Question

我试图在DB中发布一些值但由于某种原因没有插入数据。从视图返回的数组不为空。但是查询没有执行。

我的观点的一部分：

<?echo form_open_multipart('eva/evluation_questions_value_entry'); ?>

<input type="hidden" name="id" value="<?php echo htmlspecialchars($id) ?>">
<?php  foreach ($info as $row){ 
  echo "<div class='row'>";
  $i=$row->id;
  echo "<div class='firstc'>".$row->id ."</div>";

  echo "<div class='secondc'>".$row->name ."</div>";

  echo  '<input type="hidden" name="training_evaluation_entry_id'.$i.'" value='.$i.'>';
            //some codes
<input id="submit" type="submit" value="Submit" name="Submit">

  </form>

控制器：

    public function evluation_questions_value_entry() {
    $this->load->helper('url');
    $this->load->helper(array('form', 'url'));
    $this->load->model('TrainingEvaluationModel');
    $trainingEvaluation = $this->TrainingEvaluationModel->evluation_questions_value_entry();
}

模特：

$this->ip_address = isset($_SERVER['HTTP_X_FORWARDED_FOR']) ? $_SERVER['HTTP_X_FORWARDED_FOR'] . '||' . $_SERVER['REMOTE_ADDR'] : $_SERVER['REMOTE_ADDR'];        
$this->created_on = date('Y-m-d H:i:s');
$this->created_by = 101;
$this->updated_on = null;
$this->updated_by = 0;


for ($i = 1; $i <= 13; $i++) {

    $this->training_evaluation_entry_id = $_POST['training_evaluation_entry_id'.$i];
    $this->value = $this->db->escape($_POST['group'.$i]);


    $query = "INSERT INTO training_evaluation_info(training_evaluation_entry_id,value,ip_address,created_by) 
    VALUES($this->training_evaluation_entry_id,$this->value,'$this->ip_address',$this->created_by)";
    if ($this->db->affected_rows()>0)
{
    return TRUE;
}
return FALSE;

}

它没有给出任何错误，数据数组是否正确传递给模型，但为什么它没有插入DB？

Answer 1

你的引用放在一边看起来很危险而没有逃避所有字段（我建议您了解参数化查询），在这种情况下，您只需设置$query;

$query = "INSERT INTO training_evaluation_info(training_evaluation_entry_id,value,ip_address,created_by) 
VALUES($this->training_evaluation_entry_id,$this->value,'$this->ip_address',$this->created_by)";

......但从未真正执行过它。

您需要添加;

$this->db->query($query);

var dump显示结果但数据未插入DB

1 个答案: