Question

我的模型中有以下代码：

public function getSite()
{
    $this->load->library('site');
    $data = array();
    $str = "SELECT * FROM sitematrix_sites";
    $query = $this->getQuery($str);
    if($query){ 
        foreach($query->result() as $row){
            $site = new Site();
            array_push($data, $site->load($row));
        }

        foreach($data as $site){
            $site->data['database'] = $this->getDatabase($site->data['site_id']);
            $site->data['baseline'] = $this->getBaseline($site->data['site_id']);
        }

        return $data;
    }
    else{
        return false;
    }
}

我创建了一个名为site的类：

class Site{

public $data;

public function __contruct(){

    $this->data = ArrayObject(array(), ArrayObject::STD_PROP_LIST);

    $this->data['site_id']                      =   '';
    $this->data['site']                         =   '';
    $this->data['site_code']                    =   '';
    $this->data['tc_10']                        =   array();
    $this->data['tc9x']                         =   array();
    $this->data['tc8x']                         =   array();
    $this->data['tc2008']                       =   array();
    $this->data['eng_2005']                     =   array();
    $this->data['database']                     =   array();
    $this->data['baseline']                     =   array();

但是，我收到以下错误：

<p>Severity: Notice</p>
<p>Message:  Trying to get property of non-object</p>
<p>Filename: models/sitematrix_model.php</p>
<p>Line Number: 77</p>

有什么想法？我尝试将这些值作为$site->data->database和$site->data['database']访问，但两者都不起作用。

line 77:  $site->data['database'] = $this->getDatabase($site->data['site_id']);

函数getDatabase：

public function getDatabase($id){
    $database = array();
    $str = "SELECT * FROM site_database where site_id='$id'";
    $query = $this->getQuery($str);
    if($query){
        foreach($query->result() as $row){
            array_push($database, $row->name);
        }
    }

    return $database;
}

函数getQuery：

public function getQuery($str){
    $data = array();
    $query = $this->db->query($str);

    if(!$query || $query->num_rows() == 0){
        return false;
    }
    return $query;
}

由于

Answer 1

进行一些调查，我认为$site不是Site，而是数组。

Array ( [site_id] => 1 [site] => Ames [site_code] => ame [windows_ref_unit_location] => \\plm\amnas\tc_ref [unix_ref_unit_location] => /tc_ref [windows_rte_location] => \\plm\amnas\tc_rte [unix_rte_location] => /tc_rte [windows_toolbox_location] => <path-to-local-toolbox> [unix_toolbox_location] => /tc_ref/TOOLBOX [UGS_LICENSE_SERVER] => 28000@cinxflex1,28000@cinxflex2,28000@cinxflex3 [UGII_LICENSE_FILE] => 27000@cinxflex1,27000@cinxflex2,27000@cinxflex3 [unix_dev_units] => /tc_work [unix_devop_path] => /usr/site/devop_tools [netapp_filer] => \\plm\amnas [perforce_proxy_path] => /p4p/p4p_cache [primary_contact] => Philomena Siddle [secondary_contact] => Mathieu Sarrazy [num_users] => 7 )

因此，没有像$site->data['site_id']这样的东西，而是$site['site_id'];

Answer 2

我很确定您的错误消息是抱怨

$这 - ＆GT; getDatabase

该行的一部分，而不是$ site-＆gt;方面。

所有这些功能都在Model上，而不是Class站点吗？

当您在模特中工作时，$ this-＆gt;必须坚持功能和来自该班级的变种

访问PHP codeigniter中的对象变量

2 个答案: