In the context of creating a table layout for displaying images from a database, what are some common pitfalls to avoid when iterating through query results?

One common pitfall to avoid when iterating through query results to display images in a table layout is not properly handling cases where there are no images to display. This can result in errors or empty table cells. To solve this, you should check if there are any images in the query results before attempting to display them in the table. 

// Check if there are any images in the query results
if(mysqli_num_rows($result) > 0) {
    // Iterate through the query results and display images in table cells
    while($row = mysqli_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td><img src='" . $row['image_url'] . "'></td>";
        echo "</tr>";
    }
} else {
    // Display a message if there are no images to show
    echo "<tr><td colspan='2'>No images found</td></tr>";
}

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