Btukfyl

i have spent a day now but i am not able to solve the following problem.
I have this code to access my database and i want to get some values.
The weird thing is, that the same code works for the same php file on the same server in the same directory.
I wanted to create a new php file with a different sql selection, but this one always returns null instead of a json array.

Any helpful advice would be great!
I am running a wordpress site, maybe thats an important information.

<?php

if( isset( $_POST[user_id] ) )

{

    $user_id = $_POST[user_id];



    $servername = "****";

    $username = "****";

    $password = "****";

    $dbname = "****";



    // Create connection

    $conn = new mysqli($servername, $username, $password, $dbname);



    // Check connection

    if ($conn->connect_error) {

        die("Connection failed: " . $conn->connect_error);

    } 

    //echo "Connected successfully";

    //echo $user_id;





    $sql = "SELECT *

    FROM felderkonfig

    WHERE felderkonfig_user_id = '$user_id' ";



    $result = mysqli_query($conn, $sql);    //sql abfrage in array speichern

    $myArray = array();

    $myArray =mysqli_fetch_assoc($result);  //umwandeln in assoziatives array



    echo json_encode($myArray);     //echo als json array





    $conn->close();

    exit(); 

}else{

    echo "Fehler bei if isset";     

}

?>

You should include this file to the prefered php file like where ever you want to connect the database

<?php include 'connect.php' ?>

in connect.php file should have the connection settings like what you metioned in the question exluded selct query

I dont' know why but the code is working know. The behaviour was quite weird, maybe there were some server-side problems, but i don't know. It would be great if someone coulde tell me how i can make the folllwing code safer. Thanks a lot.

<?php

if( isset( $_POST[virtuellesdepot_id] ) )

{

    $virtuellesdepot_id = $_POST[virtuellesdepot_id];



    $servername = "****";

    $username = "****";

    $password = "****";

    $dbname = "****";



    // Create connection

    $conn = new mysqli($servername, $username, $password, $dbname);



    // Check connection

    if ($conn->connect_error) {

        die("Connection failed: " . $conn->connect_error);

    } 

    //echo "Connected successfully";



    //echo $user_id;





    $sql = "SELECT * FROM virtuelledepots ";   



    $result = mysqli_query($conn, $sql);    //sql abfrage in array speichern

    $myArray = array();

    $myArray = mysqli_fetch_assoc($result); //umwandeln in assoziatives array





    echo json_encode($myArray);     //echo als json array





    $conn->close();

    exit(); 

}else{

    echo "Fehler bei if isset"; 

}

?>