I have uploaded an Image in my database. The datatype is BLOB. Now I want to view this image in my browser. I have written the following code but its not generating the image. Kindly check it.
Thanks
<?php
include 'connect.php';
//$id= $_GET['product_id'];
$query_images = "SELECT image FROM product_images WHERE product_id=121";
if (!$query_images_result = mysql_query($query_images))
{
echo mysql_error();
}
else
{
$fetch_images = mysql_fetch_array($query_images_result);
$print_images = $fetch_images['image'];
header('Content-type:image/jpeg');
echo $print_images;
}
?>
File 2
<body>
<img src='single_product_image_show.php' alt="image" />
</body>
maybe something like this? You might wanna use base64 encoding to build in the image
FYI: NOT TESTED.
$sql = "SELECT `image` FROM `product_images` WHERE `product_id`=121 LIMIT 1";
$query_images_result = mysql_query($sql);
if (!$query_images_result)
{
echo mysql_error();
}
else
{
list($print_images) = mysql_fetch_array($query_images_result);
$base64source = "data:image/jpeg;base64," . base64_encode($print_images);
echo '<img src="'.$base64source.'" alt="" />';
}
There are multiple possible reasons why this could happen. First try your script directly without "header('Content-type:image/jpeg');" so you can see what it actually returns. Like http://[youhostname]/single_product_image_show.php I assume you will see some errors.
Some best practices to serve images from php:
If you can remove the end php tag (?>). You don't actually need it and you can avoid issues when you have whitespace character after the close tag which will be written to the response and so results as a corrupt image.
You should specify the Content-Length header as well. It's not mandatory but the browser
will know how big the file it has to download.
Specify some cache control headers (https://en.wikipedia.org/wiki/List_of_HTTP_header_fields) so browsers can cache your image so it won't result a full page load and mysql query every time the image appears on the site.
You can store ETag and/or Last-modified in the database which you can retrieve and put in the headers.
Related
okay so it is appearing to me my header() isn't doing anything. i tried moving it to the very top of the file instead nothing changed unless I put it in its own but then the rest of the file doesn't run but it does send the header.
EDIT: I updated my code with your suggestions, but it still isn't working correctly. Below is an image of the network tab when I inspect the page. The big thing I notice is the calls are being made but its fetching the same about of data for every image and its a very tiny amount of data.
edit: I am getting and Invalid URL Error when I inspect in chrome unless I remove data:image/png;base64,
Referrer Policy: strict-origin-when-cross-origin
I can't seem to get this working correctly. The png image should be pulling from the database and then displaying in the HTML. (and yes I know this isn't SQL injection safe) I assume there is some issue with it being called from the loop, but I am unsure. Or that it can't find the getImage.php to use it this way do i need to import it or something earlier in the code? It does correctly pull in the item_id. The result just gives me a broken image box. I know my getImage.php is outputting the correctly formatted image, because I have taken the output directly from there and inserted in place of the call for it and the image has show up. So it tells me this line below is my issue
echo ('<td><img src='"getImage.php?image_id="' . $row["item_id"].'"></td><td>');
if I edit as follows it works fine (but obviously it give me the same image for every entry) DBS is the decoded byte string that getImage.php outputs
echo ('<td><img src="data:image/png;base64,DBS"></td><td>');
PHP for the Front Facing page (blackmarket.php:
$sql="SELECT * FROM `item` WHERE 1";
$result = $link->query($sql);
echo("<table>");
if ($result->num_rows > 0) {
// output data of each row
echo("<tr>");
while($row = $result->fetch_assoc()) {
$newrow=0;
echo("<td>");
echo ('<img src="getImage.php?image_id='.$row['item_id'].'" />');
getImage.php:
<?php
require_once "connect.php";
if(isset($_GET['item_id'])) {
$sql = "SELECT item_img FROM item WHERE item_id=" . $_GET['item_id'];
$result = mysqli_query($link, $sql) or die("<b>Error:</b> I can't find the image <br/>" . mysqli_error($link));
$row = mysqli_fetch_array($result);
header("Content-type: image/png");
header("Content-Length: " . strlen($row["item_img"]));
echo $row["item_img"];
}
mysqli_close($link);
Just specify
echo "<td><img src=\"getImage.php?image_id={$row['item_id']}\"></td>";
The browser is responsible for assembling the page, and makes a separate call to the getImage.php script and reads the image data that you are returning.
Return the binary image data directly, along with the size. This might also be part of the problem.
<?php
require_once "connect.php";
if(isset($_GET['item_id'])) {
$sql = "SELECT item_img FROM item WHERE item_id=" . $_GET['item_id'];
$result = mysqli_query($link, $sql) or die("<b>Error:</b> I can't find the image <br/>" . mysqli_error($link));
$row = mysqli_fetch_array($result);
header("Content-type: image/png");
header("Content-Length: " . strlen($row['item_img']));
echo $row['item_img'];
}
mysqli_close($link);
Omit the closing php end tag ?> from this script. PHP best practices is to omit it for all scripts but it is especially important in a script like this that returns binary data, as a bit of whitespace or a UTF-8 BOM character can corrupt your output and is often hard to debug.
