Question

我正在使用CodeIgniter并收到以下错误：尝试获取非对象的属性。我100％肯定它，因为查询没有找到数据库中的现有行，因为它什么时候工作正常。我是CodeIgniter的新手，不知道如何解决这个问题？

这是我模型中的功能

function getSent(){

    $this->db->where('submit_time', $this->uri->segment(3));
    $tempsent = $this->db->get('0_request_details');  //this is the problem here. If it doesnt find the row from the database above it doesnt know what to do here

    if($tempsent){

        $data['sent'] = $tempsent->row();

    }

    return $data;
}

Answer 1

试试这个，添加一个IF

function getSent()
{
    $this->db->where('submit_time', $this->uri->segment(3));
    $tempsent = $this->db->get('0_request_details');
    if ( $tempsent->num_rows() > 0 )
    {
        // If a row is found...
        $data['sent'] = $tempsent->row();
        return $data;
    }
    else
    {
        return false;
    }
}

Answer 2

试试这个......：）

$this->db->select('field_1 , field_2');
$this->db->where('field_name' , $this->uri->segment(3));
$tempsent = $this->db->get('table_name');

$data = (object) NULL;

if($tempsent->num_rows() > 0)
{
    $data->sent = $info->result();
}

return $data;

Answer 3

这里有一些建议 -

是您的数据库表，真正称为＆＃39; 0_request_details＆＃39; ??? 有一个数字启动数据库表名称可能是一个问题。对于这个例子，我会称之为“request_details”＆＃39;
首先检查并确认$ this-＆gt; uri-＆gt; segment（3）中的值。如果它是空白或格式错误你想知道。并且您不想将其提交到数据库 - 您想要一个明确的错误，让您知道该问题。
不要将数据库结果附加到模型中的$ data。在你的控制器中做到这一点对于更复杂的任务，您将有多个数组/对象转到$ data - 您希望能够在控制器中清楚地看到它。
不要忘记说明要返回的数据库表字段。

控制器中的

   // example your model is called request 
   // check if we got a submit_time back
   // if we don't get a submit_time back then its invalid
   if ( ! $submit_time =  $this->request->returnValidatedSubmitTime ) {

       $this->showInvalidSubmitTime($submit_time) ; } 

   // $submit_time is valid, now try to get a database record
   // check if we got a record back
   elseif( ! $data['sent'] = $this->request->getSentWith($submit_time) ) {

       $this->showNoRecordsFor($submit_time) ;  }

   else {

      // Success 
      // $data['sent'] has been set and available to views    
   }

在模型中

    function returnValidatedSubmitTime(){       

    if($this->uri->segment(3) ) {

     $time = $this->uri->segment(3) ; 

     // do some validation 

    // if its valid 

     return $time; 

    } 
    else { return false ; } 

  } 

    function getSentWith($submit_time){

        // the fields that will be returned     
        $this->db->select( 'id, firstname, lastname' );
        $this->db->where('submit_time', $submit_time);
        $query = $this->db->get('request_details'); 

        // we are only expecting one row so thats what we check for       
            if ( $query->num_rows() == 1 ) {

              return $query->row(); }

             else
            // no results 
            { return FALSE; }
    }

Answer 4

这是您的代码

function getSent(){

$this->db->where('submit_time', $this->uri->segment(3));
$tempsent = $this->db->get('0_request_details');  //this is the problem here. If it doesnt find the     row from the database above it doesnt know what to do here

if($tempsent){

    $data['sent'] = $tempsent->row();

}

return $data;
}

你要做的就是拖动数组结果;你可以用两种方式做到这一点

function getSent(){

$this->db->where('submit_time', $this->uri->segment(3));
$tempsent = $this->db->get('0_request_details');  //this is the problem here. If it doesnt find the     row from the database above it doesnt know what to do here

if($tempsent){
foreach($tempsent->result() as $ts)
{
$data['sent']=$ts;
}

}

return $data;
}

其次，您可以在查询末尾附加 - ＆gt; result（）以获得指向的结果，然后将输出用作数组索引。

希望不知怎的，你有了这个想法

试图获取非对象的属性 - 数据库行不存在

4 个答案: