Oracle Blob as img src in PHP page - php

I have a site that currently uses images on a file server. The images appear on a page where the user can drag and drop each as is needed. This is done with jQuery and the images are enclosed in a list. Each image is pretty standard:
<img src='//network_path/image.png' height='80px'>
Now however I need to reference images stored as a BLOB in an Oracle database (no choice on this, so not a merit discussion). I have no problem retrieving the BLOB and displaying on it's own using:
$sql = "SELECT image FROM images WHERE image_id = 123";
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
But I need to [efficiently] get that image as the src attribute of the img tag. I tried imagecreatefromstring() but that just returns the image in the browser, ignoring the other html. I looked at data uri, but the IE8 size limit rules that out.
So now I am kind of stuck. My searches keep coming up with using a src attribute that loads another page that contains the image. But I need the image itself to actually show on the page. (Note: I say image, meaning at least one image but as many as eight on a page).
Any help would be greatly appreciated.

Well, you can do a few things. 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); ?>" />

But I need to [efficiently] get that image as the src attribute of the img tag
As Byron already says, the accepted and right way is to output the blob in an independent image resource, and to embed that using an img tag. It's the only good way. You can use data: URIs but they
fatten your HTML code
don't work in IE < 8 and are limited to 32 KB in IE 8,
expand the data volume by 33%, and
take away the brower's possibility to cache the image resource.
Almost never a good option.

The normal way to do this is with a <img src=/path/to/script?id=32> field. It will show up on the page not as a link. What is the problem with this? Do you want to embed the image data into HTML?
To make it more efficient, you can implement some type of caching ie, write the image data to a file and do a header(location...) if you find it instead of querying the db again. Also the browser caching headers should be set so the browser doesn't download the image if it has it cached locally.

You may try this :
$img = $row['IMAGE']->load();
print('<img src="data:image/png;base64,'.base64_encode($img).'" />');

<?php
if(isset($_POST['']))//get the id
$roll_no=$_POST[''];
$conn = oci_connect("", "", "");//DB connection
$query = 'SELECT image FROM TABLE where id=:id';
$stmt = oci_parse ($conn, $query);
oci_bind_by_name($stmt, ':id', $id);
oci_execute($stmt);
$arr = oci_fetch_array($stmt, OCI_ASSOC);
$result = $arr['image']->load();
header("Content-type: image/JPEG");
echo $result;
oci_close($conn);
?>

Related

Php images display from database

i created a form which store clients passports in my database. i want to retrive the images but its has not been giving me the images on the screen. instead , it write long alphabeth of different symbols .
below is my codes , pls help me out. thanks
$sql = "SELECT * FROM `file` WHERE id = 8";
$mq = mysqli_query($dbconnect, $sql) or die ("not working query");
$row = mysqli_fetch_array($mq) or die("line 44 not working");
$s=$row['data'];
echo $row['data'];
echo '<img src="'.$s.'" alt="HTML5 Icon" style="width:128px;height:128px">';
what is wrong with my codes please.
thanks
Because you say you see lots of characters instead of the filename, I assume that you don't store a path to that image (then you could use src as you tried to use), but the image content itself and that it is jpg (you can change to png or whatever type you have there). If so, use data URI syntax for src:
$src = "data:image/jpg;base64,".base64_encode($row['data']);
echo '<img src="'.$src.'" alt="HTML5 Icon" style="width:128px;height:128px">';
It is called data URI - you can read more about it for example here: https://en.wikipedia.org/wiki/Data_URI_scheme

php - Unable to display image

I'm very new to development, and I'm trying to retrieve an image from an SQL server over an odbc connection using the below:
<?php
require ('connect.inc.php');
$sql = "SELECt image FROM test.dbo.clients WHERE id = 1";
$sql_exec = odbc_exec($con, $sql);
if($sql_array = odbc_fetch_array($sql_exec)){
$image = base64_encode($sql_array['image']);
header('Content-type: image/jpeg');
echo "<img src=".$image."/>";
}
?>
The issue is that I'm getting an icon showing a broken image.
The query is correct as when I change it to the code below, it returns this string instead of the image:
Q2hyeXNhbnRoZW11bS5qcGc=
<?php
require ('connect.inc.php');
$sql = "SELECt image FROM test.dbo.clients WHERE id = 1";
$sql_exec = odbc_exec($con, $sql);
if($sql_array = odbc_fetch_array($sql_exec)){
$image = base64_encode($sql_array['image']);
echo $image;
}
?>
I know the code is not correct and might be vulenrable to SQL injection, however I'd appreciate if you can help me retrieve the image.
Many thanks in advance,
J
If you decode the base64 string you get, then you'll see that the decoded data is Chrysanthemum.jpg. This is just the filename of the image, not the image data.
You need to either store the image in the database (not the filename) or add some code to read the image from the filesystem.
BTW, Content-Type image/jpeg requires that the data is the raw image, but your content (<img src=...> ...</img>) is an HTML fragment.

