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Codeigniter和Ajax联系表

[英]codeigniter and ajax contact form

I'm trying to use ajax within my contact form in a codeigniter app. 我正在尝试在Codeigniter应用程序的联系表单中使用Ajax。 I have it to where the ajax call is made, but no post data is being sent to the server. 我将其发送到进行了Ajax调用的位置,但是没有将后数据发送到服务器。 I can't figure out why. 我不知道为什么。 Please help. 请帮忙。

I do have some returns in there, but they do nothing. 我确实有一些回报,但是他们什么也没做。 Also, $this->input->post('name') is null. 同样,$ this-> input-> post('name')为null。

Form view 表格检视

<?php
                    echo form_open('welcome/submit_contact');

                    echo form_error('name');
                    echo form_label('Name', 'name'"');
                    echo form_input('name', set_value('name'), 'id="name);

                    echo form_error('email');
                    echo form_label('Email', 'email');
                    echo form_input('email', set_value('email'), 'id="email"');

                    echo form_error('phone');
                    echo form_label('Phone', 'phone');
                    echo form_input('phone', set_value('phone'), 'id="phone"');
                    #echo '<h5>Do not start with "1" and no dashes.</h5>';

                    echo form_error('message');
                    echo form_label('Message', 'message');
                    echo form_textarea('message', set_value('message'), 'id="message"');

                    $submitData = array(
                        'name'  => 'submit',
                        'value' => 'Submit',
                        'id'    => 'button'

                    );
                    echo form_submit($submitData);

                    echo form_close();
                ?>
                <script type="text/javascript">

                $(function() {

                    $('form').click(function() {

                        // get the form values
                        var form_data = {
                            name: $('#name').val(),
                            email: $('#email').val(),
                            message: $('#message').val()
                        };

                        // send the form data to the controller
                        $.ajax({
                            url: "<?php echo site_url('welcome/submit_contact'); ?>",
                            type: "post",
                            data: form_data,
                            success: function(msg) {
                                $('form').prepend('<h5 class="good">Message sent!</h5>');
                                $('h5.good').delay(3000).fadeOut(500);
                                alert(msg);
                            }
                        });

                        // prevents from refreshing the page
                        return false;   
                    });
                });
                </script>

Controller function 控制器功能

function submit_contact()
    {
        $this->load->library('form_validation');
        $this->form_validation->set_error_delimiters('<h4 class="bad">', '</h4>');

        $name = $this->input->post('name');
        echo "name = ".$name;

        $this->form_validation->set_rules('name', 'Name', 'trim|required|alpha_dash|xss_clean');
        $this->form_validation->set_rules('email', 'Email', 'trim|required|valid_email|xss_clean');
        $this->form_validation->set_rules('phone', 'Phone' , 'trim|integer|exact_length[10]|xss_clean');
        $this->form_validation->set_rules('message', 'Message', 'trim|required|max_length[1000]|xss_clean');

        // there are validation errors
        if($this->form_validation->run() == FALSE)
        {
            return "error";
        }
        else // there are no validation errors
        {
            /*************************
            need to actually send the email 
            *************************/
            return null;
        }

    }

EDIT: I've updated the code in my question. 编辑:我已经更新了我的问题中的代码。 Basically now if there are validation errors, how would I get them to display on the form? 基本上现在,如果存在验证错误,我如何使它们显示在表单上? I'm assuming I would return a value from my controller and then have a statement in my ajax success that if msg == "error" display the errors elseif msg == null, display success message. 我假设我将从控制器返回一个值,然后在ajax成功中有一条语句,如果msg ==“ error”显示错误elseif msg == null,则显示成功消息。 But how would i tell my view to display those errors based on an ajax success variable? 但是我如何告诉我的看法基于ajax成功变量显示那些错误?

i think you should put id on input and textarea not on label ie 我认为你应该把id放在输入和textarea而不是标签上

$data = array
(
   "name"=>'message',
   "value"=>'message',
   "id"=>'message'
)


form_textarea($data);

if you set the id on the label then jquery will pick up nothing from what the user inserted and also codeigniter validation won't work correctly. 如果您在标签上设置id,则jquery不会从用户插入的内容中获取任何内容,并且codeigniter验证将无法正常工作。 This is why your post result to be NULL 这就是为什么您的帖子结果为NULL

the same for other input field 其他输入字段相同

EDIT 编辑

you are asking data via ajax so return a nice json object (remove all your debug prints first): 您正在通过ajax询问数据,因此返回一个不错的json对象(首先删除所有调试输出):

// there are validation errors
if($this->form_validation->run() == FALSE)
{
    echo(json_encode("validate"=>FALSE));
}
else // there are no validation errors
{
    /*************************
    need to actually send the email, then send you mail 
    *************************/
    echo(json_encode("validate"=>TRUE));
}

then test it on your ajax success function to display positive or negative message 然后在ajax成功函数上对其进行测试,以显示肯定或否定消息

 <script type="text/javascript">

        $(function() {

            $('form').click(function() {

                // get the form values
                var form_data = {
                    name: $('#name').val(),
                    email: $('#email').val(),
                    message: $('#message').val()
                };

                // send the form data to the controller
                $.ajax({
                    url: "<?php echo site_url('welcome/submit_contact'); ?>",
                    type: "post",
                    data: form_data,
                    dataType: "json"
                    success: function(msg)
                    {
                        if(msg.validate)
                        {
                           $('form').prepend('<h5 class="good">Message sent!</h5>');
                           $('h5.good').delay(3000).fadeOut(500);
                        }
                        else
                           //display error
                    }
                });

                // prevents from refreshing the page
                return false;   
            });
        });
        </script>

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