相关疑难解决方法(0)

我试图从MySQL表中选择数据,但我收到以下错误消息之一:

mysql_fetch_array()期望参数1是资源,给定布尔值

要么

mysqli_fetch_array()期望参数1为mysqli_result,给定布尔值

要么

在布尔/非对象上调用成员函数fetch_array()

这是我的代码:

$username = $_POST['username'];
$password = $_POST['password'];

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

while($row = mysql_fetch_array($result)) {
    echo $row['FirstName'];
}

这同样适用于代码

$result = mysqli_query($mysqli, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
    ...

和

$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
    ...

和

$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid …

所以我正在尝试创建一个可用于连接mysql数据库的类.在尝试使用我的课程之前,不应该先进,这一切都很有趣和游戏.这是我的代码:

班级:

<?php

class createCon  {
    var $host = 'localhost';
    var $user = 'root';
    var $pass = '';
    var $db = 'example';
    var $myconn;

    function connect() {
        $con = mysqli_connect($this->host, $this->user, $this->pass, $this->db);
        if (!$con) {
            die('Could not connect to database!');
        } else {
            $this->myconn = $con;
            echo 'Connection established!';}
        return $this->myconn;
    }

    function close() {
        mysqli_close($myconn);
        echo 'Connection closed!';
    }

}

?>

这是我尝试查询数据库的地方:

<?php

include 'connect.php';

$connection = new createCon();
$connection->connect();

$query = 'SELECT * FROM  `nickname`';
$result = …