php - 我的代码错误通知 : Trying to get property of non-object in

Question

我是编程新手。 我编写了一个代码来获取最后添加的记录，但它抛出了一个错误。

Notice: Trying to get property of non-object in C:\xampp\htdocs\fetch.php on line 17 0 results

获取.php

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "d4rky";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT TOP 1 * FROM data ORDER by id DESC LIMIT 1";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        echo "ID: " . $row["id"]. " - Name: " . $row["dataset"]. " " . $row["datamet"]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?>

表名称:数据

提前致谢。

Answer 1

看来您正在将 sql server 语法与 mysql 混合在一起。

SELECT TOP 1 用于sql server。在 mysql 中，您必须使用 Limit 来获取特定数量的记录，例如。

从表名中选择 field1、fileld2、field3 ORDER BY id DESC LIMIT 1;

或者只需使用即可影响所有字段

SELECT * FROM table_name ORDER BY id DESC LIMIT 1;