How to place an image inside echo and not show if it is empty

Question

I made the code to show only if you have image, but it also shows the body of empty, it says in the source code of the page that my PHP is a comment as shown in the image below.

And the code is this:

How do I not show the empty pictures the way you are here?

echo '

                  <php

                  $seleciona = mysqli_query($conexao, "SELECT * FROM postagem where status=1 ORDER BY id desc");
                  while($campo=mysqli_fetch_array($seleciona)){
                    $nome_imagem = $campo["nome_imagem"];                    
                    ?>
                    <center>
                    <div id="panel" align="left">
                    <label class="titulo">&nbsp;&nbsp;'. $result1[$i][1] . '</label><br>
                    <p class="descricao">'. $result1[$i][2] . '</p><br>

                    <?php if ($nome_imagem != null){?><p><img src="foto/'. $result1[$i][3] . '" class="foto"></p><?php } ?>

                    <span class="glyphicon glyphicon-user" aria-hidden="true"></span>&nbsp;&nbsp;Postado por: '. $result1[$i][6] . ' em '. $result1[$i][4] . ' às '. $result1[$i][5] . '</td>&nbsp;&nbsp;

                    <a href="0_excluir_postagem.php?editaid='. $result1[$i][0] . '"><span class="glyphicon glyphicon-trash"></span></a>&nbsp;&nbsp;     
                    </div></center>  

                    <?php }?> 
                    </div>' ; 

                  }

Answer 1

Why you do not create a verification method if the image is valid:

<?php 
function checkImage($img = null, $nome = null)
{
   if ($img != null && $nome != null) {
       return true;
   }
   return false;
   
} ?>


<?php if (checkImage($result1[$i][3],$nome_imagem)):?>
    <p><img src="foto/'. $result1[$i][3] . '" class="foto"></p>
<?php endif; ?>

Answer 2

Alter:

<?php if ($nome_imagem != null){?><p><img src="foto/'. $result1[$i][3] . '" class="foto"></p><?php } ?>

To:

<?php if (!empty($nome_imagem)) { ?><p><img src="foto/'. $result1[$i][3] . '" class="foto"></p><?php } ?>

I don’t understand the logic of testing a variable ($nome_imagem) and display another ($result1[$i][3]). But it’s okay. It’s not the point of the question, let alone the answer.

Back to the question, about the empty():

Returna FALSE if the tested variable exists and is not empty. Otherwise, returns TRUE.

empty() considers the following cases as emptiness:

• "" (an empty string)
• 0 (0 as an integer)
• 0.0 (0 as a floating point)
• "0" (0 as a string)
• NULL
• FALSE
• array() (an empty array)
• $var; (a declared variable but no value)

Reference: PHP - Empty

Compare the variable with null would ignore all these other mentioned cases. (if ($nome_imagem != null){ //...).