Foreign key in phpmyadmin giving error in newsfeed - php

Question

I am working on newsfeed and I hope I this right but I want that the image that is uploaded, is linked to the user who uploads it. I have 2 tables.
users in here the accounts are stored.
picas in here the uploaded images are stored.
I have in users a primary key called user_id and I have in the picas table a primary key called id.
I made in picas a new colomn called user_id and when an image is uploaded I tell PHP to get the id from the user who is logged in and then insert that within the user_id colomn from the table picas so I can see which user uploaded the image.
So my users table looks like this:
+-------------+-------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+-------------+------+-----+---------+----------------+
| user_id | int(1) | NO | PRI | NONE | AUTO_INCREMENT |
+-------------+-------------+------+-----+---------+----------------+
And my picas table looks like this:
+-------------+-------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+-------------+------+-----+---------+----------------+
| id | int(1) | NO | PRI | NONE | AUTO_INCREMENT |
+-------------+-------------+------+-----+---------+----------------+
| user_id | int(1) | NO | | NONE | |
+-------------+-------------+------+-----+---------+----------------+
For now I have a problem with the newsfeed. It displays different users with their avatar and names, but if I post something when I'm logged in on an account, it posts for every user. So if I have 5 different users, I log in, I upload an image, then I see 5 posts instant. Check my screenshot here for better display --> http://i.imgur.com/VXwlbra.png
Same goes for the title if you click the image. If I give up a title, then it shows also on the other users.
My code when I upload:
uploadfile.php here I store the uploaded file into the database.
<?php
if (isset($_POST['submit_pica'])) {
$newName = md5(time().$_FILES['pica']['tmp_name']).'.jpeg';
$postTitle = $_POST['postTitle'];
$ses_user = $_SESSION['username'];
$getuserid = mysqli_fetch_assoc($mysqli->query("SELECT id FROM users WHERE username='$ses_user'"));
$userid = $getuserid['id'];
$result = $mysqli->query("INSERT INTO picas (id, name, title, created_at, user_id)
VALUES (null, '$newName', '$postTitle', null, '$userid')");
echo "<script>
$('.upload_button').click(function() {
$('#uploadform').slideUp(300);
});
</script>";
move_uploaded_file($_FILES['pica']['tmp_name'], 'upload/'.$newName);
}
?>
showimages.php (here I output the images from the database.
<?php
$result = $mysqli->query("SELECT *
FROM picas, users
ORDER BY created_at DESC");
$ses_user = $_SESSION['username'];
while($pica = $result->fetch_assoc()) {
echo '<div class="image_post">
<div class="user_avatar"><img src="avatars/'.$pica['username'].'.jpeg" /></div>
<div class="user_name">'.$pica['username'].'</div> <br><br><br>
<div class="image_title">'.$pica['title'].'</div>
<img src="upload/'.$pica['name'].'" />
</div>';
}
?>
My friend says I had to create a foreign key so it created a relationship between the user_id from picas and user_id from users.
Apparently it didn't work. I got the latest version from everything.
I can only choose from: users or picas table (logic, because those are the only ones I have) and then in the option field right next to it, I can only choose id or user_id (depends on which table I choose).
It gives me the this error: http://i.imgur.com/2ukfKsH.png
In the imgur link you see also what I fill in. I go to the table: picas > structure > relation view
It isn't working. I hope using the foreign key is the solution to this problem with the newsfeed, because I want that when the image is uploaded, it outputs the user who uploaded it with the uploaded image and not that it does that + every other user that is stored in the database. Please help me. I struggle with this for already a week. It's driving me crazy!

Answer 1

Thanks for the care you've taken writing this question.
Pro tip: Avoid using SELECT * in queries in software. Using SELECT * allows you to be sloppy in your thinking about what you want in your result set.
Pro tip: If you want results from more than one table, use JOIN directives rather than a comma-separated list of tables. This makes you think clearly about how your tables relate to each other.
You're using this query to display your items.
SELECT * /*wrong!*/
FROM picas, users /*combinatorial explosion!*/
ORDER BY created_at DESC
This query assembles a result set using every possible combination of rows from picas and users. Lucky for you you don't have 50 users and 100 entries in picas yet, or this query would yield 5000 rows.
What you need for your query is something like this:
SELECT users.user_id, users.username,
picas.id, picas.name, picas.title, picas.created_at
FROM users
JOIN picas ON users.user_id = picas.userid
WHERE users.username = '$ses_user'
ORDER BY picas.created_at DESC
This will construct a resultset with the columns I've mentioned. It will use the criterion in the ON clause to join the data from the two tables in a way that makes sense.