Php 使用CodeIgniter从数据库返回字符串

这是我的型号：

function getAvenuName($id) {
        $this->db->distinct();
        $this->db->select('admin.username');
        $this->db->from('personal_closest');
        $this->db->join('admin', 'admin.id = personal_closest.avenu_id', 'left');
        $this->db->where('personal_closest.avenu_id', $id);
        $query = $this->db->get();
        return $query->result();        
    }

foreach ($listofcloset['postByUser'] as $key => $value) { 
            $listofcloset['postByUser'][$key]['calculatetime']   =  $this->calculatetime(strtotime($value['postdatetime']));
            $listofcloset['postByUser'][$key]['countcomments']   =  $this->countcomments($value['id']); 
            $listofcloset['postByUser'][$key]['comments']        =  $this->comments->sltpostcomments($value['personal_closest_id']); 
            $listofcloset['postByUser'][$key]['countCloset']     =  $this->closet->countCloset($value['id']);            
            $listofcloset['postByUser'][$key]['givencool']       =  $this->checkCoolIsGiven($this->session->userdata('user_id'),$value['user_id'],$value['personal_closest_id']);
            $listofcloset['postByUser'][$key]['givencloset']     =  count($this->closet->selectIdbycloset($this->session->userdata('user_id'),$value['user_id'],$value['personal_closest_id']));
            if (empty($listofcloset['postByUser'][$key]['username'])){
                $listofcloset['postByUser'][$key]['username'] = $this->personal_closest->getAvenuName($value['avenu_id']);
            }
        }

这是我的控制器：

function getAvenuName($id) {
        $this->db->distinct();
        $this->db->select('admin.username');
        $this->db->from('personal_closest');
        $this->db->join('admin', 'admin.id = personal_closest.avenu_id', 'left');
        $this->db->where('personal_closest.avenu_id', $id);
        $query = $this->db->get();
        return $query->result();        
    }

foreach ($listofcloset['postByUser'] as $key => $value) { 
            $listofcloset['postByUser'][$key]['calculatetime']   =  $this->calculatetime(strtotime($value['postdatetime']));
            $listofcloset['postByUser'][$key]['countcomments']   =  $this->countcomments($value['id']); 
            $listofcloset['postByUser'][$key]['comments']        =  $this->comments->sltpostcomments($value['personal_closest_id']); 
            $listofcloset['postByUser'][$key]['countCloset']     =  $this->closet->countCloset($value['id']);            
            $listofcloset['postByUser'][$key]['givencool']       =  $this->checkCoolIsGiven($this->session->userdata('user_id'),$value['user_id'],$value['personal_closest_id']);
            $listofcloset['postByUser'][$key]['givencloset']     =  count($this->closet->selectIdbycloset($this->session->userdata('user_id'),$value['user_id'],$value['personal_closest_id']));
            if (empty($listofcloset['postByUser'][$key]['username'])){
                $listofcloset['postByUser'][$key]['username'] = $this->personal_closest->getAvenuName($value['avenu_id']);
            }
        }

当我使用用户名时，我遇到了数组到字符串转换的问题。我正试图找到一个解决方案，只取数组中的字符串

这就是结果：

[username] => Array
            (
                [0] => stdClass Object
                    (
                        [username] => USER1
                    )

            )

---

使用以下代码您将获得直接数组

function getAvenuName($id) {
    $this->db->distinct();
    $this->db->select('admin.username');
    $this->db->from('personal_closest');
    $this->db->join('admin', 'admin.id = personal_closest.avenu_id', 'left');
    $this->db->where('personal_closest.avenu_id', $id);
    $query = $this->db->get();
    return $query->row_array();        
}

这应该是：）

---

尝试将

$query->result（）

替换为

$query->result\u array（）

这很好，但请始终为我打印一个数组，我需要一个字符串[username]=>数组（[username]=>mydocument）事实上，在视图中，我得到了以下错误：严重性：注意消息：数组到字符串的转换将结果更改为$result，以避免可能需要此转换的任何其他人产生混淆。不客气。我很高兴能帮助你！