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我得到了我的代碼小問題...因爲我嘗試SELECT
某物從數據庫中,然後INSERT
一些價值到另一個表,但在正常代碼w3school 和我遇到錯誤試圖讓非對象的屬性:使用SELECT
Tryingo獲得非對象
這裏的財產是我的代碼:
<?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";
?>
這就是側面與產品守則
變化請告訴我上線,你所面對的錯誤? – Ali
錯誤在第28行$ result = $ conn-> query($ sqluser); 注意:試着在第28行的C:\ xampp \ htdocs \ car-repair \ insert.php中獲取非對象的屬性。 – artist
@MaciekWiśniewski您的語句序列是錯誤的。使用'$ result = $ conn-> query($ sqluser); 'if'循環之前# – Apb