So I'm trying to display images from a database for a website that I'm building for a friend. He wants to be able to upload an image, and have it display on the front page.
I have the image uploading all working. Saving it in the database as a BLOB.
I began working on displaying the images on a separate file. I got it working on my test website, images displayed how he wanted them. But then I moved the code over to the actual website, got it to how it needed to be and it didn't work.
I ended up trying it again on my test website, and it all worked fine.
For some, it isn't working on the actual website, but it's working fine on my test website.
The actual website grabs more information the my test website does.
Here is the code the I've used on my test website.
image.php
$query = mysql_query("SELECT * FROM images LIMIT 1");
while($row = mysql_fetch_assoc($query)){
$image_id = $row['id'];
echo "<img src=showimage.php?id=".$image_id.">";
}
Here is how I'm grabbing the image from the database and displaying it.
showimage.php
include 'inc/db.php';
$id = addslashes($_REQUEST['id']);
$image = mysql_query("SELECT image FROM sites WHERE id = '$id'");
$image = mysql_fetch_assoc($image);
$image = $image['image'];
header("Content-type: image/png");
echo $image;
Using that code on the test website works fine, there's no problem there.
Here is my code for the actual website.
sites.php
$select = mysql_query("SELECT * FROM sites");
while($row = mysql_fetch_assoc($select)){
$image_id = $row['id'];
echo '
<tr class="bottom"><td>'.$row['host'].'</td>
<td>'.$row['currency'].''.$row['price'].' every '.$row['payment'].'</td>
<td>'.$row['domain'].'</td>
<td>'.$row['paid_currency'].''.$row['paid_domain'].'</td>
<td>'.$row['features'].'</td>
<td>
<span title="Edit '.$row['id'].'"><img src="images/edit.png" alt="Edit"></span>
<span title="Delete '.$row['id'].'"><img src="images/delete.png" alt="Delete"></span>
</td>
<td><img src=showimage.php?id='.$image_id.'></td></tr>
';
}
That using the showimage.php file to grab the image from the database, but isn't working at all.
If you're set on using the database, look into using the imagecreatefrompng() function, depending on how the img was put into the DB.
Also, instead of addslashes(), try doing $id = (int)$_GET['id']; then checking if $id > 0. Finally, +1 on storing images on the filesystem and not as a BLOB.
[And insert mysql_query-is-deprecated-use-the-PDO-extension lecture here]
Related
I am trying to query the image file names from the database using the Id inputted by the user and want to display the images with the resulting file names form the select statement. However the echo function is showing only the broken image icons instead of the image itself. How can I fix this? Thanks!
//$searchValue is the Id inputted by the user in the html page
$result = mysqli_query($con,"SELECT FileName FROM images WHERE Id='$searchValue'");
//$folder is the path where the images are currently stored
$folder = "C:/xampp/htdocs/images/";
while($row = mysqli_fetch_array($result)){
$file = $row["FileName"];
echo '<img src="'.$folder.$file.'"><br />';
}
Firstly I know I shouldn't put the image in a DB but I am still learning.
I want to show the image, where am I going wrong?
In my database I have a column Image that is a BLOB. Small sized image. And I've checked an image array is there.
After I connect the db here is the php
$result = mysql_query("SELECT Image FROM table WHERE ID = 1");
$row = mysql_fetch_array($result);
echo $row["image"];
Any help appreciated.
All I want is just the image to show. Nothing fancy as I will expand on it once I get an image up on the screen.
I think I need an extra step between fetch array to split a variable that contains just the image and to display that. This is where I get lost.
Cheers.
This is the code I use to update the DB (from a form)
$myphoto = $_FILES['MyPhoto'];
$query = mysql_query("UPDATE table SET Image = '$myphoto' WHERE ID = 1") or die(mysql_error());
You need to echo an HTML <image> tag with the src set to your query results to render an image. Use this code:
$result = mysql_query("SELECT Image FROM TABLE WHERE ID = 1");
$row = mysql_fetch_array($result);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $row['Image'] ).'"/>';
I am building an image uploading website. Images are uploaded to a directory on the server and data such as the filename is stored in a MySql table. I have created a 'gallery' page which displays all the uploaded images as thumbnails. When a user clicks on one of the images, it takes them to 'image.php' page, which will display the image full size and echo information such as the username of the person who uploaded the image etc.
