标签: codeigniter

以下nginx配置似乎无法正常工作,并404在访问时返回错误http://example.com/xml_apps/- 但是joomla网站可以从根网址正常工作http://example.com

更新 看起来请求现在发送到codeigniter应用程序,因为404错误来自CI应用程序.

http://example.com/xml_apps/index.php/xmlengine/jmprovince?category=1&count=10

我们是否需要某种配置来添加codeigniter级别？让控制器/动作工作.

server {
    listen   80;
    root /home/ubuntu/websites/example.com/public_html;
    index index.html index.htm index.php;
    server_name example.com;
location / {
    try_files $uri $uri/ /index.php;
}
location /xml_apps/ {
    if ($request_uri ~ "^system.*"){
        rewrite ^/xml_apps/(.*)$ /xml_apps/index.php?/$1 last;
    }
    if (!-e $request_filename){
        rewrite ^/xml_apps/(.*)$ /xml_apps/index.php?/$1 last;
    }
}
location ~ \.php$ {
    try_files $uri =404;
    fastcgi_pass unix:/var/run/php5-fpm.sock;
    fastcgi_index index.php;
    fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name;
    include fastcgi_params;
    fastcgi_hide_header X-Powered-By;
    fastcgi_param  REQUEST_URI      $request_uri;
    fastcgi_param  QUERY_STRING     $query_string;
    fastcgi_param  REQUEST_METHOD …

我正在尝试使用Mandrill和CodeIgniter发送电子邮件.我可以按照文档中的描述使用Mandrill API:

require_once(mandrill/Mandrill.php);

$Mandrill = new Mandrill($apikey);

$params = array(
    "html" => "<p>\r\n\tHi Adam,</p>\r\n<p>\r\n\tThanks for <a    href=\"http://mandrill.com\">registering</a>.</p>\r\n<p>etc etc</p>",
    "text" => null,
    "from_email" => "xxx@xxx.example.com",
    "from_name" => "chris french",
    "subject" => "Your recent registration",
    "to" => array(array("email" => xxx@yyy.example.com")),
    "track_opens" => true,
    "track_clicks" => true,
    "auto_text" => true
);

$Mandrill->messages->send($params, true));

这很简单,但当我尝试通过CodeIgniter发送Mandrill邮件时,我得到错误结果; 这是我的代码:

$this->load->library('mandrill/Mandrill');
$this->Mandrill->apikey($apikey);
//...
//All other options
$this->Mandrill->messages->send($params, true));

库正在加载成功,发送电子邮件是我收到错误的地方.

抛出错误:

Fatal error: Call to a member function send() on null

我当前的CI项目现在处理文件.我想上传一个xls | xlsx文件到服务器,然后将excel文件数据导入Database表.

作为第一部分,上传的文件成功,我将文件上传到相同级别的应用程序,系统,资产中的uploads文件夹.现在我想加载这个文件并将这些内容导入到DB.我正在使用PHPExcel来执行此操作

这是我的控制器

$this->load->library('phpexcel');
$this->load->library('PHPExcel/iofactory');

$objPHPExcel = new PHPExcel();

$objReader= IOFactory::createReader('Excel2007');
$objReader->setReadDataOnly(true);

$objPHPExcel=$objReader->load(BASE_URL().'uploads/Data_Report.xls');

$objWorksheet=$objPHPExcel->setActiveSheetIndex(0);

$this->load->model('datas_model');

for($i=2;$i<=77;$i++) {

     $client_name= $objWorksheet->getCellByColumnAndRow(0,$i)->getValue();
     $client_address= $objWorksheet->getCellByColumnAndRow(1,$i)->getValue();
     $data_inp=array('name'=>$client_name, 'address'=>$client_address);
     $this->datas_model->add_data($data_inp);
   }

这是我的观点

<?php echo form_open_multipart('../settings_controller/upload_data/do_upload');?>

   <div class="custom-file-upload">
        <input type="file" id="file" name="userfile" multiple/>
   </div>
   <div class="button-container-2">
        <button class="btn btn-primary" id="updown-btn" type="submit" style="height:45px;">Upload </button>
   </div>

<?php echo "</form>"?>

当我跑步时,它显示我的错误 Uncaught exception 'PHPExcel_Reader_Exception' with message 'Could not open http://localhost/myproject/uploads/Data_Report.xls for reading! File does not exist.'

当我把那个Data_Report.xls文件放在htdocs中时,它可以成功运行.

问题是当我使用BASE_URL().'uploads/Data_Report.xls'时.但该文件在实际存在,我通过粘贴url上的localhost/myproject/uploads/Data_Report.xls进行验证,并成功下载. …

我正在运行在服务器上的codeigniter中运行一个Web应用程序.这里我有一个用户注册表单,在localhost中工作正常.但是当涉及到服务器时,当我尝试注册用户时,我的页面显示错误:

mysql_escape_string() function is deprecated use mysql_real_escape_string() in mysql/mysql_driver

我尝试更改我的mysql_driver页面,但更改后的所有内容都变为空白.谁能帮我解决这个错误？

我正在尝试从我的localhost发送带有我的codeigniter项目的电子邮件,但它始终显示如下错误:

    220 mx.google.com ESMTP pp9sm11498734pbb.65 - gsmtp 

hello: 250-mx.google.com at your service, [202.43.95.33]
250-SIZE 35882577
250-8BITMIME
250-AUTH LOGIN PLAIN XOAUTH XOAUTH2 PLAIN-CLIENTTOKEN
250-ENHANCEDSTATUSCODES
250-PIPELINING
250-CHUNKING
250 SMTPUTF8
Failed to authenticate password. Error: 535-5.7.8 Username and Password not accepted. Learn more at 535 5.7.8 http://support.google.com/mail/bin/answer.py?answer=14257 pp9sm11498734pbb.65 - gsmtp 
from: 530-5.5.1 Authentication Required. Learn more at
530 5.5.1 http://support.google.com/mail/bin/answer.py?answer=14257 pp9sm11498734pbb.65 - gsmtp
The following SMTP error was encountered: 530-5.5.1 Authentication Required. Learn more at 530 5.5.1 http://support.google.com/mail/bin/answer.py?answer=14257 pp9sm11498734pbb.65 - gsmtp 
to: 530-5.5.1 …

我是CI的新手,现在我正在尝试使用CodeIgniter 3来开发我的网站.我只是解压缩框架并在config/autoload.php文件中只改变一个想法:

$autoload['libraries'] = array('database','input');

当我运行该网站时,出现错误:

Unable to load the requested class: Input

当我尝试使用CI版本2.2.0稳定时,每件事情都可以,没有错误可以解释一下为什么并帮助我解决它？

在CodeIgniter中我的安装应用程序routes.php我想检查是否存在两个文件并根据它们进行设置.

其中,如果文件存在则会升级,否则转到第一步.

有更好,更安全的方式吗？

$admin = dirname(FCPATH) . '/admin/config/database.php';
$catalog = dirname(FCPATH) . '/catalog/config/database.php';

if (file_exists($admin, $catalog)) {
    $route['default_controller'] = "upgrade/index";
    $route['404_override'] = '';
} else {
    $route['default_controller'] = "step_1/index";
    $route['404_override'] = '';
}

我通过点击该链接向我的用户的电子邮件发送链接以激活他的帐户.示例链接:http: //test.com/welcome/make_active_user/($ user_id)

然后我$user_id在我的控制器中编码,之后我的链接看起来像下面的那个

http://test.com/welcome/make_active_user/17elQOjxrOXDsZdDZVFY7JFrB0TJFojdS+5OTpiIq/XXIaVrDfVWQ7xgOXXqGsURIFe/Udvm/XHrNWtbZXFA2g==

一切都很好.现在我想解码$user_id,但" /"符号创建问题.codeigniter只在" /"符号之前使用了那些字符.如何在没有" /"的情况下获得编码的user_id的输出

我$this->encrypt->encode($user_id,$key);在我的控制器中使用" "

请帮忙

升级后Codeigniter 2到3版本,我收到此错误..为什么会这样？

遇到了未被捕获的异常

类型:例外

消息:已配置的数据库连接是持久的.中止.

文件名:/var/www/vhosts/xxx.com/app/system/libraries/Session/drivers/Session_database_driver.php

行号:94

回溯:

文件:/var/www/vhosts/xxx.com/app/application/core/MY_Controller.php行:11功能:__ construct

文件:/var/www/vhosts/xxx.com/app/application/core/MY_Controller.php行:52功能:__ construct

文件:/var/www/vhosts/xxx.com/app/application/controllers/Dashboard.php行:7功能:__ construct

文件:/var/www/vhosts/xxx.com/application/index.php行:293功能:require_once

我阅读了CodeIgniter关于迁移类的文档.我能够执行迁移但是如何在迁移中使用down()函数？

这是文档中的示例

defined('BASEPATH') OR exit('No direct script access allowed');

class Migration_Add_blog extends CI_Migration {

    public function up()
    {
        $this->dbforge->add_field(array(
            'blog_id' => array(
                'type' => 'INT',
                'constraint' => 5,
                'unsigned' => TRUE,
                'auto_increment' => TRUE
            ),
            'blog_title' => array(
                'type' => 'VARCHAR',
                'constraint' => '100',
            ),
            'blog_description' => array(
                'type' => 'TEXT',
                'null' => TRUE,
            ),
        ));

        $this->dbforge->create_table('blog');
    }

    public function down()
    {
        $this->dbforge->drop_table('blog');
    }

在我的页面中,我设置了这样的迁移:

class Migrate extends CI_Controller {

    public function __construct() {
        parent::__construct();
        $this->load->library('migration');
    }

    public function index() …