html中的Controllo dati数据库javascript

我在检查数据库中的数据以查看图像时遇到问题。 这是.php页面的代码

<?php
$servername = "";
$username = "";
$password = "";
$dbname = "";
$getarray = [];
$i=0;

$cookie_name="id";
$iden=$_COOKIE[$cookie_name];
while ($i<6){
    //Column Selection
    if ($i==0){
        $getcolumn="username";
    }elseif($i==1){
        $getcolumn="vita";
    }elseif($i==2){
        $getcolumn="sanita";
    }elseif($i==3){
        $getcolumn="soldi";
    }elseif($i==4){
        $getcolumn="guerra";
    }
    //Open connection for query
    $conn = mysqli_connect($servername, $username, $password, $dbname);
    //query
    $sql = "SELECT * FROM user WHERE id LIKE '$iden'";
    $result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$get= $row[$getcolumn];//Select the column
}
} else {
echo "0 results";
}
array_push($getarray, $get);
$i=$i+1;
}
?>
<script>
  let data = <?php echo json_encode($getarray); ?>;

  //Debug Alert
  //for(var i=0; i<data.length; i++){
  //alert(data[i]);
  //}

  //Log data to console
  //for(let i = 0; i < data.length; i++){
  //  console.log(data[i]);
  //}
  console.log(data[4])

</script>

<div id="x"></div>
<script type="module">
    var a=1;
    if(a===data[4]){
        var img = document.createElement("img");

        img.src = "img/generale/logo.png";
        var src = document.getElementById("x");

        src.appendChild(img);
    }
</script>