链接到页面动态时，仅显示所有链接中第一个表的信息

Question

In the site that I am creating presents in a page all the news with 1 dynamic link that this creates a page with the content of this news however all generated links will always stop to the first value of the database and not to the ID that is intended. Someone can help me? Thank you.

Page to generate links

https://imgur.com/9jS9mMv

Page of any displayed link shows only the first value of the table.

https://imgur.com/pWBNY2K

if(isset($_GET['id'])){

    $ID = mysqli_real_escape_string($db, $_GET['id']);

    $db = mysqli_connect("localhost", "root", "root", "base_de_dados_museu");

    $sql = "SELECT * FROM destaques where id='$ID'";
    $result = mysqli_query($db, $sql) or die ("bad query: $sql");
    $row = mysqli_fetch_array($result);
    $result = mysqli_query($db, $sql) or die ("bad query: $sql");

    if(isset($_GET['post'])){
        $titulo = mysqli_real_escape_string($db, $_GET['titulo']);
        $image = mysqli_real_escape_string($db, $_GET['image']);
        $texto = mysqli_real_escape_string($db, $_GET['texto']);

        $sql = "INSERT INTO destaques (id, titulo, texto, image) VALUES ('$titulo', '$image', '$texto')";
        $result = mysqli_query($db, $sql) or die ("bad query: $sql");
    }
}

Answer 1

There are some errors in the code. I will list some of them but maybe there are others...

Four columns here but only three values:

$sql = "INSERT INTO destaques (id, titulo, texto, image) VALUES ('$titulo', '$image', '$texto')";

The result variable is overwritten:

$sql = "SELECT * FROM destaques where id='$ID'";
    $result = mysqli_query($db, $sql) or die ("bad query: $sql");
    $row = mysqli_fetch_array($result);
    $result = mysqli_query($db, $sql) or die ("bad query: $sql");