You are confusing two things:
data: URIs allow you to embed an image directly in the source code of an HTML page, so that the browser can render them immediately without another call to the server. They use base64 to make the data consist only of text, because you can't put non-text characters in the middle of your HTML source code
Normal https: URLs tell the browser to make a request to the server, which returns the image data, with no extra encoding needed. Most commonly, the web server will load the image straight from disk, but the browser doesn't know about that, so you can just as validly have a PHP script which returns the same data, based on whatever logic you want.
In your code, you've mixed the two: you've prefixed a normal URL with data:, so the browser is expecting the data right there in the source code, and never makes a call to getImage.php.
What you need to do instead is make sure that in your getImage.php you
echo out just the image data - no whitespace or debug output mixed in
not base64 encoded (if it's stored in the database that way, decode it before sending to the browser)
tell the browser it's an image with an appropriate Content-Type header, e.g. header('Content-Type: image/png'); - again, no mention of base64 encoding or anything else, you're serving an image just as if it was a file on your server
You shouldn't need to set the Content-Length header manually, the server will sort that out just like it would if you were outputting HTML
You can then load that image URL directly in the browser and it should display, no need to even look at your HTML until that part is working right.
Well I finally figured out what was going wrong. In my connect.php I had some white space and that white space was being called before my header so my header wouldn't work
I am able to retrieve single image from mysql database using php.
Now I want to retrieve multiple images.
Here is the piece of code:
<?php
--db connection code--
$sql = "SELECT image FROM try WHERE status='0'";
$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
while ($row = #mysql_fetch_assoc($result))
{
// set the header for the image
header("Content-type: image/jpeg");
echo mysql_result($result, 0);
echo '<br>';
}
?>
executing this code, only first image in table is being displayed.
How to retrieve multiple images?
You are using Content-type: image/jpeg, which means it can only return one image, as the browser will only expect one. If you want to return multiple images to download you have to do some workaround.
One option would be to first pack the files together into one file (like a ZIP archive), and push that to the user.
Or if you don't want to push this as a download, then create a HTML page, with image links to your images. Each of the link will only return one of course.
You need to pass the loop counter in result_mysql
$count=0;
while ($row = #mysql_fetch_assoc($result))
{
// set the header for the image
header("Content-type: image/jpeg");
echo mysql_result($result,$count);
$count++;
}
Please don't use the error suppressor #
mysql_result
NOTE : Please don't use mysql* functions in your new code they are
depriciated in newer versions use mysqli* or PDO
As the Content-type: image/jpeg returns only a single image, you should consider saving the paths of the images in your database table. Later on you won't have any problems fetching all of them and echoing the source elements in your html.
I have a postgres database with stored images 'bytea' type and I try to display them into a browser with PHP. I found the way to display one of them but I can't make it for more than one. The code I use is the following:
File Name - display_image.php
$conn = pg_connect("dbname=test user=postgres password=postgres");
$temp = '/home/postgres/tmp.jpg';
$query = "select lo_export(image, '$temp') from map ";
$result = pg_query($query);
if($result)
{
while ($line = pg_fetch_array($result))
{
$ctobj = $line["image"];
echo "<IMG SRC=show.php> </br>";
}
}
else { echo "File does not exists."; }
pg_close($conn);
File Name - show.php
header("Content-type: image/jpeg");
$jpeg = fopen("/home/postgres/tmp.jpg","r");
$image = fread($jpeg,filesize("/home/postgres/tmp.jpg"));
echo $image;
The problem seems to be the "tmp.jpg" virtual file which displays only one image. If the result of the query is 7 images then it displays 7 times the same image within a while loop. How can I solve this?
Thanks for the interest!
I did this some time ago for bytea.
You need to run your bytea data through pg_unescape_bytea. See http://php.net/manual/en/function.pg-unescape-bytea.php
Basically your SQL query returns the bytea field in an escaped format.
However this is not what you are doing. And so the above is just for the next poor sap who comes here looking for bytea help. Please amend to note you are using LOB's not BYTEA's.
Also note that your code there is not concurrency safe. If two users request different images, my guess is that you will get both users getting different images. For this reason you should add the oid to the retrieval url, and name your file /tmp/$oid.jpg where $oid is the oid of the large object. You will need to retrieve that info (I believe it looks like it is stored in the image field of map?). On the other hand that assumes that all files are essentially public. if that's not the case, you want to move everything into the show_image.php and clean up when you are done.
I've create a table where I've saved images through "BLOB". I need to show those images along with other items. But I don't know how to show those images together in the same page.
Here's my php code that displays other things in form of a table. Similarily, I wanted to display images accordingly. Any help?
