Question

在我提交空表单的时候，它会在我的数据库中插入一个空的raw。我也在使用CodeIgniter表单验证。如何使用CodeIgniter避免这种情况。请帮帮我。 ex-如果从空是没有提交。

 $this->form_validation->set_rules('name', 'Name', 'trim|required|min_length[5]');
    // validate name
    $this->form_validation->set_rules('email', 'Email', 'trim|required|valid_email');
    // validate email
    $this->form_validation->set_rules('phone', 'Phone Number', 'trim|required|min_length[10]|max_length[10]');
    // validate phone number
    $this->form_validation->set_rules('comment', 'Questions or Comments', 'required|min_length[10]');
    // validate phone number
    if ($this->form_validation->run() == FALSE)// if from validation failed load following pages from views
    {
        $this->load->view('template/header.php');
        $this->load->view('pages/contacts');
        $this->load->view('template/footer.php');
    }
    else // if from validation success load following pages from views
    {
        $this->load->view('template/header.php');
        $data['result']="Thank you for contacting us! We will reply you soon... ";
        $this->load->view('pages/contacts',$data);
        $this->load->view('template/footer.php');
    }

我也在使用必要的验证规则，请帮我解决这个问题

<?php echo form_open('form/contact') ?><!--Creates an opening form
        ex :<form method="post" accept-charset="utf-8" action="localhost/index.php/form(controller)/subscribe(function)" />-->
        <!-- Contact form Start -->
        <div class="span7">
            <div class="box">
                <?php echo validation_errors('<div class="notice marker-on-bottom bg-darkRed fg-white" id="bottom_form_error">', '</div>'); ?>
                    <?php if(isset($result)){echo '<div class="notice marker-on-bottom  bg-green fg-white">'.$result.'</div>';}?>
                    <!-- Display Validation error massages  -->
                <h6>*All fields marked are required</h6>
                    <lable>Your name*</lable>
                    <div class="input-control text" data-role="input-control">
                        <input type="text" name="name" value="<?php echo set_value('name'); ?>" placeholder="type your name">
                        <button class="btn-clear" tabindex="-1"></button>
                    </div>
                    <lable>Email*</lable>
                    <div class="input-control text" data-role="input-control">
                        <input type="text" name="email" value="<?php echo set_value('email'); ?>" placeholder="type email address">
                        <button class="btn-clear" tabindex="-1"></button>
                    </div>
                    <lable>Phone*</lable>
                    <div class="input-control text" data-role="input-control">
                        <input type="text" name="phone" value="<?php echo set_value('phone'); ?>" placeholder="type phone number">
                        <button class="btn-clear" tabindex="-1"></button>
                    </div>
                    <lable>Questions or Comments*</lable>
                    <div class="input-control textarea" data-role="input-control">
                        <textarea name="comment" value="<?php echo set_value('comment'); ?>" >...</textarea>
                    </div>
                    <input name='contactus_status' type='hidden' value='no'/>
                    <input type="submit" value="Submit" class="info">
            </div>
        </div>
    <?php  echo form_close(); ?><!-- Contact form End -->

Answer 1

您正在寻找必需的Codeigniter服务器端验证

$this->form_validation->set_rules('username', 'Username', 'required');

完整文档Form validation codeigniter

如果您需要客户端验证jQuery Validation Plugin

以下是Jquery表单验证的Sample Demo

Answer 2

您可以使用JavaScript阻止仅在客户端进行实际提交，但是使用该方法您无法确定是否提交了表单（因为可能会混淆客户端脚本）。相反，您必须验证提交的值。

Answer 3

我似乎你的MySQL代码在if else语句的验证之外。一定要把它放在正确的位置。

以下是您样本中缺失的数据。

if ($this->form_validation->run() == FALSE){

}else{
    // Be sure to put your database insert code here...
    $data['name'] = set_value('name');
    $data['email'] = set_value('email');
    $data['phone'] = set_value('phone');
    $data['comment'] = set_value('comment');
    $this->db->insert('table_name',$data);
}

输入为空时禁用提交 - CodeIgniter

3 个答案: