mysqli 版本的 mysql 模式

Question

这是 PHP 中的 MySQL 模式：

$username="username";
$password="password";
$database="username-databaseName";

// Opens a connection to a mySQL server
$connection=mysql_connect (localhost, $username, $password);
if (!$connection) {
  die("Not connected : " . mysql_error());
}

// Set the active mySQL database
$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
  die ("Can\'t use db : " . mysql_error());
}

// Search the rows in the markers table
$query =  some query
$result = mysql_query($query);

我尝试将大部分内容替换为 mysqli 模式，然后将查询部分粘贴在底部，如下所示：

//Database Information
$db_host = "localhost"; //Host address (most likely localhost)
$db_name = "username-databaseName"; //Name of Database
$db_user = "username"; //Name of database user
$db_pass = "password"; //Password for database user

/* Create a new mysqli object with database connection parameters */
$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);
GLOBAL $mysqli;

if ($mysqli->connect_errno) {
        echo "<p>MySQL error no {$mysqli->connect_errno} : {$mysqli->connect_error}</p>";
        exit();
    }

// Search the rows in the markers table
$query =  some query
$result = mysql_query($query);

但是，我收到以下错误消息：

Invalid query: No database selected

我做错了什么？

Answer 1

首先，您拨打的是mysql_query 而不是mysqli_query，就像您打算的那样。

第二，由于你使用的是面向对象的形式，你需要调用mysqli_query作为方法：

$result = $mysqli->query($query);