具有2个左连接的移动应用程序的PHP脚本

I'm currently working on an Android app that needs to get and send data via PHP to a server with a MySQL database.
I'm relatively new to PHP and I couldn't find any specific help for my problem.

Here is my code

<?php
    $db_name = "testwz";  
    $mysql_user = "root";  
    $mysql_pass = "";  
    $server_name = "localhost";  
    $con = mysqli_connect($server_name, $mysql_user, $mysql_pass, $db_name);  

    $sql = "SELECT verbrauchsstelle.ID, adressen.Ort, adressen.Ortsteil,
                adressen.PLZ, adressen.Strasse, adressen.Hausnummer, zaehler.Nummer
            FROM verbrauchsstelle
            LEFT JOIN adressen ON verbrauchsstelle.adressen_id = adressen.ID 
            LEFT JOIN zaehler ON zaehler.Verbrauchsstelle_id = verbrauchsstelle.ID;
    ";

    $result = mysqli_query($con, $sql);
    $response = array();

    while($row = mysqli_fetch_array($result))
    {
        array_push($response, array(
            "vID" => $row[0],
            "Ort" => $row[1],
            "Ortsteil" => $row[2],
            "PLZ" => $row[3],
            "Strasse" => $row[4],
            "Hausnummer" => $row[5],
            "zNr" => $row[6]
        ));
    }

    echo json_encode(array("server_response" => $response));
    mysqli_close($con);
?>

The SQL itself is doing fine when used with SQLyog; when I delete one of the LEFT JOINs I get an output but using both of them, I get a blank white screen in the browser. I tried LEFT JOIN and INNER JOIN with the same result.

Maybe someone with way better PHP knowledge can help me find the error.

Thanks in advance.

You can try this:

<?php
$db_name = "testwz";  
$mysql_user = "root";  
$mysql_pass = "";  
$server_name = "localhost";  
$con = mysqli_connect($server_name, $mysql_user, $mysql_pass, $db_name);  

$sql = "SELECT verbrauchsstelle.ID, adressen.Ort, adressen.Ortsteil,
            adressen.PLZ, adressen.Strasse, adressen.Hausnummer, zaehler.Nummer
        FROM verbrauchsstelle
        LEFT JOIN adressen ON verbrauchsstelle.adressen_id = adressen.ID 
        LEFT JOIN zaehler ON zaehler.Verbrauchsstelle_id = verbrauchsstelle.ID
";

$result = mysqli_query($con, $sql);
if (!mysqli_error($con)) {
   echo json_encode(array("server_response" => mysqli_error($con)));
   mysqli_close($con);
   exit;
}

$response = array();

while($row = mysqli_fetch_array($result))
{
    array_push($response, array(
        "vID" => $row[0],
        "Ort" => $row[1],
        "Ortsteil" => $row[2],
        "PLZ" => $row[3],
        "Strasse" => $row[4],
        "Hausnummer" => $row[5],
        "zNr" => $row[6]
    ));
}

echo json_encode(array("server_response" => $response));
mysqli_close($con);

This will show you the exact error if any else it will give you the desired output what you want.