I'm trying to display a blob image from mySQL, and I'm getting weird characters instead.
To insert data I use:
if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {
// Temporary file name stored on the server
$tmpName = $_FILES['image']['tmp_name'];
// Read the file
$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));
$data = addslashes($data);
fclose($fp);
// Insert data into mysql
$sql="INSERT INTO $tbl_name( image )VALUES('$data')";
$result=mysql_query($sql);
}
To display:
$tbl_name="table"; // Table name
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
if (!$result) {
echo "There is nothing";
}
while($rows= mysql_fetch_assoc($result)){
//header('Content-type: image/jpg');
echo $rows['image'];
?>
How can I fix that?
Thank you in advance.
First, it's best practice to not store images in your database. Store them in a folder instead.
Second, you are seeing the raw data of the image in ASCII.
To answer your question if you must store images in a database, make a new php file.
header('content-type: image/png'); // Use a variable for the image type
// Code here to select a single image from the database.
// Echo the database value.
Call the picture with
<img src="link_to_php_file.php" alt="" />
You need to send an image header, like:
header("Content-Type: image/png"); // or image/jpeg or image/gif
Keep in mind that you can only show one image per time and nothing else.
If you want to show multiple images on a webpage you will need to simply link to them:
<img src="getimage.php?id=1234" />
<img src="getimage.php?id=1235" />
<img src="getimage.php?id=1236" />
Where getimage.php contains the code that will send the Content-Type header and fetch the data BLOB from MySQL.
Related
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.
I have two files;
image.php
<?php
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
$result = mysql_query("SELECT *FROM products") or die(mysql_error());
//Your code to call database for image id (from _GET['id']), for instance:
$image_id = $_GET['pid'];
$data = mysql_query("Select img where pid = $image_id");
while($data = mysql_fetch_assoc($result)){
$image=$data['img'];
header("content-type: image/png");
echo $image;
}
?>
and ay.html to view the image;
<img src="image.php?pid=IMAGE_ID" />
After many examples I tried, and every thread on this site and elsewhere, I'm still getting this error;
The image"http....../image.php?pid=IMAGE_ID" cannot be displayed because it contains errors.
Any help would be appreciated.
By the way the MySql variable is a LONGBLOB called img
I wonder why there isn't a simple IMG type for MySQL variable!
You should post how you stored the image to see exactly what it's wrong with your code but as the error says the image has some kind of error hence it was not stored correctly to database.
Generally the steps to go from an image to binary data and reverse are as below:
Image to binary conversion:
<? php
$image_data=file_get_contents('test.jpg');
$encoded_image=base64_encode($image_data);
echo $encoded_image;
?>
Binary to image conversion:
<?php
$image_data = file_get_contents('test.jpg');
$encoded_image = base64_encode($image_data);
$decoded_image = base64_decode($encoded_image);
echo $decoded_image;
?>
What you're missing altogether is the base_64 encode to prevent symbols from being incorrectly ttransformed.
/In your html page.../
/* */
I am only starting with PHP and as a part of exercise I wanted to design small website that allows you to upload image and then display all uploaded images
I got the image upload succesfully working and images are stored in database but I cant find the way to display images in table along with other data
$i=0;
while ($row = mysql_fetch_array($result))
{
echo '<td>'.mysql_result($result,$i,0).'</td>';
echo '<td>'.mysql_result($result,$i,1).'</td>';
echo '<td>'.mysql_result($result,$i,2).'</td>';
echo '<td>'.mysql_result($result,$i,3).'</td>';
echo '<td>'.base64_encode($result,$i,4).'</td>';
echo '<tr>';
$i++;
How to modify the code so the image is displayed?
this is code used to upload image
if (isset($_FILES['photo']))
{
#list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']);
// Get image type.
// We use # to omit errors
if ($imtype == 3) // cheking image type
$ext="png"; // to use it later in HTTP headers
elseif ($imtype == 2)
$ext="jpeg";
elseif ($imtype == 1)
$ext="gif";
else
$msg = 'Error: unknown file format';
if (!isset($msg)) // If there was no error
{
$data = file_get_contents($_FILES['photo']['tmp_name']);
$data = mysql_real_escape_string($data);
// Preparing data to be used in MySQL query
mysql_query("INSERT INTO {$table}
SET ext='$ext', title='$title',
data='$data'");
This is where I test it enter link description here
I was looking at Stack Overflow examples, but I couldn't find any that has the loop in it with data outputted into a table.
