Display a link if there is an entry in the table.

There is a table tbl_uploads where the paths to the images (x-rays) are stored. There is a priem table where patient records are stored. It is necessary to do so if there is a X-ray picture in the tbl_uploads table that relates to this technique, then output the link. And if not, then just the text that there is no X-ray. I do this so first I push the sessid field into the array (the field where the reception number is stored) from the tbl_uploads table

$rentresult = mysqli_query($con, "SELECT sessid FROM tbl_uploads WHERE size = $id ;" ); $rent = array(); while( $rowr = mysqli_fetch_assoc($rentresult ) ) { $rent[] = $rowr['sessid'] ; 

Then, with the second query, when I pull out all the records from the priem table, I check if there is a session number in the $rent[] array and if not, then the text is not X-ray and if there is that link

 $result2 = mysqli_query($con,"SELECT * FROM priem WHERE card_id=$id") or die("<p>Не удается выбрать запись из базы данных</p>". mysqli_error($con)); ?><?php while($row2 = mysqli_fetch_assoc($result2 )) { $has_xray='Нет рентгена'; if (in_array($row2['sessid'] , $rent)) { $has_xray='Есть рентген'; } 

And finally, this way I display

 <td> <?php echo $has_xray ; ?> </td> 

The trouble is that even when there is an x-ray in the table, it still shows that there is no x-ray. Where do I make a mistake? Can a condition, can I check it wrong?