Delete first post and show second in PHP - php

Question

I have a question in PHP.
I am creating a website with posts but I can't make the PHP to show just the second post (without show the first).
My code is like this:
<?php
$result = mysqli_query($dbc, "SELECT * FROM projects");
$x = 1;
while($row = mysqli_fetch_array($result)){
$nome = $row['name'];
$conteudo = $row['description'];
$imagem = $row['image'];
$imagem2 = $row['image2'];
?>
<?php static $count2 = 0; if ($count2 == "1") { break; } else { ?>
<div class="content justify" id="projects-<?php echo $x; ?>" >
<?php echo $conteudo; ?>
<?php if(!empty($imagem2)) { ?>
<img class="hide-for-small" src="images/contebt/project/<?php echo $image2; ?>">
<?php }; ?>
</div>
<?php $count2++; } ?>
<?php $x++;}; ?>
With this code I can show just the first post, but I want to show just the second. Can anybody help me, please? Thanks!

Answer 1

This should work with this code :
PHP
<?php
$result = mysqli_query($dbc, "SELECT * FROM projects");
$x = 0;
while($row = mysqli_fetch_array($result)) {
$nome = $row['name'];
$conteudo = $row['description'];
$imagem = $row['image'];
$imagem2 = $row['image2'];
if (!$x) {
$x = 1;
continue;
}
?>
<div class="content justify" id="projects-<?php echo $x; ?>" >
<?php echo $conteudo; ?>
<?php if(!empty($imagem2)) { ?>
<img class="hide-for-small" src="images/contebt/project/<?php echo $image2; ?>">
<?php } ?>
</div>
<?php
$x++;
}
?>
SQL
But you should directly espace the first row directly in mysql usign either limit :
$result = mysqli_query($dbc, "SELECT * FROM projects LIMIT 1, 1");
or where statement :
$result = mysqli_query($dbc, "SELECT * FROM projects where id > 1");