As you are starting with PHP, you should start with the right foot :)
Don't use mysql_* functions, these are deprecated. Instead, use mysqli_* or PDO
Storing images in the DB is "a bad idea", you better store the files in the file system (a-k.a. the HD) and reference the name to a field in the DB.
Instead of mysql_fetch_array, try with mysql_fetch_assoc so you can call the fields by name. i.e.: $row['db_field_name'];
About your problem, you will need a extra script to display the stored image. Something like this:
<?php
...
$row = mysql_fetch_assoc($result);
header("Content-type: image/jpeg"); //Set the correct image type.
echo $row['data'];
exit;
?>
And in your display html:
...
<img src="yourscript.php?image=XX" ...>
...
Hope this help :)
I have successfully added the original image into my imgs/ folder and also onto the server. But I'm wanting to add the thumbnail into the database. I've added it into the imgs/ folder but can't seem to find away to insert it into the database.
This is the final bit of code that is used to crop the img and insert it to the folder.
I need to insert it into the database also so I can call on it for the $_SESSION User and the Users friend as I have profiles.
if (isset($_POST["upload_thumbnail"]) && strlen($large_photo_exists)>0) {
//Get the new coordinates to crop the image.
$x1 = $_POST["x1"];
$y1 = $_POST["y1"];
$x2 = $_POST["x2"];
$y2 = $_POST["y2"];
$w = $_POST["w"];
$h = $_POST["h"];
//Scale the image to the thumb_width set above
$scale = $thumb_width/$w;
$cropped = resizeThumbnailImage($thumb_image_location, $large_image_location,$w,$h,$x1,$y1,$scale);
//Reload the page again to view the thumbnail
header("location:".$_SERVER["PHP_SELF"]);
exit();
}
if(isset($_GET['a'])){
if ($_GET['a']=="delete"){
if (file_exists($large_image_location)) {
unlink($large_image_location);
}
if (file_exists($thumb_image_location)) {
unlink($thumb_image_location);
$creator_id = $_SESSION['id'];
$sql = "UPDATE users SET user_pic_small='".$img."' WHERE id=$creator_id";
$sql2 = "INSERT INTO userphotos(photo_ownerid,photo_ispublic, photo_name, photo_caption, photo_imagedata) VALUES ($creator_id,1,'Profile Picture','Profile Picture','$img')";
// insert the image
if(!mysql_query($sql)) {
echo "Fail. It broke.";
}else{
$c=mysql_query($sql2);
echo "<script> parent.alert('Image Uploaded','',1000);</script>";
}
}
}
}
Hope someone can help. Thanks.
If you want to add in your database the path of thumbnail ($thumb_image_location), just add the code that inserts the path before unlink().
If you want to store the whole image into database, you need the column to be MEDIUMBLOB type, then, before unlink() read the code of the file that contains the image, for example with:
$img = file_get_contents($thumb_image_location);
Then, INSERT data stored in $img into your database.
Ideally, you don't want to be adding the thumbnail itself to the database, just a reference (filepath) to the file. So, while I don't know what your database looks like, you need to go through the following steps:
Create a field in your table called 'thumbnail' or similar. This will hold the name which the thumbnail file is saved as.
Add the filepath to the database immediately after you crop the large image (ie between the lines '$cropped = ...' and 'header("location....' in your code)
Whenever a user or user's friend is logged in, check this field and pull any thumbnail images referenced in the table.
And that is basically it.
If you want to store only the path of the Image into Database ,than it's fine to insert only the path and serve it with HTML.
Otherwise if you want to store the raw data of the image into Database than you have to encode the Image into base64 String .
Sending/Displaying a base64 encoded Image - Here is how you encode and image at base64.
And store this huge string into a Blob Field Type into databse.
You can use this:
// Read the file
$fp = fopen($file, 'r');
$data = fread($fp, filesize($file));
$data = addslashes($data);
fclose($fp);
// Create the query and insert into our database.
// image is an BLOB field type
$query = "INSERT INTO tbl_images ";
$query .= "(image) VALUES ('$data')";
$results = mysql_query($query, $link);
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;?>'>