OOPS with OOP, PHP database connection

Hey guys I am developing a database connection with OOP PHP and I am stuck on this error; Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:Users...function.php on line 26. I am new to OOP so any help greatly appreciated

带有OOP的OOPS,PHP数据库连接

嘿,我正在开发与OOP PHP的数据库连接,我卡住了这个错误; 警告:mysqli_query()期望参数1是在第26行的C: Users ... function.php中给出的mysqli,null。 我是OOP的新手,因此非常感谢大家的帮助

get values from txt file in php

i know this will be easy for you but not for me :) i need to get the contents of txt file "file.txt" file.txt contetnt is word1 word2 word3 each in a new line or can be as word1, word2, word3, or "word1", "word2", "word3", whichever is easier to do. i have config.php file which i need it to read all and every word in file.txt and process as $value

在php中从txt文件获取值

我知道这对你来说很容易,但对我来说不是这样:) 我需要得到txt文件的内容“file.txt” file.txt文件是 word1 word2 word3 每一行放在一个新的行中,或者可以是word1, word2, word3,或者"word1", "word2", "word3",以较容易的方式执行。 我有config.php文件,我需要它来读取file.txt所有和每个单词,然后处理为$value ,然后由script.php处理,该文件处理$value作为file.txt中所有单词的单

Php: simple code validation failure

I'm currently working on a simple php function for an entity in the framework Symfony2. That function keeps generating errors, although I cannot see what is wrong. Here's the code : public function getForm(array $reqFields){ var $formHelper = new FormHelper; var $fields = new array; if($reqFields == null){ foreach ($this->getArray() as $k => $v) {

PHP:简单的代码验证失败

我目前正在为Symfony2框架中的一个实体开发一个简单的php函数。 该函数不断产生错误,虽然我看不出有什么问题。 代码如下: public function getForm(array $reqFields){ var $formHelper = new FormHelper; var $fields = new array; if($reqFields == null){ foreach ($this->getArray() as $k => $v) { array_push($fields, $formHelper->getTextField($v, $k)); }

Declaring class variables question

What do these var declarations actually do inside of a PHP class: class VbForumsPageMain extends BxDolPageView { var $_oMain; var $_oTemplate; var $_oConfig; var $_oDb; function VbForumsPageMain(&$oMain) { $this->_oMain = &$oMain; $this->_oTemplate = $oMain->_oTemplate; $this->_oConfig = $oMain->_oConfig; $this->_oDb

声明类变量的问题

这些var声明实际上在PHP类中做了什么: class VbForumsPageMain extends BxDolPageView { var $_oMain; var $_oTemplate; var $_oConfig; var $_oDb; function VbForumsPageMain(&$oMain) { $this->_oMain = &$oMain; $this->_oTemplate = $oMain->_oTemplate; $this->_oConfig = $oMain->_oConfig; $this->_oDb = $oMain->_oDb; pa

Php: about echoing a null variable

I'm currently learning php using Murach (the book, published in 2010). One of the exercises has these statements:

Php:关于回显一个null变量

我目前正在使用Murach学习php(该书于2010年出版)。 其中一个练习有这些陈述:

Expression is not allowed as field default value

I am trying to make $app available for the whole class. First, I get: "Expression is not allowed as field default value" Second, on line 5, I get: Unidentified variable $app How can I achieve my goal? class UserController extends XController { var $app = Yii::app();; public function init() { $test = $app; You can not call a method to set a default value f

表达式不允许作为字段默认值

我正在尝试为全班提供$app 。 首先,我得到: “表达式不允许作为字段默认值” 其次,在第5行,我得到: 未知变量$ app 我怎样才能实现我的目标? class UserController extends XController { var $app = Yii::app();; public function init() { $test = $app; 即使它是静态方法,您也无法调用方法来为PHP中的变量设置默认值。 将其更改为在构造函数中设置: use Yii; class UserController ex

Should I place variables in class or constructor? PHP

My question(s) is one of best practices for OOP. Im using Codeigniter framework/PHP. I have a class: class Test() { var $my_data = array(); function my_function() { //do something } } Is it ok to declare $my_data in the class like that? or should it go in the constructor? Basically every function will be writing to $my_data so in a sense it will be a class-wide var

我应该在类或构造函数中放置变量吗? PHP

我的问题是OOP的最佳实践之一。 我使用Codeigniter框架/ PHP。 我有一堂课: class Test() { var $my_data = array(); function my_function() { //do something } } 像这样在类中声明$ my_data可以吗? 或者它应该在构造函数中? 基本上,每个函数都会写入$ my_data,因此在某种意义上它将是一个全局变量(全局?,不确定术语) 另外,我应该使用var还是private? 是var弃用赞成声明变量范

PHP class: Global variable vs Accessbale varibales

This question already has an answer here: What does PHP keyword 'var' do? 7 answers The keyword before variable name defines the visibility of the variable. It defines the access rights for the particular variable. var When using var , it will be publicly accessible through out your project same as public . protected When using protected , variable is only accessible for the

PHP类:全局变量与Accessbale varibales

这个问题在这里已经有了答案: PHP关键字'var'做什么? 7个答案 变量名前面的keyword定义了变量的visibility 。 它定义了特定变量的access权限。 VAR 当使用var ,它可以通过与public相同的项目公开访问。 保护 当使用protected ,变量只能被扩展到特定页面的父类的类访问。 注意:由于兼容性原因(作为public关键字的同义词),仍然支持使用var关键字声明变量的PHP 4方法。 在5.1.3之前的PHP 5中,它

$total = 0 and var $total = 0. Which is the correct one?

Possible Duplicate: What does PHP keyword 'var' do? I know this can sound like a strange question, but it's something I can't get out of my head. I'm defining a class to handle invoice lines and their respective taxes and my current 'prototype' looks like this: public function getRate() { if (!empty($this->lines)) { var $total = 0;

$ total = 0和var $ total = 0.哪一个是正确的?

可能重复: PHP关键字'var'做什么? 我知道这听起来像一个奇怪的问题,但这是我无法摆脱我的头脑的东西。 我正在定义一个类来处理发票行和他们各自的税收,我目前的“原型”如下所示: public function getRate() { if (!empty($this->lines)) { var $total = 0; foreach ($this->lines as $line) { $total += $line->subtotal; }

Serve webpage in firewalled ip through portal machine

I have house.dynamicdnsservice.com pointing to my house IP. My firewall allows incoming https connections to go to 192.168.0.22, where I put google oauth login. That part is working fine. I'm able to access my house from anywhere and run the google login. My next step is to provide access to logged in users to a webpage in IP 192.168.0.34. The firewall does NOT allow that IP to be acce

通过门户机器在防火墙ip服务网页

我有指向我的房子IP的house.dynamicdnsservice.com。 我的防火墙允许传入的https连接转到192.168.0.22,在那里我把谷歌oauth登录。 那部分工作正常。 我可以从任何地方访问我的房子并运行谷歌登录。 我的下一步是提供登录用户访问IP 192.168.0.34中的网页的权限。 防火墙不允许从WAN端访问该IP。 我如何使用.22机器来服务/重定向/入口防火墙.34雅博国际顶级? (当然,登录后) 我在两台机器上运行ubuntu / apache / php。