我的代码没有什么问题...因为我尝试从数据库中SELECT
某事,然后INSERT
一些值到另一个表,但在w3school的正常代码中
我收到错误
Tryingo to get property of non-object
这是我的代码:
<?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";
?>
这是产品侧面的代码
最佳答案
请查看 IF 循环附近的更改
$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";
关于php - 尝试获取非对象的属性: Using SELECT,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34923955/