I am unsure as to what would be the correct way of displaying the image. The images in my MySql table have unique ID's which I'm guessing will have to be manipulated in some way, but how do I would I get the ID of the photo that has been clicked on into the 'image.php' MySql query?
Hope this has been explained well enough. Thanks in advance.
gallery.php page... (exclusing database connections etc.)
//Retrieves data from MySQL
$data = mysql_query("SELECT * FROM photos");
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
?>
<section class="thumbnails group">
<?php Echo "<img src=http://.../thumbs/tn_".$info['filename'] .">"; }?>
</section>
image.php page...
//Retrieves data from MySQL
$data = mysql_query("SELECT * FROM photos WHERE 'id' = ??");
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
?>
<section class="main-image group">
<?php echo "<img src=http://.../images/".$info['filename'] .">"; }?>
</section>
You can simply pass the ID of the image through in the querystring
<img src="/path/to/thumbnail">
In your image.php page, you can retrieve the ID like this (assuming it's an integer):
$imageID = intval( $_GET["image"]);
You should then be able to retrieve the path to the image and display it.
# liquorvicar answer is correct one, there's no ambiguity in his answer! well,
try this;
<?php Echo "<img src=http://...thumbs/tn_".$info['filename'] .">"; }?>
on image.php page
$id=intval($_REQUEST['id'])
now you have id for that specific image show its related info.
SOLVED!
so ive created an image upload page which saves the image to a folder and sends the file name to the DB, ive checked to see if they are actually being added to the DB and folder which they are, but when i call the data to another page to display the images i get broken images.
below is my code to call the images, its some code ive scraped together from various tutorials as they gave me the same problem as im having now.
UPDATE:
ive managed to get the images showing but now im faced with being shown the same image for each row of data called, the id and img_name and right for each row but the image is always the same as the first listed.
UPDATED CODE:
<?php
//connect to database
include ('connect.php');
//save the name of image in table
$query = mysql_query("select * from tbl_img") or die(mysql_error());
//retrieve all image from database and store them in a variable
while ($row = mysql_fetch_assoc($query))
{
$img_name = $row['img'];
}
?>
<?php
include ('connect.php');
$img_id = mysql_query("SELECT * FROM tbl_img");
while ($row = mysql_fetch_assoc($img_id))
{
$id = $row['img_id'];
echo "
$id<br>
$img_name<br>
<img src='http://localhost/testarea/include/site_images/$img_name' />
";
echo "<br><br><br></p>
";
}
?>
If you start the path of image with / you mean an absolute path where / is the DocumentRoot folder (or the directory of virtualhost)
With src ="includes/xxx/image.png" you mean that includes is in the same folder with the php script. If it is not you can use relative path like
src="../includes/xxx/image.png" for example
<?php
include ('connect.php');
$img_id = mysql_query("SELECT * FROM tbl_img");
while ($row = mysql_fetch_assoc($img_id))
{
$img_name = $row['img'];
$id = $row['img_id'];
echo "
$id<br>
$img_name<br>
<img src='http://localhost/testarea/include/site_images/$img_name' />
";
echo "<br><br><br></p>
";
}
How can i use php to get an image url from database (based on user id) and display it out as an image.
http://mysite.com/img.php?id=338576
The code below shows a php script goes to a database , check whether the specific user(using the id in the link above) exist then get the image url out of that user's row.
<?php
//connect to database
include ("connect.php");
//Get the values
$id = mysql_real_escape_string($_GET['id']);
//get the image url
$result = mysql_query("SELECT * FROM users WHERE id='$id'")
or die(mysql_error());
$row = mysql_fetch_array($result);
if($row)
{
$img_url = row['imgurl'];
mysql_close($connect);
}
else{
}
?>
The $Img_url is an actual image link www.asite.com/image.jpg
The problem is how can i use php to make an image from the image link($img_url)? By which http://mysite.com/img.php?id=338576 will turn into an image.
I hope someone could guide me
Thanks!
The easiest way:
header('Location: url/to/image');
You could also proxy the request, which uses your bandwidth:
echo(file_get_contents('url/to/image'));
This is pretty basic stuff. Am I understanding you correctly? I would just do this:
<?php
$img_html = '<img src="' . $img_url . '" />';
echo $img_html;
?>
Or check this answer:
How do I script a php file to display an image like <img src="/img.php?imageID=32" />?