你能得到一个调用类的变量吗？
I have a class Page that creates an instance of DB, which is named $db.
我有一个创建DB实例的类Page,名为$ db。
In the __construct() of Page, I create the new $db object and I pull a bunch of config data from a file.
在Page的__construct()中,我创建了新的$ db对象,并从文件中提取了一堆配置数据。
Now the DB class has a method _connectToDB() which (attempts) to connect to the database.
现在,DB类有一个方法_connectToDB(),它(尝试)连接到数据库。
Is there a way in the DB class to call the parent class's config array? I don't want to make global variables if I don't have to and I don't want to grab the config data twice.
DB类中有没有办法调用父类的配置数组?如果我不需要,我不想创建全局变量,我不想两次获取配置数据。
Pseudo code might look something like this...
伪代码可能看起来像这样......
$dbUsername = get_calling_class_vars(configArray['dbUserName']);
6 个解决方案
#1
6
I find that it's often easier to initialise all the "important" objects close to whatever variables they need to know. You could try it this way:
我发现通常更容易初始化所有“重要”对象,接近他们需要知道的任何变量。你可以这样试试:
/* Code to get config variables here */
$DB = new DB($config);
/* You might want to delete the database password from $config here */
$Page = new Page($config, $DB);
Doing it this way means you can also do type-checking on the database object if you want to:
这样做意味着您还可以对数据库对象进行类型检查,如果您想:
class Page {
function __construct(array $config, DBClass $db) { }
}
更多相关文章
- 传入连接对象或在类中创建它？ （OOP）
- 在MySQL数据库中搜索与多个Tag记录相关/包含的书签
- 使用jquery和php手动在数据库中增加一个字段[复制]
- 当将带有撇号的单词保存到数据库时，mysql_real_escape_string会添
- php扫马、内网查mysql数据库账号密码（内网webshell渗透）
- thinkphp5 数据库和模型详解 之1 数据库基础
- Wordpress查询另一个wordpress数据库for循环
- CakePHP2 Request请求对象中文教程
- 电商网站项目总结（面向对象编程篇）