我的代码通知错误:试图获取[重复]中非对象的属性

Question

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<?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:data

提前谢谢你。

Answer 1

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$sql = "SELECT * FROM data ORDER by id DESC LIMIT 1";

Answer 2

似乎您正在将sql server语法与MySQL混合在一起。

用于sql Server。在mysql中，您必须使用 来获取特定数量的记录，例如。

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或者只需使用