MySQL数据库存储图像的URL和PHP页面显示吗？

Question

I was wondering if there's any way I could store an image URL in a database with a unique numeric id, and a php page display the image by getting data (the image URL) from the database and displaying it to the user on the image.php page?

An example of this would be: - Having the image URL http://www.example.com/images/image10.jpg (where this URL would be stored in the database, alongside an id of 10) - Having an image.php page which retrieves the image URL from the database and displays the image on the same page (the image.php page) (and doesn't redirect to the image URL) - an example result being image.php?id=10 where the image.php file would source the image URL from the database with an image id of 10 (using my example: www.example.com/images/image10.jpg) and display it (so it would be like viewing the image as if you were viewing www.example.com/images/image10.jpg but instead the URL would be http://www.example.com/image.php?id=10 )

Is there any way to do this? I'm new to php and MySQL so not really good with code - trying to experiment :)

Thanks in advance!

Edit: I think people are getting confused by my bad description (sorry abot that). I've noticed in your answers that you have included the actual image URL in your code - I dont understand why though.

EXAMPLE of how I'd like this to work:

My table would have these feilds: img_id img_url

The image.php file would GET the img_url FROM the table using the id on the end of the image.php url. FOR EXAMPLE: - image.php?id=10 - the image.php file would use the id (which is 10), and display the image by retreiving the img_url (which would be: www.example.com/images/image10.jpg) from the table and displaying it - the image id would be img_id in the table. - image.php?id=12 - the image.php file would use the id (which is 12) and display the image by retreiving the img_url (which would be: www.example.com/images/image12.jpg) from the table and displaying it - the image id would be img_id in the table.

Answer 1

You can try this:

//File: image.php?id=10 
//you query   
$img = 'http://example.com/sample.jpg'; // Image path from database
$getInfo = getimagesize($img);
header('Content-type: ' . $getInfo['mime']);
readfile($img);

Answer 2

I think what you need is this :

$image_path="http://example.com/image10.jpg";' 'mysql_query("INSERT INTO table VALUES ('',$image_path)");

in the php have this :

$id=$_GET['id'];if (isset($_GET['id'])){' '$result = mysql_query("SELECT path FROM table WHERE id=$id");' '$row = mysql_fetch_row($result);' '$path=$row[0];echo "Image number ".$id." is <img src='".$path."'>';}

Answer 3

It all depends on your code organisation, let me explain my self, right now in my mind i can think of two ways of organisation:

First, using a folder containing all images, with the same name + number for each image; eg: images/img1.jpg - images/img2.jpg - images/img3.jpg etc

Using this method you'll not have to use even database, just within your page you go for this code, exemple for page http://exemple.com/images.php?id=10

$id = (isset($_GET["id"]))? $_GET["id"] : NULL; //Get the ID

Then use the method given by user3410107

$img = "http://example.com/img$id.jpg"; // Image path
$getInfo = getimagesize($img);
header('Content-type: ' . $getInfo['mime']);
readfile($img);

Second method is indeed using database, in this case there should be at least 2 columns within db, a unique ID + Images path, in this case you'll go for:

$query = mysql_query("SELECT id, imagePath FROM table WHERE id = $id");
while($data = mysql_fetch_assoc($query)){
$img = $data["imagePath"];
}

Once you get the image path, the same process as before:

$getInfo = getimagesize($img);
header('Content-type: ' . $getInfo['mime']);
readfile($img);

For security reasons add this line after $id variable

$id = (isset($_GET["id"]))? $_GET["id"] : NULL; //Get the ID
$id = mysql_real_escape_string($id);