dongzhouji4021 2014-03-25 09:06
浏览 6
已采纳

传递数组以在PHP中查看和显示数据

 private function displayallsurveys() {
    $this->load->database();
    $sql = "SELECT * from survey";
    $query = $this->db->query($sql);
    $data = array();
    if ($query->num_rows() > 0) {

       $data = $query->result_array();

       return $data;


    } else {
        return $data;
    }
}

Passing to the View

    $data = $this->displayallsurveys();
    $this->load->view('SurveyPage', $data);

When i show it on the from end

<ul class="nav nav-sidebar">
        <li class="active"><a href="#">Home</a></li>
        <li><a href="#">Add new Survey</a></li>
       <?php  foreach ($data as $row) {
            $SurveyTitle = $row['SurveyTitle'];
            $SurveyId = $row['SurveyId'];


                  echo'<li id=' . $SurveyId . '><a href=' . $SurveyTitle . '>' . $SurveyTitle . '</a><li>';
       }

          ?>

      </ul>

I am getting undefined variable $data

Please someone help

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4条回答 默认 最新

  • douzhen1234 2014-03-25 09:09
    关注

    if you wanted to use $data variable in the view then you would need to pass it as

    $this->load->view('SurveyPage', array('data' => $data));
    

    Otherwise with the following the array keys becomes the variable names

    $this->load->view('SurveyPage', $data);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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