Select first image from user SQL

Question

I have a query which shows all images from a user. (In this case the user_id = 3.)

I want to return the results of the query and display near the user record, a static picture which I have uploaded. This picture is a small icon.

Here is my sql query:

$sql = "SELECT username as user, p.image as user_image, i.image, i.id as image_id, i.description as text, UNIX_TIMESTAMP(i.date) as image_date, COALESCE ( imgcount.cnt, 0 ) as comments
        FROM users u
        LEFT JOIN images i ON i.user_id = u.id
        LEFT JOIN images p ON p.id = (SELECT b.id FROM images AS b where u.id = b.user_id ORDER BY b.id DESC LIMIT 1)
        LEFT JOIN (SELECT image_id, COUNT(*) as cnt FROM commentaries GROUP BY image_id  ) imgcount ON i.id = imgcount.image_id
        WHERE i.user_id = 3
        ORDER BY i.date DESC";

And here are the results:

[images_list] => Array
    (
        [0] => Array
            (
                [user] => 3333
                [user_image] => http://127.0.0.1/auth_system_1/upload_images/24/24_nsm5rixy14lexm9cy15wzyg9u_224.jpg
                [image] => http://127.0.0.1/auth_system_1/upload_images/224/224_nsm5rixy14lexm9cy15wzyg9u_224.jpg
                [image_id] => 5
                [text] => 
                [image_date] => 7 hours, 1 minute
                [comments] => 2
            )

        [1] => Array
            (
                [user] => 3333
                [user_image] => http://127.0.0.1/auth_system_1/upload_images/24/24_nsm5rixy14lexm9cy15wzyg9u_224.jpg
                [image] => http://127.0.0.1/auth_system_1/upload_images/224/224_gfbyjh6zf66g914e28bsfdkuf_f4d.jpg
                [image_id] => 3
                [text] => 
                [image_date] => 20 hours, 50 minutes
                [comments] => 0
            )

        [2] => Array
            (
                [user] => 3333
                [user_image] => http://127.0.0.1/auth_system_1/upload_images/24/24_nsm5rixy14lexm9cy15wzyg9u_224.jpg
                [image] => http://127.0.0.1/auth_system_1/upload_images/224/224_80jrg3z0xrh9isskc3yuhtqh1_163.jpeg
                [image_id] => 1
                [text] => test1
                [image_date] => 1 day, 22 hours
                [comments] => 0
            )

    )

I do not have any idea how to place a small icon near the first image that is returned from the query. If I had one more field with a 1 indicating it is the first record, or a 1 indicating it is not the first record, I think I could do what I want.

By the way, the fields of my images table are as follows:

• id
• user_id
• image
• date

Is it possible to add one more field to the query, which will indicate with a 1 or 0 whether or not it is the latest record in the result set when sorting by date?

Maybe you can show me an easier way to do this?

Answer 1

It would be silly to make your SQL database add this field when you can do it very easily in PHP.

Just create a boolean variable that is set to false on the first iteration of your loop.

$first = true;

for( $data as $key => $value )
{
    if( $first )
    {
        echo "I'm the first row";
        $first = false;
    }

    // Do other stuff
}

---

EDIT: How to identify first and last item in an array

$keys = array_keys( $data );
$values = array_values( $data );

$count = count( $data );

for( $a = 0; $a < $count; $a++ )
{
    $key = $keys[ $a ];
    $value = $values[ $a ];

    if( $a == 0 )
    {
        // First
    }
    else if( $a+1 == $count )
    {
        // Last
    }

    // All
}

Answer 2

Do you mean you want the first image that appears in the array or the first image with the smallest id?

The first image in the array returned in print_r should be accessible using

$images_list[0]['image'].     

Does this not work?

If you wanted to find the first image in the array using a foreach statement, you could add a flag before the loop, like:

$firstImage = 0;    
foreach ($images_list as $key)     
{    
  $firstImage++;    
  if (firstImage == 1 )     
  {    
    $imageSrc = $key['image'];    
  }    
}    

Please note the code above has not been tested, just a guide.