来自AJAX请求的CodeIgniter中的表单验证失败

Question

An error occurs with my PHP and jQuery code as follows:

jQuery:

$('#form').submit(function(e){
    e.preventDefault();
    var me = $(this);
    var data = me.serialize();
    $.ajax({
        method :'POST',
        url:me.attr('action'),
        data:{data},
        dataType:'json',
        async:false,
        success:function(data){
            alert(data.success);
            if(data.success == true){
                alert('true');
            }
            else{
                alert('false');
            }
        }
    });
});

HTML form:

    <?= form_open('practice/check', array('id'=>'form'))?>
    <input type="text" name="username">
    <input type="password" name="password">
    <button type="submit" name="button" id="submit">click</button>
    <?= form_close();?>

PHP controller:

public function check(){
    $this->load->library('form_validation');
    $data= array('success'=>false, 'message'=>array());
    $this->form_validation->set_rules('username','Username','trim|required');
    $this->form_validation->set_rules('password','Password','required');

    if($this->form_validation->run() == false){
      foreach ($_POST as $key => $value) {
        $data['message'][$key] = form_error($key);
      }
    }
    else{
      $data['success'] = true;
    }
    echo json_encode($data);
}

when alert event happens in jquery code always false return even fields are filled.. please help me and thanks in advance..

Answer 1

You need to change

data: { data }

to

data: data

in your AJAX code.

What your version is doing is wrapping the data inside an extra object. This results in a single variable called "data" being submitted to the server in the request body, like this:

data=username%3Duser123%26password%3Dabcdef

When what you actually want is this:

username=user123&password=abcdef

Therefore, PHP does not recognise the values you're sending it. It can only see a single variable called "data" which, as far as it's concerned, contains a random string. It needs to be able to see the "username" and "password" variables separately.

If you look at this JSFiddle http://jsfiddle.net/jcx2Ly5e/3/ with your Network tools open - submit the form and watch for the two requests to "test.php". The first is incorrect and the second is correct. Look inside the request body of each one to see the difference.

PS As you'll notice, none of this has anything to do with JSON. Your PHP returns JSON, but the form data is submitted in standard form-url-encoded format. The JSON is not part of the issue so I edited your question to remove that.