<?php
$display_query = mysql_query("SELECT * FROM eportal");
echo "<table id='pageTable'><thead><tr><th>Item code</th><th>Description</th><th>Cost</th></tr></thead>";
echo "<tbody>";
while($row = mysql_fetch_array($display_query)){
print "<tr><td>".$row['itemid']."</td><td>".$row['description']."</td><td>";
print "₹".$row['cost']."</td></tr>";
}
echo "</tbody>";
echo "</table>";
mysql_close($connection);
?>
Saving images to the DB is not a good idea but if You think You need to it this way, then You can retrieve the data from DB table, encode it to base64 (http://php.net/base64_encode) and then in HTML print it in this way:
<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAUAAAAFCAYAAACNbyblAAAAHElEQVQI12P4//8/w38GIAXDIBKE0DHxgljNBAAO9TXL0Y4OHwAAAABJRU5ErkJggg==" alt="Red dot">
Using PHP You would write:
echo '<img src="data:'.$image_mime_type.';base64,'.base64_encode($image_data_from_db).'" alt="My image alt" />';
As other people mentioned, storing the images in the database is usually a bad idea.
Images are not transmitted in the same HTTP response with another page data.
To show images from the database, you would need to implement a script which, given the item id, would read the field and send the image's binary data; and provide the path to the script in your form's <img src="">:
while($row = mysql_fetch_array($display_query)){
print "<tr><td>".$row['itemid']."</td><td>".$row['description']."</td><td>";
print "₹".$row['cost']."</td><td>";
print "<img src=\"image.php?id=".$row['id']."\"></td></tr>";
}
image.php is another script which outputs the value of image_blob given the eportal.id. You would also need to provide correct size and mime type in the headers.
You better just store the images in a file (accessible by the web server) and store the path fo the file in the database.
Read the Blob data and write it into the file with header type image.. and try to print it, It should display the image file.
And yes saving image or any file in DB is really a bad habit as you are increasing DB size and it slowdown the performance also.. I suggest you to just try to convert you Blob into Image but don't apply in your work. Just save the image at desired location and keep its location path into DB to fetch and save next time.
The debate of storing blobs versus storing a path to the image file on disk has been debated over and over again. Microsoft provides a research paper comparing the pros and cons of each here. With that said, to display a blob as an image you need to make a call to a separate page and output header information that tells the browser what type of image is stored.
For example:
connectToDatabase();
$sql = "SELECT image_blob FROM eportal;";
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);
header("Content-type: image/jpeg");
echo $row['image_blob'];
$db->close();
In case you still want to save your images in database, you will need second script which will get those images from database and pass them to browser with correct headers.
header("Content-type: image/jpeg");
#
# Replace this with database read:
# readfile('myimage.jpg');
Also, you will need to store what kind of image u use. There will be different header for JPEG, GIF or PNG file.
I'd like for my app to be able to tell if an image hasn't been viewed in the last 30 days and remove it (along with data in the DB associated with it). I know you can have PHP read and output the image dynamically but I've heard its quite taxing on the system. Is there a way to for me to track these hits even when the image is viewed directly? (would htaccess be able to do this?) Thanks in advance.
For .htaccess something like this...
RewriteRule ^image/(.*)$ image.php?id=$1
And for PHP...
$id = $_GET["id"]; //don't forget to sanitize
mysql_query("UPDATE images SET views = views + 1 WHERE image = '$id'"); //update no of views
header("Content-Type: image/jpeg"); //send image header
readfile($id); //output file to browser
You can parse out the HTTP logs and do this analysis post-factum.
That said, I'd still recommend going with a "dynamic PHP file outputting an image" approach - note that the PHP file can simply stream out a file. Yes, it'll be slower than not going through PHP at all, but it won't be a significant performance hit on your system.
You can create a script that fetches images and pings your database to show that the image has been viewed. By setting the return type of your script to image/jpg you can just use this script instead of accessing images directly:
<?php
//simplified image display script
$imagepath="phpimages/$id.jpg";
$db->pingImage($id);
$image=imagecreatefromjpeg($imagepath);
header('Content-Type: image/jpeg');
imagejpeg($image);
?>
Then use:
<img src='fetchimage.php?id=478' alt='tracked image' />
As far as I know, you'll need to have a database for this.
Sample schema:
Images
id (int, PK)
path (text)
hits (int)
PHP:
// get_image.php
// note, this is just an example and I'm not considering security or hacking attempts
$sql = mysql_query('SELECT * FROM images WHERE id = '.$_GET['id']);
$img
if(mysql_num_rows($img) == 1)
while($row = mysql_fetch_assoc($sql))
$img = $row;
else
// handle some error
mysql_query("UPDATE images SET hit = hits + 1 WHERE id = ".$_GET['id']);
$img = file_get_contents($img['path']);
header('image/jpg');
echo $img;
There could be errors here, I'm just giving you a general idea...