the blog geets printed out.
I've got a blob saved in my mysql database and im trying to turn it back into an image.
the imageName, imageType, imageSize, imageContect all receive values when i run my code. The problem is that the imageContent variable displays a load of random characters rather then an image. it seems that the reason for this is the headers but i've no idea what to do. can anyone help me to recode the image. thanks
while($row = mysql_fetch_array($result))
{
echo "name ".$row['imageName'].'<BR>';
echo "type ".$row['imageType'].'<BR>';
echo "size ".$row['imageSize'].'<BR>';
echo '<B>'.$row['blogTitle'].'</B><br />';
echo '<A HREF = http://www.alcaeos.com/blog/displayblogProcess.php?mode=edit&blogID='.$row['blogID'].'>Edit</A> ';
echo '<A HREF = http://www.alcaeos.com/blog/displayblogProcess.php?mode=delete&blogID='.$row['blogID'].'>Delete</A><BR />';
echo $row['blog'].'<br />';
header("Content-length:".$row['imageSize']);
header("Content-type:.".$row['imageType']);
header("Content-Disposition: attachment; filename=".$row['imageName']);
echo $row['imageContent'].'---------<br /><br /><br />';
}
Here is a fairly simple example of displaying an image stored as a blob.
<?php
require_once ('./includes/db.inc.php'); // Connect to the db.
//let the browser know its an image
header("Content-type: image/jpeg");
// Make the query.
$query = "SELECT thumbnail FROM items where item_id=" . $_GET['item_id'];
$result = #mysql_query ($query);
if ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo $row['thumbnail'];
}
mysql_close(); // Close the database connection.
?>
Skip the Content-length and Content-disposition, just use Content-type and set it to a valid MIME type.
Assuming they're JPEGs, you'd do this:
header( 'Content-type: image/jpeg' );
Check out this link too