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Showing results for tags 'rotate'.
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I am working on trying to get php code to take a record set from mysql table and rotate the records, so when you refresh the page, the first record would show at the top of the list, then record 2 and so on down the line. If you were a new person to come to the page, record 2 would show at the top of the list, then record 3 and down the line, and record 1 would now be at the bottom. I have a position column in the table and when the page loads it will update the position field of each record to be one less then it was before, so when the next person to come to the page, it will be up one spot in the table. My issue is that when the page gets refreshed to quickly or I have had it not finish updating the records and then there is multiple records with the same position number. I am looking to only have 9 records only for the example, so if Position is set to 1 it will be set to 9, if not it will take Position - 1. if(isset($_GET["page"])){ } else{ $sql = "SELECT ID,Name,Position FROM tableA ORDER BY Position DESC"; $result = $conn->query($sql); while ($row = $result->fetch_assoc()) { if($row["Position"] == 1){ $p = 9; } else{ $p = $row["Position"]-1; } $conn->query("UPDATE tableA SET Position = $p WHERE ID = $row[ID]"); } mysqli_free_result($result); } Am I going down the right path? It just seems like it will not work 100% of the time. Hopefully I explained what is going wrong fully and clearly. Thank in advance.
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Hi, New coder here! I'm making a php If loop that will post my image a set number of times (depending on user input in a text box), and I have that all figured out. However, I can't figure out how to rotate the image a random amount of degrees and have a random width/height for each image. Will rand generate new numbers for each time the image is echo-ed, or will it give the same number for all images? This is what I have: <?php function pic($picture) { $degree = rand(0,360) $num = rand(50,250) $total = 0; while ($total < $picture){ echo '<img src=blabla.gif width="$num" height="$num">'; $total = $total + 1; } } $picture = $_GET['input']; $a = pic($picture); } ?> Thanks in advance!