The images from my database are appearing as little small question marks

// connects to database here
if(isset($_GET['id']))
{
$id = mysql_real_escape_string($_GET['id']);
$query = mysql_query("SELECT * FROM `blob` WHERE `id`='$id'");
$data = mysql_fetch_array($query);
header("content-type: image/jpeg");
echo $imageData;
}
else
{
echo "Error!";
}
<img src="showimage.php?id=1">
the images are then being shown through and image tag shown above but they just appear as small icons (the correct number for those listed in the database but not the actual image)
First, remove the <img src="showimage.php?id=1"> and put it on a different page; you can't keep it in the same page since it causes the web server to send headers before your script is even executed; your showimage.php should only contain PHP code, nothing else.
Then, change your code like this :
$data = mysql_fetch_array($query);
header("content-type: image/jpeg");
echo $data["image"];
I assume that your row is called "image" from your previous question.
Also, don't ask multiple questions for the same problem, you already got several answers on your previous question.

How to display through php an image stored in a folder considering that my image's link is stored in MySQL database

As good practice, I'm only storing my image's link in my database, the questios are:
How should I store my image's link? (let's say it's on c:)
c://image.jpg?
Which peace of PHP code should I use to display that image?
I'm only displaying the path, what should I do to display the image?
Can I use this:
$query = "SELECT ImageURL from WhateverTable";
$result = mysql_query($query) or die(mysql_error());
$Image = mysql_fetch_row($result);
echo "<img src='$Image[0]' alt='This is an image'>";
Thank you all lads
You only want to store the relative path, not the absolute path, as linking to something like
<img src="/var/www/vhosts/website.com/images/file.jpg">
would return a 404 on any real website. store your files in the database via a relative path (/images/file.jpg) or by only the filename if they are all in the same directory.
alternatively, you can learn MongoDB and it allows you to actually store files in the database itself.
I would strongly suggest that you use PDO instead.
Use relative URLs to your image folder in case you need to move them one day.
Here is an example.
// relative to your public webroot
$publicImageUrl = '/images/in/here';
// Pull up some record, maybe of a product
$select = 'SELECT imageFilename FROM products WHERE id = 2332';
$results = mysql_query($select);
if(!$results) {
// issue with query. deal with it here
} else {
if( mysql_num_rows($result) ) {
// record not found. deal with it here
}
$row = mysql_fetch_array($result);
$imageSrc = $publicImageUrl . '/' . $row['imageFilename'];
}
And then your HTML would be as follows
<img src="<?php echo $imageSrc; ?>" />
use PDO for php <-> mysql connection
post mysql query output

show image from blob mysql

Hi I have a image table in my database. These are stored as blob along with details such as image type & name.
I am having a problem showing the image, all I get is a white box with a red cross in it.
code:
<?php
include '../connection.php';
$ID = $_GET['id'];
$query = "SELECT * FROM `images` WHERE `image_id` = '$ID'";
$result=mysql_query($query);
$row = mysql_fetch_array($result);
$image = $row['image'];
$image_type= $row['image_type'];
header("Content-type: $image_type");
print $image;
exit;
?>
Thanks
Well here is a short answer.
<?php
include '../connection.php';
$id = (int)$_GET['id'];
$query = "SELECT * FROM `images` WHERE `image_id` = '$id'";
$result=mysql_query($query);
$row = mysql_fetch_array($result);
$image = $row['image'];
$image_type= $row['image_type'];
$size = $row['image_size'];
//alternative
/* list($image, $image_type, $size) = array(
$row['image'],
$row['image_type'],
$row['image_size']
);
*/
$ext = explode('/', $image_type);
$name = $id . '.' . $ext[1];
header("Content-type: $image_type");
header("Content-length: $size");
header("Content-Disposition: attachment; filename=$name");
print $image;
exit;
Check your blobtype to be a least MEDIUMBLOB which is able to store data up to 16M
To debug this, I'd suggest commenting-out the Content-type header line, then hitting the url directly in the browser. This will allow you to see any errors, warnings or notices that PHP might be emitting.
Couple of things to try
maybe something is failing and it is returning an error msg in html instead of the expected image
is the content_type stored correctly in the database or are you just storing the file extension of the image
content-type should look something like this
image/gif
image/jpeg
That looks like it might work, what's going wrong?
Try to specify the fields explicitly:
SELECT image, image_type FROM ...
What happens when you run the query from the database?
Are you loading the image like:
<img src="image.php?id=12">
Or do you load the PHP as its own page?
maybe you could try
$row = mysql_fetch_assoc($result);
instead of
$row = mysql_fetch_array($result);
You said in a comment that the table initially said "[BLOB - 64.0 KiB]" but you then changed it to "MEDIUMBLOB". This will expand the size that you can store, but all of your existing data will still be truncated to 64KiB.
Make sure that the field type you use is large enough to store the data you want to store (16mb in a MEDIUMBLOB or ~4gb in a LONGBLOB I'm pretty sure) and then re-insert all of your data.
Other than the security problems mentioned, I don't see why the code shouldn't work other than the database problem.
Or if you don't want to create a separate php file, you can inline it
<?php
// retrieve blob into $img
?><img src='data:image/png;base64,<?php echo base64_encode( $img );?>' alt='Image <?php echo $id;?>'>

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