【问题标题】:CI Active record not work in Google Chrome?CI 活动记录在 Google Chrome 中不起作用?
【发布时间】:2015-07-29 08:18:01
【问题描述】:

我在 CI 中编写了这个查询。

$this->db->select("initiation.*, project.projectname,subproject.subprojectname, concat(employee.firstname,' ',employee.lastname) as 'name'")
                ->from('initiation')
                ->join('project','initiation.projectid=project.id')
                ->join('employee','initiation.leaderid = employee.id')
                ->join('subproject','initiation.subprojectid=subproject.id','left');
$query = $this->db->get();
return $query->result();

它在 Firefox 中运行良好,但在 Chrome 中出现此错误消息。

致命错误:在非对象上调用成员函数 result()

有什么想法吗?

【问题讨论】:

  • 在您解决问题后接受某人的回答。它会帮助别人。

标签: php codeigniter


【解决方案1】:

尝试使用 result_array();确保你有自动加载的数据库库

$this->db->select("initiation.*, project.projectname,subproject.subprojectname, concat(employee.firstname,' ',employee.lastname) as 'name'")
                ->from('initiation')
                ->join('project','initiation.projectid=project.id')
                ->join('employee','initiation.leaderid = employee.id')
                ->join('subproject','initiation.subprojectid=subproject.id','left');
$query = $this->db->get();

if ($query->num_rows() > 0) {
return $query->result_array();
} else {
return false;
}

【讨论】:

    【解决方案2】:
    $this->db->select("initiation.*, project.projectname,subproject.subprojectname, concat(employee.firstname,' ',employee.lastname) as 'name'")
        ->from('initiation')
        ->join('project','initiation.projectid=project.id')
        ->join('employee','initiation.leaderid = employee.id')
        ->join('subproject','initiation.subprojectid=subproject.id','left');
    $query = $this->db->get();
    $result = $query->result_array();//changed
    return $result;//changed
    

    【讨论】:

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