PHP查询不通过表单在db中更新

Question

i'm working on a project that includes a minimail/chat function that has the option to tick off when u have read the message and flag a user.

The flag and tick off options doesnt work anymore after i eddited something but i cannot find the problem anymore.

The game doesnt support newer versions of PHP for now.

• $m['id']; = 22
• $m['read'] = 0

Query 1 = array(3) { ["id"]=> string(2) "28" ["mark_read_x"]=> string(2) "10" ["mark_read_y"]=> string(2) "10" }

Query 2 = array(3) { ["id"]=> string(2) "28" ["flag_user_x"]=> string(1) "8" ["flag_user_y"]=> string(1) "4" }

<?php
   $mid = $m['id'];

   if(isset($_POST['mark_read'])){
    $mid2 = $_POST['id'];
    $query2 = "UPDATE `user_messages` SET `read` = '1' WHERE `read` = '0' AND `user_messages`.`id` = '$mid2'";
    mysql_query($query2) or die(mysql_error());
    header("Refresh:0");
    }
?>

<?php if($m['read'] === '0'){ ?>

<form method="post">
   <input type="hidden" name="id" value="<?php echo $mid; ?>">
   <input src="tick.png" title="Mark as read" type="image" alt="submit" name="mark_read">
</form>

<?php } else { echo ''; } ?>

<?php
   if(isset($_POST['flag_user'])){
    $mid2 = $_POST['id'];
    $query3 = "UPDATE `user_messages` SET `flagged` = '1' WHERE `flagged` = '0' AND `user_messages`.`id` = '$mid2'";
    mysql_query($query3) or die(mysql_error());
    header("Refresh:0");
   }
?>

<form method="post">
   <input type="hidden" name="id" value="<?php echo $mid; ?>">
   <input src="exclamation.png" title="Flag message" type="image" alt="submit" name="flag_user">
</form>

Answer 1

Changed

<input src="tick.png" title="Mark as read" type="image" alt="submit" name="mark_read">

to

<button type="submit" name="mark_read" style="padding: 0; border: none;"><img title="Mark as read" src="tick.png" /></button>

Input type failed to submit data because of the image type. works fine now.