我如何将此表单转换为codeigniter [关闭]

Question

I have the form below for a login page

<form action = "http://<?php echo $_SERVER['HTTP_HOST']; ?>/user/login" method = "post">
    <fieldset>
        <label for="email" class="block clearfix">
            Email Address:
            <span class="block input-icon input-icon-right">
                <input type="text" class="form-control" placeholder="Email" name="email" id = "email"/> <i class="icon-user"></i>
            </span>
        </label>
        <label for ="password" class="block clearfix">
            Password:
            <span class="block input-icon input-icon-right">
                <input type="password" name="password" id = "password" class="form-control" placeholder="Password"/> <i class="icon-lock"></i>
            </span>
        </label>
        <div class="clearfix">
            <button type="submit" class="button width-35 pull-right btn btn-sm btn-primary">
                <i class="icon-key"></i>
                Login
            </button>
        </div>
    </fieldset>
</form>

I'd like to convert this to use the form_helper packaged with CodeIgniter. My issue though is the documentation only showed the most basic of examples. I'm not sure how I would put so much extra data inside each label (including the form_input). What is the easiest way to convert this?

Answer 1

Attributes can be provided as an array for form_input() calls.

If I were to recreate your email text box:

<input type="text" class="form-control" placeholder="Email" name="email" id = "email"/>

I would create it as:

$data = array(

'name' => 'email' , 
'id'   => 'email' ,
'class' => 'form_control' ,
'placeholder' => 'Email'

);

echo form_input($data);