1
我有一個表單,用於顯示調查中的問題和答案。用戶可以選擇5個單選按鈕 - 選擇問題答案。他們是陣列,關鍵是question_id
- 每個問題都有這5個單選按鈕,它們在這個鍵上有所不同。你能幫我進行表單驗證嗎? 現在,如果您選擇其中一個問題的答案,請提交表單。只有在所有問題都有答案的情況下才能提交。這是我的觀點:如何使用CodeIgniter中的廣播組驗證表單
<html>
<head></head>
<body>
<?php
$survey_id = $this->uri->segment(3);
$question_id = $this->uri->segment(4);
$att=array('id'=>'form');
echo form_open('index/survey_fill/' .$survey_id .'/'. $question_id , $att);
echo "<table border="0" id='questionsTable' >";
echo "<tr><th>Въпрос</th></tr>";
\t echo validation_errors();
$index = 0;
\t foreach ($question as $row)
\t {
\t \t echo "<tr id='$index'>";
\t \t $index++;
\t ?>
\t \t
\t \t <td>
\t \t <?php echo "$row->question"; ?><br/>
\t \t <?php echo "<input type='hidden' name='question_id' value='$row->question_id' />"; ?>
\t \t <?php
\t \t
\t \t $data=array(
\t \t \t 'name' => 'answer['.$row->question_id.']',
\t \t \t 'value' => '5',
\t \t \t 'class' => 'answer'
\t \t);
\t \t echo "<input type='hidden' name='survey_id' value='$row->survey_id'>";
\t \t
\t \t echo form_radio($data);
\t \t echo " 5 ";
\t \t $data=array(
\t \t \t 'name' => 'answer['.$row->question_id.']',
\t \t \t 'value' => '4',
\t \t \t 'class' => 'answer'
\t \t);
\t \t echo form_radio($data);
\t \t echo " 4 ";
\t \t $data=array(
\t \t \t 'name' => 'answer['.$row->question_id.']',
\t \t \t 'value' => '3',
\t \t \t 'class' => 'answer'
\t \t \t
\t \t);
\t \t echo form_radio($data);
\t \t echo " 3 ";
\t \t $data=array(
\t \t \t 'name' => 'answer['.$row->question_id.']',
\t \t \t 'value' => '2',
\t \t \t 'class' => 'answer'
\t \t);
\t \t echo form_radio($data);
\t \t echo " 2 ";
\t \t $data=array(
\t \t \t 'name' => 'answer['.$row->question_id.']',
\t \t \t 'value' => '1',
\t \t \t 'class' => 'answer'
\t \t);
\t \t echo form_radio($data);
\t \t echo " 1 ";
\t \t ?>
\t \t </td></tr>
\t <?php
\t }
?> \t \t
\t
</table>
<?php echo "<input type='hidden' name='question_id' value='$row->question_id' />"; ?>
<?php echo '<input type="submit" id="button" name = "submit" value="Submit" class="btn btn-success">';
?>
</form>
</div>
</body>
</html>
我的控制器:
public function survey_fill()
{
$this->form_validation->set_rules('answer[]', 'Answer', 'required');
if ($this->form_validation->run()==FALSE)
{
$this->survey_show();
}
else
{
\t \t if ($this->user_model->survey_fill())
\t {
header('Refresh: 2; url=/survey/index.php/index/surveys_show'); \t \t
\t }
else
{
$this->load->model('user_model');
$data['survey'] =$this->user_model->survey_show();
$data['dynamic_view'] = 'survey_show';
$data['menu']=$this->menu_model->get_menu();
$this->load->view('templates/main',$data);
\t \t }
}
}
謝謝!它起作用了,我用了第二個,因爲它給了我第一個例子中的undefined for question_id。我在控制器中編寫表單驗證並使用示例2並在那裏寫入query $ questions。 :) –
很高興我能幫忙!還有未定義?你還在傳遞你的其他價值觀嗎?我有興趣知道什麼對你來說第一個沒用。 –
Becaurse在我寫表單驗證的函數中,我沒有這個查詢,這就是爲什麼'question_id'是未定義的。我添加了:$ query = $ this-> user_model-> getSurvey($ this-> uri-> segment(3),$ this-> uri-> segment(4)); $ questions = $ query-> result(); foreach($ questions as $ question){ $ this-> form_validation-> set_rules('answer ['。$ question-> question_id。']','Question'。$ question-> question_id,'required') ; }現在沒關係。 :) –