Question

我将Database.php和Config.php放在同一个位置。在Database.php中，如果我执行以下操作，则可以使用

<?php
class Database
{
    const DB_ADAPTER = 'mysql';
    const DB_USER = 'something';
    const DB_PASSWORD = 'something';
    const DB_DATABASE = 'something';
    const DB_HOST = "localhost";
    const DB_CHARSET = "utf8";

    /* All my code */

     $dns = self::DB_ADAPTER . ':host=' . self::DB_HOST . ';dbname=' . self::DB_DATABASE;
            self::$dbLink = new \PDO($dns, self::DB_USER, self::DB_PASSWORD);

}

如果我将常量移动到Config.php并将上面的内容更改为以下

<?php

include('Config.php');

class Database
{
    /** @var PDO The shared database link. */
    protected static $dbLink;

    /* All my code */

     $dns = Config::DB_ADAPTER . ':host=' . Config::DB_HOST . ';dbname=' . Config::DB_DATABASE;
            self::$dbLink = new \PDO($dns, Config::DB_USER, Config::DB_PASSWORD);

}

它失败了，未定义的类常量＆＃39; DB_ADAPTER＆＃39;在Database.php中（如果我删除DB_ADAPTER，它将在下一个常量上失败）。

为什么会这样？

由于

更新

<?php

/**
 * Class containing configuration details
 */
class Config
{
    const DB_ADAPTER = 'mysql';
    const DB_USER = 'something';
    const DB_PASSWORD = 'something';
    const DB_DATABASE = 'something_portal_db';
    const DB_HOST = "localhost";
    const DB_CHARSET = "utf8";
}

Answer 1

你不应该在没有对象创建的情况下使用 ~~static和~~ public吗？

<?php

/**
 * Class containing configuration details
 */
class Config
{
    public const DB_ADAPTER = 'mysql';
    public const DB_USER = 'something';
    public const DB_PASSWORD = 'something';
    public const DB_DATABASE = 'something_portal_db';
    public const DB_HOST = "localhost";
    public const DB_CHARSET = "utf8";
}

PHP未定义的配置文件类

1 个答案: