标签: codeigniter

我正在尝试使用url帮助程序获取文件名列表,但是我收到以下错误:

调用未定义的函数get_filenames()

我自动加载了url帮助器,然后在Controller中显式加载它,我仍然得到错误.为什么我收到此错误？

控制器:

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Downloads extends CI_Controller {

    public function index()
    {   
        $this->load->helper("url");
        $this->template->write_view('content', 'download_view');
        $this->template->render();
    }
}

视图:

<? get_filenames("/somefolder"); ?>

这非常令人沮丧!我尝试将简单的ajax帖子发送到我的本地Web应用程序,并且该参数为空.为了调试它,我打开了小提琴手但是同样的结果正在发生!

我有这个控制器

class Site extends CI_Controller {

public function __construct() {
    parent::__construct();

} //etc

有了这个功能

    public function hello() {
    var_dump($_SERVER['REQUEST_METHOD']);
    var_dump($_SERVER['REQUEST_URI']);
    $hello = $this->input->post("hello");
    echo "hello is $hello";
}

现在我尝试在Fiddler中测试这个,我去Composer,选择一个新的Post请求并输入正文hello = wtsgppp.

这是fiddler中hello函数的结果:

string(4) "POST"
string(25) "/cmd/index.php/site/hello"
hello is 

你可以看到你好是空的.任何帮助都非常感谢!!

让事情变得更加烦人 - 一旦我将fiddler中的http请求更改为GET并在url字符串中添加hello(并将控制器更改为echo get而不是post),它就可以了!那么为什么要上班但是帖子没有？

图书馆

class getLang {
   public function getText($en, $ar) {
      $CI =& get_instance();
      $lang = $CI->session->userdata('language');
      if ($lang == 'en') {
         $string = $en;
      } else if ($lang == 'ar') {
         if (!empty($ar)) {
            $string = $ar;   
         } else if (empty ($ar)) {
            $string = $en;
         }
      }
      return $string;
   }
}

调节器

class Home extends MY_Controller {
   public function test() {
      $this->load->library('getLang');
      return $this->getLang->getText('text', '');
   }
}

当我跑回家/测试时,我收到此错误:

A PHP Error was encountered

Severity: Notice

Message: Undefined property: Home::$getLang

Filename: …

我使用下面的函数将我的图像文件上传到我的服务器(Localhost).很好,图像正在上传.但是我需要两个图像字段,因此一旦点击提交按钮,就应该上传两个图像.我使用了这里描述的Codeigniter多文件上传功能,但它无法正常工作.

我收到此错误消息

A PHP Error was encountered

Severity: Warning

Message: is_uploaded_file() expects parameter 1 to be string, array given

Filename: libraries/Upload.php

Line Number: 161

好吧,我无法理解错误在哪里.我用来上传单张图片的功能是

function do_upload()
    {
        $config['upload_path'] = './uploads/';
        $config['allowed_types'] = 'jpeg|png|gif';
        $config['max_size'] = '0';
        $config['max_width']  = '0';
        $config['max_height']  = '0';

        $this->load->library('upload', $config);

        if ( ! $this->upload->do_upload())
        {
            $error = array('error' => $this->upload->display_errors());
           //failed display the errors
        }
        else
        {
            //success

        }
    }

另外我还想问我可以更改我选择的输入字段名称.即我总是需要实施<input type="file" name="userfile"/>.是否可以更改名称？我尝试更改它,我收到消息没有选择文件,所以这必须意味着我无法改变它.

我在config文件夹中有一个pagination.php文件.代码如下

$config['num_links'] = 5;
$config['use_page_numbers'] = TRUE;
$config['query_string_segment'] = 'page';

$config['full_tag_open'] = '<div class="pagination"><ul>';
$config['full_tag_close'] = '</ul></div>';

$config['first_link'] = '&laquo; First';
$config['first_tag_open'] = '<li class="prev page">';
$config['first_tag_close'] = '</li>';

$config['last_link'] = 'Last &raquo;';
$config['last_tag_open'] = '<li class="next page">';
$config['last_tag_close'] = '</li>';

$config['next_link'] = 'Next &rarr;';
$config['next_tag_open'] = '<li class="next page">';
$config['next_tag_close'] = '</li>';

$config['prev_link'] = '&larr; Previous';
$config['prev_tag_open'] = '<li class="prev page">';
$config['prev_tag_close'] = '</li>';

$config['cur_tag_open'] = '<li class="active"><a class="navigation" href="">';
$config['cur_tag_close'] = '</a></li>';

$config['num_tag_open'] = '<li class="page">';
$config['num_tag_close'] = …

以下此功能是否能够阻止XSS攻击和SQL注入？

require_once('security.class.php');
function secure_data($data) {
    $data = mysql_real_escape_string($data);
    $filtered_data = filter_var($data, FILTER_SANITIZE_STRING);
    $secure_class = new security_class();
    $clean_data = $secure_class->xss_clean($filtered_data);         
    return $clean_data;
}

该安全类是笨.

构建自定义实体存储库时遇到一些问题.

尝试自定义实体存储库时,我有以下致命错误

Fatal error: Uncaught exception 'BadMethodCallException' with message 'Undefined method 'getByParentId'. The method name must start with either findBy or findOneBy!' in C:\Users\user\Desktop\projects\interview\application\libraries\Doctrine\ORM\EntityRepository.php:215 Stack trace: #0 C:\Users\user\Desktop\projects\interview\application\controllers\CommentController.php(58): Doctrine\ORM\EntityRepository->__call('getByParentId', Array) #1 C:\Users\user\Desktop\projects\interview\application\controllers\CommentController.php(58): Doctrine\ORM\EntityRepository->getByParentId('1') #2 [internal function]: CommentController->viewCommentsListByParentId('1') #3 C:\Users\user\Desktop\projects\interview\system\core\CodeIgniter.php(359): call_user_func_array(Array, Array) #4 C:\Users\user\Desktop\projects\interview\index.php(203): require_once('C:\Users\user\D...') #5 {main} thrown in C:\Users\user\Desktop\projects\interview\application\libraries\Doctrine\ORM\EntityRepository.php on line 215

我的项目结构是Mappings Repositories Entities Proxies模型

我的doctrine.php有这样的存储库加载器

// load the repositories
$repositoryClassLoader = new \Doctrine\Common\ClassLoader('Repositories', APPPATH.'models');
$repositoryClassLoader->register();

我的模型类有以下声明

namespace Entities;

use Doctrine\ORM\Mapping as ORM;

/**
 * Entities\Comment
 * @Entity (repositoryClass="Repositories\CommentRepository") …

我试图将两个变量从链接传递给我的jquery ajax函数.但我不知道如何做这些ajax的东西.

我有两个ID的链接.1. rowid&2nd.BOOKID.我必须将这两个id传递给我的jquery函数.

请帮帮我怎样才能做到这一点.

`//cart item displaying in a for each loop , each row have a 'remove' link with two id 
//say i have $id='4dsf2323' & $bid='43' now i have to pass these two ids to my jquery function on click of a link     


<a id="removeid" >Remove item from cart link</a>`

我的jquery函数

    <script>
        $('removeid').click(function(e) {
      // prevent the default action when a nav button link is clicked
      e.preventDefault();

              //HOW TO GET HERE THOSE TWO IDS FROM MY ABOVE …

我尝试umat_id使用此SQL查询:

SELECT umat_id FROM msumat WHERE nama = $nama

我将此SQL查询转换为CI的Active Record:

$this->db->select('umat_id');
$terdaftar = $this->db->get_where('msumat', array('nama' => $nama));

所以这个查询应该返回一个字符串(例如:"John").

但是当我试图回应它时出现错误: 类CI_DB_mysql_result的对象无法转换为字符串

我尝试过这样的事情:echo (string)$terdaftar;但是它不起作用.

我只想回应"约翰"

编辑 刚才说我想将"John"插入变量中.怎么做？

$john = ????

我目前正在使用CodeIgniter的框架,我在HTML5中遇到数据属性问题,我不知道如何从表单中的下拉输入传递数据属性.

我知道如何使用jQuery,但我不知道是否有任何"解决方案"来获取数据属性并识别我的数据类型.

这是我的代码:

<select name="username">
<option value="1" data-attributes="users">Simon</option>
<option value="1" data-attributes="collectives">My super crew</option>
</select>

如果你有一些灯,请事先提前.

西蒙