doujia4041 2012-11-14 23:17
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使用Codeigniter处理ajax响应

I've been handling success/error messages by returning json encoded arrays as a response, but it suddenly occurred to me that this probably isn't the correct way of handling notifications.

For example, my controller will look like this:

public function controller_name() {
    //validate form input
    $this->form_validation->set_rules('id', 'id', 'required|is_natural_no_zero');

    // if validation was successful with no errors
    if ($this->form_validation->run() && $this->model_name->method()) {

        $this->data['status'] = 'success';
        $this->data['message'] = 'This is the success message';
        echo json_encode($this->data);

    } else {
        $this->data['status'] = 'error';
        $this->data['message'] = validation_errors();
        echo json_encode($this->data);
    }
}

Then the jQuery:

$.ajax({
    url: url,
    type: 'POST',
    data: data,
    success: function (r) {

        json = $.parseJSON(r);

        if (json.status == 'success') {
            if (json.message == 'added') {
                $this.addClass('success');
            } else {
                $this.removeClass('success');
            }
        } else {
            console.log('There was an error')
        }

What's the best practice way to do this? Can I throw exceptions to use the ajax error?

  • 写回答

1条回答 默认 最新

  • donglugou6652 2012-11-14 23:29
    关注

    Sending erroneous http status code should trigger the jQuery ajax error handler:

    public function controller_name() {
        //validate form input
        $this->form_validation->set_rules('id', 'id', 'required|is_natural_no_zero');
    
        // if validation was successful with no errors
        if ($this->form_validation->run() && $this->model_name->method()) {
            $this->data['message'] = 'This is the success message';
        } else {
            $this->output->set_status_header('400'); //Triggers the jQuery error callback
            $this->data['message'] = validation_errors();
    
        }
        echo json_encode($this->data);
    }
    

    JS:

    $.ajax({
        url: url,
        type: 'POST',
        data: data,
        success: function (r) {
            var json = $.parseJSON(r);
    
        },
    
        error: function( jqXhr ) {
            if( jqXhr.status == 400 ) { //Validation error or other reason for Bad Request 400
                var json = $.parseJSON( jqXhr.responseText );
            }
        }
    });
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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