PHP &

Is this correct? If not what is the correct syntax I am new to php hence trying to learn.

PHP&

它是否正确? 如果不是什么是正确的语法 我是新来的PHP,因此试图学习。

Cannot connect via PHP to docker mysql

I got a docker-compose setup with two containers: One is the php/apache service and the other container is the database (mysql). Here is my docker-compose.yml version: '2' services: app: depends_on: - db links: - db:mysql build: . image: app ports: - "80:80" restart: always links: - db:db volumes: - ../:/var/www/html

无法通过PHP连接到docker mysql

我有两个容器的docker-compose设置:一个是php / apache服务,另一个容器是数据库(mysql)。 这是我的docker-compose.yml version: '2' services: app: depends_on: - db links: - db:mysql build: . image: app ports: - "80:80" restart: always links: - db:db volumes: - ../:/var/www/html/ db: image: mysql:latest rest

Open batch file php

Before everyone starts butting in with "security risks" "cant be done" stop there and read the ENTIRE post I have a web server set up from a home laptop which is serving as a games web server im trying to create a GUI so its easier for us to maintain the server and im trying to use batch files to do the actions on the computer So to put this into perspective I have my inde

打开批处理文件php

在所有人开始接触“安全隐患”之前,“不能做”停在那里并阅读整篇文章 我有一个从家用笔记本电脑设置的网络雅博国际顶级,它充当游戏网络雅博国际顶级,我尝试创建一个GUI,以便我们更容易维护雅博国际顶级,并尝试使用批处理文件在计算机上执行操作 因此,为了使这个透视我有我的索引文件index.php

2018-07-04 17:03:18

Under what conditions does PHP 7's self refer to the base class?

As noted on Reddit's LOL PHP sub, PHP 7 may use either the extended class or the base class when referring to self , in contrast to PHP 5 which always refers to the extended class.

在什么情况下,PHP 7的自我指向基类?

正如Reddit的LOL PHP sub中所提到的,PHP 7在引用self时可以使用扩展类或基类,而PHP 5通常指扩展类。

Calling functions from within each other not working

这里的代码!不工作,请帮助。 class user{ var idgen; function uid(){ //return uniqid (rand(), true); return "123"; } function gen() { $this->idgen=$this->uid(); //$this->idgen=udi();//even this dint work } function pint() { echo "The id is :".$this->idgen; } } $use= new user(); $use->gen(); $use->pint(); Change the second line of your c

从对方内部调用功能不起作用

这里的代码!不工作,请帮助。 class user{ var idgen; function uid(){ //return uniqid (rand(), true); return "123"; } function gen() { $this->idgen=$this->uid(); //$this->idgen=udi();//even this dint work } function pint() { echo "The id is :".$this->idgen; } } $use= new user(); $use->gen(); $use->pint(); 将代码的第二行更改为 private $i

PHP input GET vars sanitizing

For my application, written in PHP 5+, I have a common.php which is included from all other pages. Within that I have an include sanitize.php which aims to sanitise any input vars used in the URL. So, targetting $_GET[] values. This is just to have one place where I can tidy any vars, if used, and use them in the code later. There seems to be no tidy way, I've seen, to sanitise based on

PHP输入GET变量消毒

对于我用PHP 5+编写的应用程序,我有一个common.php ,它包含在所有其他页面中。 在那里我有一个include sanitize.php ,它旨在清理URL中使用的任何输入变量。 所以,定位$_GET[]值。 这只是为了让我有一个地方可以整理任何变量(如果使用的话),然后在代码中使用它们。 基于预期/期望的投入,似乎没有一种整洁的方式来净化。 我最初看到的方法是这个sanitize.php有一个foreach循环遍历所有变量,查找所需的清理类型,然

how to get Factory Class to create object instances in PHP?

I have an XML file with many member entries, formatted like so:

如何让工厂类在PHP中创建对象实例?

我有一个包含许多成员条目的XML文件,格式如下: &

Put PHP img src inside string

I'm looking at a PHP logic problem for me, but I can't fix it, so I ask you guys here. How can I call/combine the underneath PHP img with the var string method? I would like to call an image with PHP: 2018-07-04 16:25:14

将PHP img src放入字符串中

我正在为我寻找一个PHP逻辑问题,但我无法解决它,所以我在这里问你们。 我如何调用/结合下面的PHP img与var string方法? 我想用PHP调用一个图像: n"+ "n
n"+ "

What is wrong with this php line?

var $foo = array('foo' => $bar); I am getting an UNEXPECTED T_VARIABLE error. I can't use variables when creating arrays? This declaration is inside a class, and I am running PHP v5.3.2 When removing the var, I get another error Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION Thanks var is not PHP syntax... A simple $foo = array('foo' => $bar); would su

这个PHP行有什么问题?

var $foo = array('foo' => $bar); 我收到一个意外的T_VARIABLE错误。 创建数组时,我不能使用变量? 这个声明是在一个类里面的,我正在运行PHP v5.3.2 当删除var时,我得到另一个错误Parse error: syntax error, unexpected T_VARIABLE, expecting T_FUNCTION 谢谢 var不是PHP语法...一个简单的 $foo = array('foo' => $bar); 就足够了。 关键字var仅用于在类中声明变量(即实例变量),但即使这是PHP4语法

Expecting statement after if

It is my code: And I get error: expecting statement before var $c. How can I solve my problem? The var keyword is deprecated and was for use in classes, which this code snippet is not. No

在if之后期望陈述

这是我的代码: 我得到错误: 期望 var $ c之前的声明 。 我如何解决我的问题? var关键字被弃用,并用于类中,而这段代码片段不是。 注意:由于兼容性原因(作为public关键字的同义词),仍然支持使