Question

这是一个更新表中字段值的脚本。查询后我收到错误：

致命错误：在非对象上调用成员函数close（）

<?php
    error_reporting(E_ALL);ini_set('display_errors','On');

    define('DB_HOST', 'xxxxxx');
    define('DB_NAME', 'xxxxxx');
    define('DB_USER','xxxxxx');
    define('DB_PASSWORD','xxxxxxx');

    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die("Failed to connect to MySQL: " . mysqli_error());

     if ($mysqli->connect_errno) {
        echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
    }

    $result = $mysqli->query("update people_table set FirstName = 'Jon' where LastName = 'Smith'") or die($mysqli->error);
    $result->close();

    ?>

$ result是我视图中的一个让我困惑的对象

Answer 1

我认为这是你需要写的而不是结果对象

$mysqli->close();

我在非对象上调用成员函数close（）时收到致命错误

1 个答案: