2010-07-05 94 views
0

我有一个有趣的事情与我的本地开发设置进行。使用codeigniter我有一个窗体应该模拟存储用户在数据库中的一些信息。在这一点上,我没有数据库设置,但我只希望表单在验证完成时将我发送到成功页面。但是当我提交表单时,它会回来,并告诉我名称字段为空,我从来没有获得成功页面。我在这里错过了什么吗?请看看这个,并告诉我,如果我错过了一些东西! 这是处理表单的控制器中的函数。Codeigniter表单验证类

function showsupport() 
{ 
    $this->form_validation->set_rules('supportername','Name','trim|required|max_length[30]'); 
    $this->form_validation->set_rules('supporteremail','Email Address','trim|required|valid_email'); 
    $this->form_validation->set_rules('pledgedate','Pledge Date','trim|required'); 
    $this->form_validation->set_rules('messagetxt','Your Message','trim|required|xss_clean'); 

    if($this->form_validation->run() == FALSE) 
    { 
     $this->template->write_view('content','supportus'); 
     $this->template->render(); 
    } else { 
     $this->template->write('content','Your submission was a success'); 
     $this->template->render(); 
    } 
} 


<div id="supportform" class="formblock"> 

       <?php 
        $formdata = array('id'=>'suppform'); 
         echo form_open('homepage/showsupport',$formdata); 
        $namedata = array('name'=>'supportname','id'=>'supportname','size'=>'30','max_length'=>'25'); 
        echo '<label for="supportername">Your Name:'.form_input($namedata,set_value('supportname')).'</label><br /><br />'; 
        $emaildata = array('name'=>'supporteremail','id'=>'supporteremail','size'=>'30','max_lenth'=>'25'); 
        echo '<label for="supporteremail">Email Address:'.form_input($emaildata,set_value('suppoteremail')).'</label><br /><br />'; 
        $pledgedata = array('name'=>'pledgedate','id'=>'pledgedate','size'=>'30','max_length'=>'20'); 
        echo '<label for="pledgedate">Today\'s Date:'.form_input($pledgedata,set_value('pledgedate')).'</label><br /><br />'; 
        $msgdata = array('name'=>'messagetxt','id'=>'messagetxt','col'=>'2','rows'=>'8'); 
        echo '<label for="messagetext">Your Pledge:'.form_textarea($msgdata).'</label><br />'; 
        $submitdata = array('name'=>'submitbtn','id'=>'submitbtn','value'=>'Send'); 
        echo '<label for="submitbutton">'.form_submit($submitdata).'</label><br />'; 
        echo form_close(); 
       ?> 
      </div> 
     <div id="errorsechoed"> 
      <div class="ui-widget"> 
     <div class="ui-state-error ui-corner-all" style="padding: 0 .7em; border: none;">       
         <?php echo validation_errors('<p><span class="ui-icon ui-icon-alert" style="float: left; margin-right: .3em;"></span>');?> 

这是它自己的形式。我必须在几个小时内进行演示,而这种形式的工作是关键。感谢你们的帮助。

Ë

回答

2

还有就是你的表单验证规则和查看HTML元素之间的名称不匹配。

的验证规则

$this->form_validation->set_rules('supportername','Name','trim|required|max_length[30]'); 

的HTML

$namedata = array('name'=>'supportname','id'=>'supportname','size'=>'30','max_length'=>'25'); 
echo '<label for="supportername">Your Name:'.form_input($namedata,set_value('supportname')).'</label><br /><br />'; 

要么更改验证规则名称为 'supportname'

或者

将'name'=>'supportname'更改为'name'=>'supportername'

+0

谢谢你,太多的咖啡!谢谢 – 2010-07-06 14:23:55