0
<?php
$result = ExibirDados();
while ($resultado = mysqli_fetch_array($result))
{ ?>
<tr>
<td><img <?php echo "src=upload/".$resultado.">"; ?> </td>
<td>
<?php } ?>
is giving this error Notice: Array to string Conversion in C: xampp htdocs Constructesfree Dozero index.php on line 150 src=upload/Array>
Would not be
$resultado["nome_do_campo_imagem_na_tabela"]
?– Isac
correct, thank you worked out !
– Hamilton Ventura