仅显示查询表中存在的MySQL列

Question

I have several MySQL tables that are dynamically generated into a html table one table at a time through the code below. However, the tables don't have the same columns. i.e. One table has a description column, whereas the other does not.

Is the following code the best way to have all the possible MySQL columns among the various tables in the script but only show the MySQL columns that exist for the selected table? I feel like I'm redundant by writing "isset" for every column. Thanks!

<?php
$query = " SELECT * FROM $tablename ";

$query_select = mysqli_query($con,$query);
while($row = mysqli_fetch_array($query_select)) {
?>

<table>
  <tr>

    <?php if(isset($row['name'])){ ?>
      <td><?php echo $row['name'];?></td>
    <?php } ?>

    <?php if(isset($row['description'])){ ?>
      <td><?php echo $row['description']?></td>
    <?php } ?>

  </tr>
</table>

Answer 1

You could just loop through the results. However, that would not perform any checks. Most likely, you'll have to do something like this, depending on what you're actually getting from the database.

<?php foreach ($row as $field): ?>
    <?php if ($field): ?>
        <td><?php echo $field; ?></td>
    <?php endif; ?>
<?php endforeach; ?>

Edit: In keeping in line with the comment you added above, you could simply remove the if clause.