Question

我的PHP代码有问题，我已经阅读了很多答案，但我找不到答案。我有两个文件：db_connect.php

<?php

/**
* A class file to connect to database
*/
class DB_CONNECT {

// constructor
function __construct() {
    // connecting to database
    $this->connect();
}

// destructor
function __destruct() {
    // closing db connection
    $this->close();
}

/**
 * Function to connect with database
 */
function connect() {
    // import database connection variables
    require_once __DIR__ . '/config.php';

    // Connecting to mysql database
$mysqli = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_DATABASE);


    // Selecing database
    $mysqli ->select_db(DB_DATABASE);

    // returing connection cursor
    return $mysqli; 
}



/**
 * Function to close db connection
 */
function close() {
    // closing db connection
    mysql_close();
$mysqli->close();
}

}

?>

和get_courses_details.php

<?php

/*
 * Following code will list all the products
 */



// array for JSON response
$response = array();

// include db connect class
require_once __DIR__ . '/db_connect.php';

// connecting to db
$db = new DB_CONNECT();

// get all products from products table
$result = $db -> query("SELECT * FROM corsi");

// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// products node
$response["corsi"] = array();

while ($row = mysql_fetch_array($result)) {
    // temp user array
    $corsi = array();
    $corsi ["corso_id"] = $row["corso_id"];
    $corsi ["corso_nome"] = $row["corso_nome"];


    // push single product into final response array
    array_push($response["corsi"], $corsi);
}
// success
$response["success"] = 1;

// echoing JSON response
echo json_encode($response);
} else {
// no products found
$response["success"] = 0;
$response["message"] = "No courses found";

// echo no users JSON
echo json_encode($response);
}
?>

但是当我尝试运行它时显示以下消息：致命错误：在第19行的get_courses_details.php中调用未定义的方法DB_CONNECT :: query（）

你知道我该如何解决它吗？

Answer 1

您正在使用$db = new DB_CONNECT();制作新的类对象。然后，在您尝试运行$result = $db -> query("SELECT * FROM corsi");之后，但返回语法错误，因为类对象中没有方法query()。

您要做的是创建一个公共属性并在connect()方法中定义它，例如：

class DB_CONNECT {
    public $sth;

    ...

    public function connect() {
        ...

        $this->sth = $mysqli;
    }
}

现在您可以使用$db->sth->query()，因为$db->sth现在是您的MySQL对象。

调用未定义的方法DB_CONNECT :: query（）

1 个答案: