Php - 添加到数组失败，添加参考

Question

我的 php 环境有一些奇怪的问题：

代码如下：

public static function getAllVotesOfGroup($groupid){
    $con = new \SYSTEM\DB\Connection(new \DBD\uVote());
    $res = $con->prepare(   'selVoteByGrp',
                            'SELECT * FROM `uvote_votes` WHERE `group` = ?;',
                            array($groupid));
    $result = array();

    //$r = array();
    while($r = $res->next()){                        
        //print_r($r);                        
        $result[] = $r;
        print_r($result);            
        echo "</br></br>";
    }           
    //print_r($result);
    return $result;
}


public function next($object = false, $result_type = MYSQL_BOTH){        
    if($object){
        $this->current = mysqli_fetch_object($this->res);
    } else {
        $this->current = mysqli_fetch_assoc($this->res);
    }
    return $this->current;
}

在我的另一台机器上，此代码返回我希望这台机器返回的所有值：-> getAllVotesOfGroup(1) 的回显代码

Array ( [0] => Array ( [ID] => 1 [group] => 1 [title] => Test [text] => Testabstimmung [time_start] => 2013-06-12 [time_end] => 2015-06-13 ) ) 

Array ( [0] => Array ( [ID] => 2 [group] => 1 [title] => Test2 [text] => Testabstimmung [time_start] => 2014-06-13 [time_end] => 2012-06-16 ) [1] => Array ( [ID] => 2 [group] => 1 [title] => Test2 [text] => Testabstimmung [time_start] => 2014-06-13 [time_end] => 2012-06-16 ) ) 

Array ( [0] => Array ( [ID] => 3 [group] => 1 [title] => Test3 [text] => bla [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) [1] => Array ( [ID] => 3 [group] => 1 [title] => Test3 [text] => bla [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) [2] => Array ( [ID] => 3 [group] => 1 [title] => Test3 [text] => bla [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) ) 

Array ( [0] => Array ( [ID] => 4 [group] => 1 [title] => Test4 [text] => blub [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) [1] => Array ( [ID] => 4 [group] => 1 [title] => Test4 [text] => blub [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) [2] => Array ( [ID] => 4 [group] => 1 [title] => Test4 [text] => blub [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) [3] => Array ( [ID] => 4 [group] => 1 [title] => Test4 [text] => blub [time_start] => 0000-00-00 [time_end] => 0000-00-00 ) ) 

如您所见，之前的值已被替换，可能是因为 $r 是一个引用，并且只有引用被添加到数组中，而不是实际值。

这是为什么呢？我可以设置一些 php.ini 选项来改变这种行为吗？

补充： 这很好用！但这不是我想要的 ;-)

    $result = array();

    while($r = $res->next()){                        
        $result[] = array('title' => $r['title'],'text' => $r['text']);
    }           
    return $result;

好的，我可以解决它;-)

见http://www.php.net/manual/en/mysqli-stmt.fetch.php#83284

问题是 mysqli-stmt-fetch im 使用它将返回引用而不是数据。

Answer 1

就我所见，代码运行良好（除非我遗漏了什么）。

$result[] = $r;

这一行将新行添加到数组中，而不是覆盖旧值。老实说，我会质疑其他服务器为您返回的内容。