面向对象的 PHP 返回数据库行并访问数组元素

Question

我是面向对象 PHP 概念的新手（遵循 www.killerphp.com 上的教程），我计划将我所有的 php 应用程序迁移到 OO PHP。
我开始构建我的第一个类，它根据 "setSecurity()" 方法设置的 "where" 条件中的对象属性从数据库中读取授权级别。

我设法返回了数组 "getSecurity()" 但打印输出会给出：

security Object
(
    [secArray] => Array
        (
            [from_date1] => 1992-01-01
            [to_date1] => 0000-00-00
            [from_date2] => 1992-01-01
            [to_date2] => 0000-00-00
            [view] => 1
            [insert] => 0
            [update] => 1
            [delete] => 1
            [valid] => 1
        )

)
/*"Array 1"*/

我的问题是我不熟悉普通数组的打印输出（如下）。

Array
(
    [from_date1] => 1992-01-01
    [to_date1] => 0000-00-00
    [from_date2] => 1992-01-01
    [to_date2] => 0000-00-00
    [view] => 1
    [insert] => 0
    [update] => 1
    [delete] => 1
    [valid] => 1
)
/*"Array 2"*/

我的问题是： 1) 如何从 getSecurity() 方法（从数组 1）访问数组的元素？
2) 如何让我的方法正确返回数组（与数组 2 相同）？

以下代码片段。

非常感谢您的支持...

'test.php'

<?php
include("connect.php");  
include("security.php");
$secArray=new security();
$secArray->setSecurity('test_user',1,1,1,$link);
$secArray->getSecurity();
echo "<pre>"; print_r($secArray); echo "</pre>";
?>

'security.php'

<?php
class security
{
    public $secArray = array();

    function setSecurity($user,$appid,$funid,$objid,$conn='')
    {
        $query="SELECT lu.DATE1 as from_date1,
                    lu.DATE2 as to_date1,
                    ga.DATE1 as from_date2,
                    ga.DATE2 as to_date2,
                    ga.VIEW as view,
                    ga.INSERT as insert,
                    ga.UPDATE as update,
                    ga.DELETE as delete,
                    ob.VALID as valid
                FROM
                    user as lu
                    inner join group as ug on lu.GRP_ID = ug.ID
                    inner join privileges as ga on lu.GRP_ID = ga.GRP_ID
                    and ug.ID = ga.GRP_ID
                    inner join level1 as ob on ob.APP_ID = ga.APP_ID
                    and ob.FUN_ID = ga.FUN_ID
                    and ob.ID = ga.OBJ_ID
                where
                    USERID = '$user'
                and ga.APP_ID = $appid
                and ga.FUN_ID = $funid
                and ga.OBJ_ID = $objid";
        $result = mysql_query($query,$conn);
        $row = mysql_fetch_assoc($result);
        $this->secArray=$row;
    }

    function getSecurity()
    {
        return $this->secArray;
    }     
}
?>

Answer 1

getter 返回一个值，因此调用$secArray->getSecurity(); 并不能真正帮助您，除非您对返回的值执行某些操作。 $mySecurity=$secArray->getSecurity();use $mySecurity...

请阅读有关访问数组和对象的 PHP 文档。