Trying to get property of non-object: Using SELECT

Question

I got little problem with my code... because I try to SELECT sth from database and then INSERT some value to another table, but in normal code from w3school and I got error

Tryingo to get property of non-object

Here is my code:

<?php 
    session_start();

    function connectionDB(){
        $host = "localhost";
        $username = "root";
        $password = "";
        $db_name = "project";

        $conn = new mysqli($host, $username, $password, $db_name);
        if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
        } 
        return $conn;
    }
    function getID(){
        $id = $_GET['id'];

        return $id;
    }
    $login =$_SESSION['login'];
    $conn =connectionDB();
    $idCar = getID();
    echo $idCar;
    $sqluser = "SELECT ID_USER FROM login_table WHERE LOGIN = $login";

    if($result->num_rows > 0)
    $result = $conn->query($sqluser);
    {
        while($row = $result->fetch_assoc())
        {
            $sql = "INSERT INTO cart (id_user) VALUES ('".$row['ID_USER']."')";
        }
    }else echo"error";
?>

THIS IS THE CODE OF SIDE WITH PRODUCT

The error is in line 28 $result = $conn->query($sqluser); Notice: Trying to get property of non-object in C:\xampp\htdocs\car-repair\insert.php on line 28 — artist
– artist, Commented Jan 21, 2016 at 12:39
@MaciekWiśniewski your statement sequence is wrong. Use $result = $conn->query($sqluser); before if loop — Apb
– Apb, Commented Jan 21, 2016 at 12:42
@Abp Yeah i changed it but is still the same error, in line with if line if($result->num_rows >= 0) — artist
– artist, Commented Jan 21, 2016 at 12:46

Apb · Accepted Answer · 2016-01-21 12:44:00Z

2

Please see change near your IF loop

    $sqluser = "SELECT ID_USER FROM login_table WHERE LOGIN = $login";
    $result = $conn->query($sqluser);  //check result first

    if($result->num_rows > 0)  //get number of rows
    {
        while($row = $result->fetch_assoc())
        {
            $sql = "INSERT INTO cart (id_user) VALUES ('".$row['ID_USER']."')";
        }
    }else echo"error";

answered Jan 21, 2016 at 12:44

Apb

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