PHP Ico-Image stored in Database not shown in Browser - php

i want to read .ico-Images from a Website, store the Information to the Database and show the Images later on a global Website.
I have managed to save the images in a string into the Database. The step to show the Images on a Website is my problem.
For reading the Contents:
$data=file_get_contents("http://www.google.com/favicon.ico");
$data = base64_encode($data);
What is the right way to show that image in a single div on a website?
Fesp

You basically need to tell your script to output the content as an ico type.
<?php
//Getting your image
$data=file_get_contents("http://www.google.com/favicon.ico");
$data = base64_encode($data);
//If your storing in the db then you do the query ect togo get the data string
//Then echo out like this
header('Content-Type: image/vnd.microsoft.icon');
echo base64_decode($data);
?>
Remember you cant output anything before setting the header so perhaps you will need a seperate script to grab the file and ouput so
<img src="get_ico.php?id=1"/>
Then in get_ico.php
<?php
//Connect db ect
//Query db for $_GET['id']
//Then echo out like this
header('Content-Type: image/vnd.microsoft.icon');
echo base64_decode($row['image_data']);
?>

Related

not completed image after decoding it php Android

PHP:
<?php
require "conn.php";
$cid = "6";// $_POST["cid"];
$mysql_qry = "select image2 from cities where ID = '$cid'";
$result = mysqli_query($conn,$mysql_qry);
$row = mysqli_fetch_array($result);
header('Content-Type: image/png');
echo base64_decode($row["image2"]);
$conn->close();
?>
and this is the result
How to solve that? my photo is almost black!
As i can see you are collecting image source from the database table. You must check that weather your entire image source is saved properly after encoding it. So check the database Column Data Type in which you have stored it i doubt that its not saving the full image source hence further you are not able to get the full image.
Advice : Most probably Best Practice is that you should convert it to image and save it on disk and just save the IMAGE SOURCE PATH on your Database Table Column. And then fetch it to display when required.
Otherwise Your DB will exausted and will start taking time in retrieving the records. Also problem will start in DB Backup and migrations.
If you dont want to follow the advise then change your column type to Either Blob depending on your image size.

Retrieving Image From Database w/sql PDO

So i think this question has been addressed before but none of the answers seem to help me.
I have uploaded a .jpeg file to my database by extracting the content from the file and uploaded that to a BLOB field in my database. When it comes to retrieving the data I search the database using an id set by the query string ?id=146 (I've inserted a value that relates to a specific entry in my database just to check the functionality). When I then echo the image['image'] the binary data displays fine (well the browser renders the data how it interprets it) so its finding the entry I want, and it displays the data fine. When I add header("Content-type: image/jpeg"); and reload the page the browser tells me the image is missing. ![screen shot of missing image display][1]
The entire functionality will work as i will reference the php file in the src of an tag on the page I want the image to display. But I can't even get the image to load onto the php page when I type in its URL(inc the correct query string).
Here is a my code for the php page to find the image:
$id = intval(addslashes($_REQUEST['id']));
header("Content-type: image/jpeg");
$result = $dbh->prepare("SELECT * FROM `cefk_profile` WHERE `id`=$id");
$result->execute();
$image = $result->fetch(PDO::FETCH_ASSOC);
echo $image['img'];
And here is a shot of the page that will display the picture:
echo '<img src="php/get_img.php?id=146">';

Display image from sql

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.

In PHP, how I show various images from a BLOB field in the database with the HTML content?

I stored it images in the database using an BLOB field (I'm using SQLite). Now I want to recover this image to a HTML page and show the images there.
I can retrieve the binary data from the image from the database, but what I can do to transform this data in an image and show in the page? Currently I want to show the images inside a field in a table.
You could abuse the data: protocol, but trust me, you don't want that if you can avoid it. Normally, you create a separate php-script that serves images, so in script 1:
<img src="/myimagescript.php?id=1234">
In myimagescript.php:
//get the data from the database somehow (mysql query et al.)
//let's assuma the data is in $data
header('Content-Type: image/jpeg');//alter for png/gif/etc.
echo $data;
#uscere90 is right, but an example might help (example of a PNG image):
<?php
header("Content-type: image/png");
echo $image_data;
?>
Typically this is done by creating a wrapper script or function that retrieves the BLOB and delivers it with the appropriate content headers to be used as an <img src=''>
Doing it this way also gives you the benefit of being able to deliver or not deliver the image based on other authentication factors determined by your PHP. If, for example, a user doesn't have permission to see an image, you can instead show some default or blocking image in its place.
// File getimg.php
// Retrieve the image blob specified by $_GET['imgid'] from the database
// Assuming your blob is now in the variable $image...
header("Content-type: image/jpeg");
// Just echo out the image data
echo $image;
exit();
Now in your html:
<img src='getimg.php?imgid=12345' alt='this is your img from the database' />
You can create a simple image.php page that queries your database, then prints out a content-type relevant to the image and vomits the binary data to screen. So, in your table, you'd have <img src=image.php?id=something />, and then you'd use that id in your image.php page to do your database lookup, retrieve the binary data, and print it to screen after printing the content-type header.
image.php:
<?php
header('Content-type: image/jpeg');
//DO SQL NINJA STUFF HERE
echo mysql_result($result,0,"file_content");
?>
There are two options I would say:
You create a script that returns the image data. The <img src="-field then calls that script.
You offer the data of the images directly via a data url.
Both have it's pros and cons. For the first solution you must create a new script for the images. The second method will bloat your page if the images are large.
As there are examples for the image script method already, here is some code fragment for data URIs:
<?php
function data_uri($content, $mime)
{
$base64 = base64_encode($content);
return ('data:' . $mime . ';base64,' . $base64);
}
?>
<img src="<?php echo data_uri($content,'image/png'); ?>" />
You need to set the mime-type according to your image, image/png for PNG images, image/jpeg for JPG files etc., see here for a list.

How to Retrieve image from database?

In my Database, the images(jpeg,bmp format) are stored in bytea datatype, showing up in binary code in the database. Now I want to retrieve the images from the database. But i could not get the image in the web page. When i retrieve using this code given below, it shows the binary code value.(ie combination of numbers,characters,symbols). My code was
$dbconn = pg_connect("host=localhost user=xxxx password=xxxx dbname=xxxx")
or die('Could not connect: ' .pg_last_error());
$rs = pg_query($dbconn, "select scan_image from image where cno='4' and imageno='1'");
$image = pg_escape_bytea(pg_fetch_result($rs, 0));
echo $image;
Am i correct with this code? Please help me to find the solution.
Before you echo out the image content, you need to set the headers like:
header('Content-type: image/jpeg');
Then you can call your script in an image tag of the page where you want the fetched image to be displayed:
<img src="name_of_your_script.php">
This link will help you: Managing Images With a Web Database Application
U can use Blob to retrieve image from database. And to store image u have convert image to byte using filestream.
This link may help you - http://www.itpian.com/Coding/6298-How-to-retrieve-the-image-to-database-coding.aspx

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