PHP-MySQL: Insert and Display Photos

It would be good if you could share your results with us. So that we can solve our problems as well.

Hello,

To maintain the structure of your table, you need to proceed as follows:

- Create a new file image.php, containing:

<?php $id = $_GET['id']; // Place your database connection parameters here // We retrieve the image $sql = "SELECT photo FROM objects WHERE id='".$id."'"; $result = mysql_query($sql); // We set the headers to indicate that this is indeed a photo. header('Content-type: image/jpeg'); // cf (1) header('Content-transfer-encoding: binary'); // And we send the sauce echo $result["photo"]; ?>

(1) : this will work for a jpeg image. If you have multiple image types, I recommend saving the MIME type in the database by adding a "type" column to your table.

- Then, in your main page, add

<form action="furniture.php" method="POST"> <table cellspacing="0" cellpadding="0" border="1"> <?php if($result) { while($data = mysql_fetch_assoc($result)) { printf ( "<tr> <td><img src='image.php?id=%s' /></td> <td>%s</td> <td>%s</td> <td>%s</td> </tr>", $data["id"], $data["name"], $data["description"], $data["country"] ); } // end while if(mysql_num_rows($result) == 0) { $msg.="No information available"; } } ?> </table> </form>

Well, there you go, I don't have anything to test it with, I hope it's okay ^^

Xavier

Hi,

In my opinion, the best way would be to put all your photos in a folder, like "images" or "photos", in the same directory as your file, so as not to make things more complicated....

Then, for the display, you add a field in your database like "path", in relative form, e.g.: "photos/image1.jpg"

Finally, you insert the photo like this: <img src="$data["path"]"...

P.S.: I use mysql_fetch_array, which puts the fields in an array... so check for mysql_fetch_assoc...

--
All right, bye!!!

Je ne peux pas traiter ce type de demande.

Hello,

What does your database contain? Does it only contain the address of the photo?
In the case of the message I published, the database did not contain the address, but the photo itself encoded in binary. So echo $result["photo"] will indeed return the image. However, if, as I believe, the database only contains the address of the photo, that same line will return... just the address of the image!

Is that correct?

Xavier

My problem is actually to upload the image selected by browsing on a web page. I would like your help in order to accomplish this task.

Xavier,

I made the changes following your recommendations, but when I run my main file, the table displays without photos (in fact, a small icon appears as if the image in question was not found).

If I run the "image.php" file, then I receive the following error messages:

Notice: Undefined index: id in c:\program files\easyphp1-8\www\test.php on line 3 Warning: Cannot modify header information - headers already sent by (output started at c:\program files\easyphp1-8\www\test.php:3) in c:\program files\easyphp1-8\www\test.php on line 19

What could this be due to?

Thank you.

Hello,

The error message when you launch the image.php file is because you are calling it without putting ?id=idImage after it.

Try calling it like this, with an image id that you have, and let me know how it goes.

Xavier

To continue the discussion, I don't have any error messages, and the only thing that appears is the URL of the address when I go to
http://localhost/image.php?id=3

The field I’m targeting is indeed of type blob and contains binary data. (By the way, I initially added it "hardcoded" directly via phpMyAdmin).

...

Well, I just modified a few lines that I didn't think would have any impact, but the result is that it's working :)

I also had another issue with the upload, but I solved that too.

Thanks for the response, and sorry for the inconvenience :)

<img src="<? echo($data['field_9']); ?>">
or
<img src="<?= $data['field_9'] ?>">